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MAGICAL BOOK ON

QUICKER MATHS

Miraculous for Banks, LIC, GIC, UTI, SSC, CPO, Management, Railways and other competitive exams Stimulating for general use

M. TYRA Revised and Enlarged Edition

BSC PUBLISHING CO. PVT. LTD. C-37, GANESH NAGAR, PANDAV NAGAR COMPLEX DELHI-110092 Phone: 011-22484910, 22484911, 22484912, 22484913 e-mail: [email protected]; Website: www.bsccareer.com

BSC PUBLISHING CO. PVT. LTD. © Author First Edition — 1995 Second Edition — 1999 Third Edition — 2000 Fourth Edition — 2013 Fifth Edition — 2018

ISBN: 978-81-904589-2-4 DISCLAIMER Every effort has been made to avoid errors or omissions in the publication. In spite of this, some errors might have crept in. The publisher shall not be responsible for the same. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the next edition. It is notified that neither the publisher nor the author or sellers will be responsible for any damage or loss of action of any one, of any kind, in any manner, therefrom. All rights reserved. No part of this publication may be reproduced in any form without the prior permission of the publisher.

CONFIRM THE ORIGINALITY OF THE BOOK Published by

BSC PUBLISHING CO. PVT. LTD. C-37, Ganesh Nagar, Pandav Nagar Complex Delhi-110092

Ch. 560

Dedicated to Sri Jagdeo Singh (Dadaji)

.

Contents Chapters 1. How to Prepare for Maths 2. Addition

Page No. 1 5

3. Multiplication

11

4. Division

17

5. Divisibility

21

6. Squaring

25

7. Cube

27

8. HCF and LCM

31

9. Fractions

41

10. Decimal Fractions

49

11. Elementary Algebra

59

12. Problems on Comparison of Quantities

89

13. Surds

97

14. Number System

101

15. Binary System

125

16. Permutation & Combination

129

17. Probability

143

18. Ratio and Proportion

157

19. Partnership

183

20. Percentage

191

22. Average

215

22. Problems Based on Ages

223

23. Profit and Loss

233

24. Simple Interest

265

25. Compound Interest

277

26. Alligation

293

27. Time and Work

313

28. Work and Wages

333

29. Pipes and Cisterns

337

30. Time and Distance

343

31. Trains

359

32. Streams

373

33. Elementary Mensuration-I

381

34. Elementary Mensuration-II

401

35. Series

415

36. Data Sufficiency

433

37. Data Analysis

467

38. Caselets

533

39. Trigonometry

545

Foreword When did you get up this morning? At 6.35 am. Did you reach your destination on time? Yes, I arrived 10 minutes earlier even though I got up five minutes late. I usually walk but today I took a bus. Why not a taxi? Isn’t it faster? Faster, yes. But, in my case the speed of the bus was sufficient. Besides, I compared the fares of the two means of transport and concluded that taking a taxi would mean incurring unnecessary expenditure. Wow! what a clever guy! Saves money, saves time. Does he know some magic? Is he a student of occult sciences? No, dear. He is an ordinary mortal like any of us. But he exudes an uncommon confidence, thanks to his ability to compute at a magical pace. In other words, Quicker Maths is an asset at every step of one's life. By Quicker Maths, the book means speed and accuracy at both simple numerical operations and complex problems. And it is heartening that the book takes into account all kinds of problems that one may encounter in the ordinary run of life. There are several books one comes across which take care of problems asked in examinations. But they are conventional and provide solutions with the help of detail method. This book provides you both the detail as well as the quicker method. And it is the latter that makes the book irresistible. That the book provides you both the methods is a pointer to the fact that you are not being led into a faith where you have to blindly follow what the guru says. The conclusions have been rationally drawn. The book, therefore, serves as a guide—the true function of a guru—and once you get well-versed with the book, you will feel empowered enough to evolve formulas of your own. I think the real magic lies there! At the same time, a careful study of the book endows you with the magical ability to arrive at an answer within seconds. There may be some semi-educated persons who sneer at this value of the book. Can you beat a computer?— a contemptuous question is asked. Absurd question. For one, the globe is yet to get sufficiently equipped with computers. Two, even when we enter a full-fledged hi-tech age, let us hope it is not at the cost of our minds. Let not computers and calculators become the proverbial Frankenstein's monster. The mind is a healthy organ and computing a healthy function. Computers and calculators were devised by the brain to aid it, not to consume it. That the mathematical ability of the brain be intact is a concern of every individual. In most of the examinations even calculators are not allowed, let alone computers. And it is here that this magical book proves immensely useful. If you are a reader of this book, you can definitely feel more confident—miles ahead of others. There is no doubt that there has been a lot of labour involved in the book. For all those students who are gearing up to drive their Mathematics Marutis at an amazing 200 kmph, the book definitely provides the requisite infrastructure. And what is more, the methods are accident-free with proper cautions at necessary places. Here is a book that will help exam-takers glide and enthusiastic students enjoy the ride. In an age when speed is being maniacally pursued, a careful study of the book will serve as a powerful accelerator. At the same time, its simple language makes it easily accessible across the linguistic barriers. Besides, the fact that you vividly see the Quicker Method makes things very interesting. And so the book provides you speed not at the cost of joy. It is not merely a mechanical device, but has an organic charm. One concludes: fast driving is fun. Chetananand Singh Editor, Banking Services Chronicle

Author’s Preface We, at Banking Services Chronicle (BSC), analyse students' problems. If a student is not able to perform well in an exam, our research group members try to penetrate the student's psyche and get at the roots of the problems. In the course of our discussions we found that the mathematics section often proves to be the Achilles' heel for most of the students. Letters from our students clearly indicated that their problem was not that they could not solve the questions. No, the questions asked in general competitions are in fact so easy that most of the students would secure a cent per cent score, if it were not for the time barrier. The problem then is: INABILITY TO SOLVE the question IN TIME. Unfortunately, there was hardly any book available to the student which could take care of the time aspect. And this prompted the BSC members to action. We decided to offer a comprehensive book with our attention targeted at the twin advantage factors: speed and accuracy. Sources were hunted for: Vedic Mathematics to computer programmings. Our aim was to get everything beneficial from wherever possible. The most-encountered questions were categorised. And Quicker Methods were intelligently arrived at and diligently verified. How does the book help save your time? Probably all of you learnt by heart the multiplication-tables as children. And you have also been told that multiplication is the quicker method for a specific type of addition. Similarly, there exists a quicker method for almost every type of problem, provided you are well-versed with some key determinants and formulas. For the benefit of understanding we have also given the detail method and how we arrive at the Quicker Method. However, for practical purposes you need not delve too much into the theory. Concentrate on the working formulas instead. For the benefit of non-mathematics students, the book takes care to explain the oft-used terms in an ordinary language. So that even if you are vaguely familiar with numbers, the book will prove beneficial for it is self-explanatory. The mathematics students are relatively in a comfortable position. They do not have to make an effort to understand the concepts. But even in their case, there are certain aspects of questions asked in the competitive exams which have been left by them untouched since school days. So, a revision is desirable. In the case of every student, however, the unique selling proposition of the book lies in its ability to increase the student's problem-solving speed. Due caution has been observed to proceed methodically. Gradual progress has been made from simple to complex examples. There are theorems and solved examples followed by exercises. A systematic, chapter-by-chapter study will definitely result in a marked improvement of the student's mathematical speed. The students are requested to send their responses to the book and suggestions for further improvement. And, finally, I would extend my thanks to all those who have played a role in making the book available to the reader. I specially thank Mr Madhukar Pandey for having played a key role in promoting the endeavour, Mr Chetananand Singh for the meticulous editing of the book and Mr Niranjan Bharti for having carefully verified the results. Mr Niranjan Singh's all-round assistance cannot be forgotten. Friends kept on encouraging me at every step. The inspiration I received from Mr Sanjay, Mr Deven Bharti, Mr Nagendra Kumar Sinha, Mr Sandeep Varma, Mr Manoj Kumar, Mr Vijay Kumar, Mr Rajeev Raman, Mr Anil Kumar and Mr JK Singh, to name a few, has been invaluable. And thanks to Mr Pradeep Gupta for printing.

Preface to the Second Edition It is a great pleasure to note that Magical Book on Quicker Maths continues to be popular among the students who are looking for better results in this world of cut-throat competition. This book has brought the new concept of timesaving quicker method in mathematics. So many other publications have tried to publish similar books but none could reach even close to it. The reason is very simple. It is the first and the original book of its kind. Others can only be duplicate and not the original. Some people can even print the duplicate of the same book. It will prove dangerous to our publication as well as to our readers. So, we suggest our readers to confirm the originality of this book before buying. The confirmation is very simple. You can find a three-dimensional HOLOGRAM on the cover page of this book. This edition has been extensively revised. Mistakes in its first edition have been corrected. Some new chapters like Permutation-Combination, Probability, Binary System, Quadratic Expression etc. have been introduced. Some old chapters have been rewritten. Hope you will now find this book more comprehensive and more useful.

Preface to the Third Edition The pattern of question paper as well as the standard of questions have changed over the past couple of years. Besides, Permutation-Combination and Probability, and questions from trigonometry—in the form of Height and Distance—have also been introduced. In chapters like Data Analysis, Data Sufficiency and Series, new types of questions are being asked. With the above context in mind, a few new chapters have been introduced and a few old ones enlarged. Important Previous Exam questions have been added to almost all the chapters. But they have been added in larger numbers in the chapters specially mentioned above. An introductory chapter has been added on “How to Prepare for Maths”. I suggest going through this chapter before setting any targets. A revision in the cover price was long due. The first edition (1995), which cost Rs 200, had only 612 pages. The price remained the same even in the second edition (1999) in spite of the number of pages being increased to 749. But the third edition (2000) has gone into 807 voluminous pages. So, the price is being increased to Rs 280. Kindly bear with us.

Preface to the Fourth Edition Another edition of this book had long been overdue. No matter how good a book — well, that has been the verdict of generations of readers — there is always scope for improvement. And this edition is an endeavour in this direction. The chapter on “Division” has been introduced once again. Besides, simplicity has been the hallmark of this book for decades. I have tried to further simplify the methods wherever I could. Hope you will benefit more from this book after the incorporation of these changes.

Preface to the Fifth Edition Questions in competitive exams have changed quite a lot over the past few years. Being the pioneer in Mathsbased competitive exams, it was incumbent upon us to guide the students in the changed environment. Hence the Fifth Edition of the book that has been so dear to generations of readers. We have tried to add questions based on the latest patterns to the chapters of this book. Besides, two new chapters have been introduced: Comparison of Quantities; and Caselets. Some other chapters have been specially enhanced. In Mensuration and Percentage, we have added an Exercise each after Solved Examples. In Data Interpretation (DI), an Exercise has been added with the latest questions. This includes DI questions based on missing data as well as those based on Arithmetic. With so many additions the content became voluminous. In order to address this issue, we have changed the format of the book. Hope the book in its new format and with its revised content serves you adequately.

Chapter 1

How to Prepare for MATHS (Using this book for Competitive Exams) 1. Importance of Maths paper (PO) Quantitative Aptitude is a compulsory paper. You can't neglect. So make sure you are ready to improve your mathematical skills. Each question values 1.2 marks whereas each question of Reasoning values only 1.066 marks in PO exam. So, if you devote relatively more time on this paper you get more marks. Also, the answers of Maths questions are more confirmed than answers of Reasoning questions, which are often confusing. Most of you feel it is a more time-consuming paper, but if you follow our guidelines, you can save your valuable time in examination hall. Other exams: There are very few competitive exams without Maths paper. SSC exams have different types of Maths paper. The mains exam of SSC contains Subjective Question paper. Keeping this in mind, I have also given the detail method of each short-cut or Quicker Method given in this book. Each theorem, which gives you a direct formula also contains proof of the theorem, which is nothing but a general form (denoting numerical values by letters say X, Y, Z etc) of detail method.

2. Preparation for this paper (A) How to start your preparation Maths is a very interesting subject. If you don't find it interest-ing, it simply means you havn't tried to understand it. Let me assure you it is very simple and 100% logical. There is nothing to be assumed and nothing to be confused about. So, nothing to worry if you come forward with firm determination to learn maths. The most basic things in Maths are: (a) Addition - Subtraction (b) Multiplication - Division All these four things are most useful. At least one of these four things is certainly used in any type of mathematical question. So, if we do our basic calculations faster we save our valuable time in each question. To calculate faster, I suggest the following tips:

(i) Remember the TABLE upto 20 (at least): You should know that tables have been prepared to make calculations faster. You can see the use of table in the following example: Evaluate: 16 × 18 If you don’t remember the table of either 16 or 18 you will proceed like this: 16 18 128 16 288 But if you know the table of 16, your calculations would be: 16 × 18 = 16(10 + 8) = 16 × 10 + 16 × 8 = 160 + 128 = 288 Or, if you know the table of 18; your calculation would be 16 × 18 = 18(10 + 8) = 18 × 10 + 18 × 6 = 180 + 108 = 288 If you can, you remember the table upto 30 or 40. It will be precious for you. Note: (1) You should try the above two methods on some more examples to realise the beauty of tables. Try to evaluate: 19 × 13; l7 × 24; 18 × 32 (18 × 30 + 18 × 2 = 540 + 36 = 576); 19 × 47; 27 × 38; 33 × 37 etc. (2) All the above calculations should be done mentally. Try it. (ii) LEARN the one-line Addition or Subtraction method from this book In the first chapter we have given some methods of faster addition and subtraction. Suppose you are given to calculate: 789621 – 32169 + 4520 – 367910 = .... If you don’t follow this book you will do like:

Quicker Maths

2 789621 +4520 794141

32169 +367910 400079

794141 –400079 394062 The above method takes three steps, i.e. (i) add the two +ve values; (ii) add the two -ve values; (iii) subtract the second addition from the first addition. But you can see the one-step method given in the chapter. Have mastery over this method. It takes less writing as well as calculating time. (iii) Learn the one-line Multiplication or Division method from this book Method of faster multiplication is given in the second chapter. I think it is the most important chapter of this book. Multiplication is used in almost all the questions, so if your multiplication is faster you can save at least 35% of your usual time. You should learn to use the faster one-line method. It needs some practice to use this method frequently. The following example will show you how this method saves your valuable time. Ex. Multiply: 549 × 36 If you don't follow the one-line method of multiplication, you will calculate like: 549 36 3294 1647 19764 If this method takes 30 seconds I assures: you that one-line method given in this book will take at the most 15 seconds. Try it. One-line method of calculation for Division is also very much useful. You should learn and try it if you find it interesting. But, as division is less used, some of you may avoid this chapter.

(iv) Learn the Rule of Fractions In the chapter Ratio and Proportion on Page No. 269, I have discussed this rule. It is the faster form of unitary method. It is nothing but simplified form of Rule of Three and Rule of Proportional Division. No doubt, it works faster and is used in almost all the mathematical questions where unitary method (Aikik Niyam) is used. See the following example: Ex. If 8 men can reap 80 hectares in 24 days, how many hectares can 36 men reap in 30 days? Soln: I don't know how much time you will take to answer the question but if we follow the rule of fraction our calculation would be:

80

36 30 450 hectares. 8 24

In this book this method is used very frequently. It is only when you go through the various chapters of this book that you will find how wonderful the method is. It saves at least half of your usual time. I think this method should be adopted by all of you at any cost. First learn it and then use it wherever you can. So, these four points are necessary for your strong and firm start. And only strong and firm start is the key to sure success.

(B) Clear the Fundamentals behind each chapter There are 39 chapters in this book. Each chapter has some important basic fundas. Those fundas should be clear to you. Any doubt with the basics will hamper your further steps. Now the question arises - what are those basic fundas? Naturally, for each chapter, there are different fundas. I will discuss one chapter and its basic fundas. You can understand and find the same with different chapters. My chapter is PARTNERSHIP (on page 309)

Partnership

Simple (Different investments for the same period) Ratio of profit = P : Q.

Compound [Different investments (P & Q); Diff periods (t 1 & t 2) Ratio of profit = P × t1: Q × t 2

How to Prepare for MATHS

3

It is natural that your fundas of ratio should be clear before going through this chapter. Now, you can understand what I actually mean by the fundas. In a similar way, you can collect all the fundas and basic formulae at one place.

sources: Guides, Books, Magazines etc. The most standard and reliable sets are available in the magazine Banking Services Chronicle. Also, with our Correspondence Course we give at least 60 sets of Maths papers separately and 60 sets of Maths with full-length Practice Sets.

(C) More and More Examples

(C) Now start your practice:

You are suggested to go through as many examples as possible. Each question given in examples has some uniqueness. Mark it and keep it in mind. To collect more examples of different types you may consult different books available in the market.

From the beginning to the end, the complete session of practice should be divided into five parts. Part (i): Take your first test with previous paper without taking time into consideration. Try to solve all questions. Note down the total time and score in your performance diary. Also note down the questions which took more than one minute. Now you have to find out the reason of your low performance, if it is so. Naturally, you would find the following reasons: Some questions were difficult and time consuming. Some questions were unsolvable for you. You lost your concentration. You lost your patience. You did more writing job. You could not use Quicker Methods. Try to find out the solutions to all the above problems. If any of the questions was difficult for you, it means your initial preparation was not good. But don't worry. Go through that chapter again and clear your basic concepts. Because the standard of a question is always within your reach. If you have passed your 10th exam with maths, you can solve all the questions. You should take at least 10-12 tests in this part. Part (ii): This time the paper may be either previous or model (sample). Fix your alloted period (say 50 minutes for PO). And solve as many questions as possible within that period. Once you have completed your test, count the number of correct questions. Note down the number of questions solved by you and the no. of correct solutions in your performance diary. Now, you can find the reasons for your low scoring. If the reasons are the same as in Part (i), you try to resolve the problem again. Take at least 5-6 tests in this part. After analysing your performance and problems you should be ready for your third part of test. Part (iii): This time you try to solve all the questions within a time period fixed by you in advance. This period should be less than that for the tests in part (ii). If you couldn't do it, try it again on another test paper. Part (iii) should not be

(D) Use of Quicker Formulae Before going for quicker formula I suggest you to know the detail method of the solution as well. So, see the proofs of all the formulae carefully. Once you get familiar with the detail solution, you find it easier to understand the quicker method. Direct Formulae or Quicker Methods save your valuable time but they have very high potential of creating confusion in their usage. So you should know where the particular formula should be used. A little change in the questions may lead you to wrong solution. So be careful before using them. In case of any confusion, you are suggested to solve the questions without using direct formula or quicker method. Only frequent use of the quicker methods can make you perfect in Quicker Maths. After covering all the chapters and knowing all the methods, you should be prepared for practice.

3. Practice of Maths Paper (A) Pattern of paper You should know the pattern and style of the question paper of the exam for which you are going to appear. Suppose you are preparing for Bank PO exam. You should know that maths paper consists of 35 questions in prelims exam, 35 questions main exam and 50 questions in various PGDBF courses. Out of which 15-20 questions are from Data Analysis, 5 Questions from Data Sufficiency, 5-10 are from Numericals (Calculation based), and 20-25 are from Mathematical Chapters (like Profit & Loss, Percentage, Partnership, Mensuration, Time & Work, Train, Speed etc.). In other exams it may be different. The pattern can be known from previous papers.

(B) Collection of Previous Papers as well as sample papers If you can, you should arrange as many as possible numbers of previous and sample papers. There are many

4 considered completed unless you have achieved your goal. Part (iv): After completion of part (iii), you need to increase your speed. This part is penultimate stage of your final achievement. You should try to solve the complete paper of maths within a minimum

Quicker Maths possible time (say, 30-35 minutes for PO paper). This part may take 3 to 4 months. Keep patience and go on practicing. Part (v): This is the last part of your practice. After part (iv), you should take your test with complete fulllength paper for PO.

Chapter 2

Addition In the problem of addition we have two main factors (speed and accuracy) under consideration. We will discuss a method of addition which is faster than the method used by most people and also has a higher degree of accuracy. In the latter part of this chapter we will also discuss a method of checking and double-checking the results. In using conventional method of addition, the average man cannot always add a fairly long column of figures without making a mistake. We shall learn how to check the work by individual columns, without repeating the addition. This has several advantages: 1) We save the labour of repeating all the work; 2) We locate the error, if any, in the column where it occurs; and 3) We are certain to find error, which is not necessary in the conventional method. This last point is something that most people do not realise. Each one of us has his own weaknesses and own kind of proneness to commit error. One person may have the tendency to say that 9 times 6 is 56. If you ask him directly he will say “54”, but in the middle of a long calculation it will slip out as “56”. If it is his favourite error, he would be likely to repeat it when he checks by repetition.

Totalling in columns As in the conventional method of addition, we write the figures to be added in a column, and under the bottom figure we draw a line, so that the total will be under the column. When writing them we remember that the mathematical rule for placing the numbers is to align the right-hand-side digits (when there are whole numbers) and the decimal points (when there are decimals). For example: Right-hand-side-digits Decimal alignment alignment 4234 13.05 8238 2.51 646 539.652 5321 2431.0003

350 49.24 9989 The conventional method is to add the figures down the right-hand column, 4 plus 8 plus 6, and so on. You can do this if you wish in the new method, but it is not compulsory; you can begin working on any column. But for the sake of convenience, we will start on the righthand column. We add as we go down, but we “never count higher than 10”. That is, when the running total becomes greater than 10, we reduce it by 10 and go ahead with the reduced figure. As we do so, we make a small tick or checkmark beside the number that made our total higher than 10. For example: 4 8 4 plus 8,12: this is more than 10, so we subtract 10 from 12. Mark a tick and start adding again. 6 6 plus 2, 8 1 1 plus 8, 9 0 0 plus 9, 9 9 9 plus 9, 18: mark a tick and reduce 18 by 10, say 8. The final figure, 8, will be written under the column as the “running total”. Next we count the ticks that we have just made as we dropped 10’s. As we have 2 ticks, we write 2 under the column as the “tick figure”. The example now looks like this: 4234 8238 646 5321 350 9989 running total: 8 ticks: 2 If we repeat the same process for each of the columns we reach the result:

Quicker Maths

6 4234 8238 646 5321 350 9989 running total: 6558 ticks: 2222 Now we arrive at the final result by adding together the running total and the ticks in the way shown in the following diagram, running total: 0 6 5 5 8 0 ticks: 0 2 2 2 2 0 Total: 2 8 7 7 8 Save more time: We observe that the running total is added to the ticks below in the immediate right column. This addition of the ticks with immediate left column can be done in single step. That is, the number of ticks in the first column from right is added to the second column from right, the number of ticks in the 2nd column is added to the third column, and so on. The whole method can be understood in the following steps. 4 23 4 8 2 3 8 64 6 5321 350 9 9 8 9 Total: 8 [4 plus 8 is 12, mark a tick and add 2 to 6, which is 8; 8 plus 1 is 9; 9 plus 0 is 9; 9 plus 9 is 18, mark a tick and write down 8 in the first column of total-row.]

Step I.

Step II.

Total:

4234 8 2 3 8 6 4 6 5321 3 50 9 9 8 9 78

[3 plus 2 (number of ticks in first column) is 5; 5 plus 3 is 8; 8 plus 4 is 12, mark a tick and carry 2; 2 plus 2 is 4; 4 plus 5 is 9; 9 plus 8 is 17, mark a tick and write down 7 in 2nd column of total-row.] In a similar way we proceed for 3rd and 4th columns.

4 2 3 4 8 2 3 8 6 4 6 53 2 1 3 5 0 9 9 8 9 Total:

287 7 8

Note:We see that in the leftmost column we are left with 2 ticks. Write down the number of ticks in a column left to the leftmost column. Thus we get the answer a little earlier than the previous method. One more illustration : Q: 707.325 + 1923.82 + 58.009 + 564.943 + 65.6 = ? Solution: 707.325 1923.82 58.009 564.943 65.6 Total: 3319.697 You may raise a question:is it necessary to write the numbers in column-form? The answer is ‘no’. You may get the answer without doing so. Question written in a row-form causes a problem of alignment. If you get command over it, there is nothing better than this. For initial stage, we suggest you a method which would bring you out of the alignment problem. Step I. “Put zeros to the right of the last digit after decimal to make the no. of digits after decimal equal in each number.” For example, the above question may be written as 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 Step II. Start adding the last digit from right. Strike off the digit which as been dealt with. If you don’t cut, duplication may occur. During inning total, don’t exceed 10. That is, when we exceed 10, we mark a tick anywhere near about our calculation. Now, go ahead with the number exceeding 10. 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = __ _ _ . __7

5 plus 0 is 5; 5 plus 9 is 14, mark a tick in rough area and carry over 4; 4 plus 3 is 7; 7 plus 0 is 7, so write down 7. During this we strike off all the digits which are used. It saves us from confusion and duplica-tion. Step III. Add the number of ticks (in rough) with the digits in 2nd places, and erase that tick from rough. 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = _ _ _ . _97

1 (number of tick) plus 2 is 3; 3 plus 2 is 5; 5 plus 0

Addition is 5; 5 plus 4 is 9 and 9 plus 0 is 9; so write down 9 in its place. Step IV. 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = _ _ _ _ .697

3 plus 8 is 11; mark a tick in rough and carry over 1; 1 plus 0 is 1; 1 plus 9 is 10, mark another tick in rough and carry over zero; 0 plus 6 is 6, so put down 6 in its place. Step V. Last Step: Following the same way get the result: 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = 3319.697

Addition of numbers (without decimals) written in a row form Q. 53921 + 6308 + 86 + 7025 + 11132 = ? Soln: Step I: 53921 + 6308 + 86 + 7025 + 11132 = ___ _ 2 Step II: 53921 + 6308 + 86 + 7025 + 11132 = _ _ _ 72 Step III: 53921 + 6308 + 86 + 7025 + 11132 = _ _ 472 Step IV: 53921 + 6308 + 86 + 7025 + 11132 = _ 8472 Step V: 53921 + 6308 + 86 + 7025 + 11132 = 78472 Note: One should get good command over this method because it is very much useful and fast-calculating. If you don’t understand it, try again and again. Addition and subtraction in a single row Ex. 1: 412 – 83 + 70 = ? Step I: For units digit of our answer add and subtract the digits at units places according to the sign attached with the respective numbers. For example, in the above case the unit place of our temporary result is 2–3+0=–l So, write as: 412 – 83 + 70 = _ _ (–1) Similarly, the temporary value at tens place is 1 – 8 + 7 = 0. So, write as: 412 – 83 + 70 = _ (0) (–1) Similarly, the temporary value at hundreds place is 4. So, we write as: 412 – 83 + 70 = (4) (0) (–l) Step II: Now, the above temporary figures have to be changed into real value. To replace (–1) by a +ve digit we borrow from digits at tens or hundreds. As the digit at tens is zero, we will have to borrow from hundreds. We borrow 1 from 4 (at hundreds) which becomes 10 at tens leaving 3 at hundreds. Again we borrow 1 from tens which becomes 10 at units place, leaving 9 at tens. Thus, at units place 10 – 1 = 9. Thus our final result = 399. The above explanation can be represented as

7 (–1) (10) (–1) (10) (4) (0) (–1) (3) (9) (9) Note: The above explanation is easy to understand. And the method is more easy to perform. If you practise well, the two steps (I & II) can be performed simultaneously. The second step can be performed in another way like: (4) (0) (–l) = 400 – 1= 399 Ex. 2: 5124 – 829 + 731– 435 Soln: According to step I, the temporary figure is: (5) (– 4) (0) (– 9) Step II: Borrow 1 from 5. Thousands place becomes 5 – 1= 4.1 borrowed from thousands becomes 10 at hundreds. Now, 10 – 4 = 6 at hundreds place, but 1 is borrowed for tens. So digit at hundreds becomes 6 –1 = 5.1 borrowed from hundreds becomes 10 at tens place. Again we borrow 1 from tens for units place, after which the digit at tens place is 9. Now, 1 borrowed from tens becomes 10 at units place. Thus the result at units place is 10 – 9 = 1. Our required answer = 4591 Note: After step I we can perform like: 5 (– 4) (0) (– 9) = 5000 – 409 = 4591 But this method can’t be combined with step I to perform simultaneously. So, we should try to understand steps I & II well so that in future we can perform them simultaneously. Ex. 3: 73216 – 8396 + 3510 – 999 = ? Soln: Step I gives the result as: (7) (–2) (–5) (–16) (–9) Step II: Units digit = 10 – 9 = 1 [1 borrowed from (–16) results –16 –1 = –17] Tens digit = 20 –17 = 3 [2 borrowed from (–5) results –5 – 2 = –7] Hundreds digit = 10 – 7 = 3 [1 borrowed from –2 results –2 –1 = – 3] Thousands digit = 10 – 3 = 7 [1 borrowed from 7 results 7–1=6] So, the required value is 67331. The above calculations can also be started from the leftmost digit as done in last two examples. We have started from rightmost digit in this case. The result is the same in both cases. But for the combined operation of two steps you will have to start from rightmost digit (i.e. units digit). See Ex. 4. Note: Other method for step II: (–2) (–5) (–16) (–9) = (–2) (–6) (–6) (–9) = (–2669) Ans = 70000 – (2669) = 67331

Quicker Maths

8 Ex. 4: 89978 - 12345 - 36218 = ? Soln: Step I: (4) (1) (4) Step II: 4 1 4

(2) 1

(–5) 5

Single step solution: Now, you must learn to perform the two steps simultaneously. This is the simplest example to understand the combined method. At units place: 8 - 5 - 8 = (-5). To make it positive we have to borrow from tens. You should remember that we can’t borrow from -ve value i.e., from 12345. We will have to borrow from positive value i.e. from 89978. So, we borrowed 1 from 7 (tens digit of 89978): (-1) 8 9 9 7 8 - 12345 - 36218 = _ _ _ _ 5 Now digit at tens: (7 - 1 =) 6 - 4 - 1 = 1 Digit at hundreds: 9 - 3 - 2 = 4 Digit at thousands: 9 - 2 - 6 = 1 Digit at ten thousands: 8 - 1 - 3 = 4 the required value = 41415 Ex. 5: 28369 + 38962 - 9873 = ? Soln: Single step solution: Units digit = 9 + 2 – 3 = 8 Tens digit = 6 + 6 – 7 = 5 Hundreds digit = 3 + 9 – 8 = 4 Thousands digit = 8 + 8 - 9 = 7 Ten thousands digit = 2 + 3 = 5 required value = 57458 Ex. 6: Solve Ex. 2 by single-step method. Soln: 5124 – 829 + 731– 435 = Units digit: 4 – 9 + 1 – 5 = (–9). Borrow 1 from tens digit of the positive value. Suppose we borrowed from 3 of 731. Then –1 5124 – 829 + 731 – 435 = _ _ _ 1 Tens digit: 2 – 2 + 2 – 3 = (–1). Borrow 1 from hundreds digit of +ve value. Suppose we borrowed from 7 of 731. Then –1 –1 5124 – 829 + 731 – 435 = _ _ 9 1 Hundreds digit: 1 – 8 + 6 – 4 = (–5). Borrow 1 from thousands digit of +ve value. We have only one such digit, i.e. 5 of 5124. Then -1 -1 -1 5 1 2 4 – 829 + 731 - 435 = 4591 (Thousands digit remains as 5 - 1 = 4) Now you can perform the whole calculation in a single step without writing anything extra.

Ex. 7: Solve Ex. 3 in a single step without writing anything other than the answer. Try it yourself. Don’t move to next example until you can confidently solve such questions within seconds. Ex. 8: 10789 + 3946 – 2310 – 1223 = ? Soln: Whenever we get a value more than 10 after addition of all the units digits, we will put the units digit of the result and carry over the tens digit. We add the tens digit to +ve value, not to the –ve value. Similar method should be adopted for all digits. +1 +l +1 1 0 7 8 9 + 3946 – 2310 – 1223 = 11202 Note: 1. We put +1 over the digits of +ve value 10789. It can also be put over the digits of 3946. But it can’t be put over 2310 and 1223. 2. In the exam when you are free to use your pen on question paper you can alter the digit with your pen instead of writing +1, +2, –1, –2 .... over the digits. Hence, instead of writing 8, you should write 9 over 8 with your pen. 1

Similarly, write 8 in place of 7 . Ex. 9: 765.819 – 89.003 + 12.038 – 86.89 = ? Soln: First, equate the number of digits after decimals by putting zeros at the end. So, ? = 765.819 – 89.003 + 12.038 – 86.890 Now, apply the same method as done in Ex. 4, 5, 6, 7 & 8. -1 -1 -l -l +1 7 6 5. 8 1 9 - 89.003 + 12.038 – 86.890 = 601.964 Ex 10: 5430 - 4321 + 3216 - 6210 = ? Soln: The above case is different. The final answer comes negative. But as we don't know this in the beginning, we perform the same steps as done earlier. Step I: (-2) (1) (1) (5) Step II: The leftmost digit is negative. It can't be made positive as there is no digit at the left which can lend. So, our answer is

2000 115 1885 Note: The second step should be done mentally keeping in mind that except the leftmost digit all the other digits are positive. So, the final answer will be -ve but not (–)2115. It should be -2000 +115= –1885.

Addition

9

Ex 11: 2695 - 4327 + 3214 - 7350 = ? Soln: Step I: (-6) (2) (3) (2) So, required answer =

6000 232 5768

Method of checking the calculation: Digit sum Method This method is also called the nines-remainder method. The concept of digit-sum consists of this : I. We get the digit-sum of a number by “adding across” the number. For instance, the digit-sum of 13022 is 1 plus 3 plus 0 plus 2 plus 2 is 8. II. We always reduce the digit-sum to a single figure if it is not already a single figure. For instance, the digit-sum of 5264 is 5 plus 2 plus 6 plus 4 is 8 (17, or 1 plus 7 is 8). III. In “adding across” a number, we may drop out 9’s. Thus, if we happen to notice two digits that add up to 9, such as 2 and 7, we ignore both of them; so the digit-sum of 990919 is 1 at a glance. (If we add up 9’s we get the same result.) IV. Because “nines don’t count” in this process, as we saw in III, a digit-sum of 9 is the same as a digit-sum of zero. The digit-sum of 441, for example, is zero. Quick Addition of Digit-sum: When we are “adding across” a number, as soon as our running total reaches two digits we add these two together, and go ahead with a single digit as our new running total. For example: To get the digit-sum of 886542932851 we do like: 8 plus 8 is 16, a two – figure number. We reduce this 16 to a single figure: 1 plus 6 is 7. We go ahead with this 7; 7 plus 6 is 4 (13, or 1+3=4), 4 plus 5 is 9, forget it. 4 plus 2 is 6. Forget 9 .... Proceeding this way we get the digit-sum equal to 7. For decimals we work exactly the same way. But we don’t pay any attention to the decimal point. The digitsum of 6.256, for example, is 1. Note: It is not necessary in a practical sense to understand why the method works, but you will see how interesting this is. The basic fact is that the reduced digit-sum is the same as the remainder when the number is divided by 9. For example: Digit-sum of 523 is 1. And also when 523 is divided by 9, we get the remainder 1.

Checking of Calculation Basic rule: Whatever we do to the numbers, we also do to their digit-sum; then the result that we get

from the digit-sum of the numbers must be equal to the digit-sum of the answer. For example: The number: 23 + 49 + 15 + 30 = 117 The digit-sum: 5 + 4 + 6 + 3 = 0 Which reduces to : 0 =0 This rule is also applicable to subtraction, multiplication and upto some extent to division also. These will be discussed in the coming chapters. We should take another example of addition. 1.5 + 32.5 + 23.9 = 57.9 digit-sum: 6 + 1 + 5 =3 or, 3 =3 Thus, if we get LHS = RHS we may conclude that our calculation is correct. Sample Question: Check for all the calculations done in this chapter. Note: Suppose two students are given to solve the following question: 1.5 + 32.5 + 23.9 = ? One of them gets the solution as 57.9. Another student gets the answer 48.9. If they check their calculation by this method, both of them get it to be correct. Thus this method is not always fruitful. If our luck is against us, we may approve our wrong answer also. Addition of mixed numbers 1 4 1 Q. 3 4 9 ? 2 5 3 Solution:A conventional method for solving this question is by converting each of the numbers into pure fractional numbers first and then taking the LCM of denominators. To save time, we should add the whole numbers and the fractional values separately. Like here,

1 4 1 1 4 1 3 4 9 (3 4 9 ) 2 5 3 2 5 3 = 16

5 24 10 19 16 1 30 30

= (16 1)

19 19 19 17 17 30 30 30

2 1 3 1 Q. 5 4 2 1 3 6 4 4

Soln:

2 1 3 1 (5 4 2 1) 3 6 4 4

12 8 2 9 3 2 = 2 = 2 + 1 =3 12 12

10

Quicker Maths

Chapter 3

Multiplication Special Cases We suggest you to remember the tables up to 30 because it saves some valuable time during calculation. Multiplication should be well commanded, because it is needed in almost every question of our concern. Let us look at the case of multiplication by a number more than 10.

Multiplication by 11 Step I: The last digit of the multiplicand (number multiplied) is put down as the right-hand figure of the answer. Step II: Each successive digit of the multiplicand is added to its neighbour at the right.

The answer is 64812. As you see, each figure of the long number is used twice. It is first used as a “number”, and then, at the next step, it is used as a neighbour. Looking carefully, we can use just one rule instead of three rules. And this only rule can be called as “add the right neighbour” rule. We must first write a zero in front of the given number, or at least imagine a zero there. Then we apply the idea of adding the neighbour to every figure of the given number in turn: 05892 11 As there is no neighbour on the right, so 2 we add nothing.

05892 11 -------- As we did earlier 4812

Ex. 1. Solve 5892 × 11 = ? Soln: Step I: Put down the last figure of 5892 as the right-hand figure of the answer: 5892 11 2 Step II: Each successive figure of 5892 is added to its right-hand neighbour. 9 plus 2 is 11, put 1 below the line and carry over 1. 8 plus 9 plus 1 is 18, put 8 below the line and carry over 1. 5 plus 8 plus 1 is 14, put 4 below the line and carry over 1.

5892 11 (9 + 2 = 11, put 1 below the line and 12 carry over 1)

5892 11 (8 + 9 + 1 = 18, put 8 below the line and 812 carry over 1)

5892 11 (5 + 8 + 1 = 14, put 4 below the line and 4812 carry over 1) Step III: The first figure of 5892, 5 plus 1, becomes the left-hand figure of the answer: 5892 11 64812

05892 11 -------- Zero plus 5 plus carried-over 1 64812 is 6 This example shows why we need the zero in front of the multiplicand. It is to remind us not to stop too soon. Without the zero in front, we might have neglected the last 6, and we might then have thought that the answer was only 4812. The answer is longer than the given number by one digit, and the zero in front takes care of that. Sample Problems: Solve the following: 1) 111111 × 11 2) 23145 × 11 3) 89067 × 11 4) 5776800 × 11 5) 1122332608 × 11 Ans: 1) 1222221 2) 254595 3) 979737 4) 63544800 5) 12345658688

Multiplication by 12 To multiply any number by 12, we “Double each digit in turn and add its neighbour.” This is the same as multiplying by 11 except that now we double the “number” before we add its “neighbour.”

Quicker Maths

12 For example: Ex 1: Solve:5324 × 12 Soln: 05324 12 Step I. (double the right-hand figure 8 and add zero, as there is no neighbour)

05324 12 88 05324 12 Step III. 888 05324 12 Step IV. 3888 Step II.

Last Step.

05324 12 63888

(double the 2 and add 4) (double the 3 and add 2) (double the 5 and add 3 (=13), put 3 below the line and carry over 1) (zero doubled is zero, plus 5

plus carried-over 1) The answer is 63,888. If you go through it yourself you will find that the calculation goes very fast and is very easy. Practice Question: Solve the following: 1) 35609 × 12 2) 11123009 × 12 3) 456789 × 12 4) 22200007 × 12 5) 444890711 × 12 Ans: 1) 427308 2) 133476108 3) 5481468 4) 266400084 5) 5338688532

Multiplication by 13 To multiply any number by 13, we “Treble each digit in turn and add its right neighbour.” This is the same as multiplying by 12 except that now we “treble” the “number” before we add its “neighbour”. If we want to multiply 9483 by 13, we proceed like this: 09483 13 Step I. (treble the right-hand figure and 9 write it down as there is no neighbour on the right) 09483 13 Step II. (8 × 3 + 3 = 27, write down 7 79 and carry over 2) 09483 13 Step III. (4×3+8+2 = 22, write down 2 279 and carry over 2)

Step IV.

09483 13 (9×3 + 4 + 2 = 33, write down 3 3279

and carry over 3) 09483 13 Last Step. 123279 (0 × 3 + 9 + 3 = 12, write it down) The answer is 1,23,279. In a similar way, we can define rules for multiplication by 14, 15,....But, during these multiplications we will have to get four or five times of a digit, which is sometimes not so easy to carry over. We have an easier method of multiplication for those large values. Can you get similar methods for multiplication by 21 and 31? It is not very tough to define the rules. Try it.

Multiplication by 9 Step I: Subtract the right-hand figure of the long number from 10. This gives the right-hand figure of the answer. Step II: Taking the next digit from right, subtract it from 9 and add the neighbour on its right. Step III: At the last step, when you are under the zero in front of the long number, subtract one from the neighbour and use that as the left-hand figure of the answer. Ex. 1: 8576 × 9 = ? 08576 9 77184 Step I. Subtract the 6 of 8576 from 10, and we have 4 of the answer. Step II. Subtract the 7 from 9 (we have 2) and add the neighbour 6; the result is 8. Step III. (9 – 5) + 7 = 11; put 1 under the line and carry over 1. Step IV. (9 – 8) + 5 + l (carried over) = 7, put it down. Step V (Last step). We are under the left-hand zero, so we reduce the left-hand figure of 8576 by one, and 7 is the left-hand figure of the answer. Thus answer is 77184. Here are a few questions for you. 1) 34 × 9 = ? 2) 569 × 9 = ? 3) 1328 × 9 = ? 4) 56493 × 9 = ? 5) 89273258 × 9 = ? Answers: 1) 306 2) 5121 3)11952 4) 508437 5) 803459322 We don’t suggest you to give much emphasis on this rule. Because it is not very much easy to use. Sometimes it proves very lengthy also.

Soln:

Multiplication

13

Another method: Step I: Put a zero at the right end of the number; i.e., write 85760 for 8576. Step II: Subtract the original number from that number. Like 85760 – 8576 = 77184

Suppose you are given a large number like 125690258. And someone asks you to multiply that number by 25, What will you do? Probably you will do nothing but go for simple multiplication. Now, we suggest you to multiply that number by 100 and then divide by 4. To do so remember the two steps: Step I: Put two zeroes at the right of the number (as it has to be multiplied by 100). Step II: Divide it by 4. So, your answer is 12569025800 ÷ 4 = 3142256450. Is it easier than your method?

General Rule for Multiplication Having dealt in fairly sufficient detail with the application of special cases of multiplication, we now proceed to deal with the “General Formula” applicable to all cases of multiplication. It is sometimes not very convenient to keep all the above cases and their steps in mind, so all of us should be very much familiar with “General Formula” of multiplication.

Multiplication by a two-digit number Ex. 1: Solve (1) 12 × 13 = ? (2) 17 × 18 = ? (3) 87 × 92 = ? Soln: (1) 12 × 13 = ? Step I. Multiply the right-hand digits of multiplicand and multiplier (unit-digit of multiplicand with unit-digit of the multiplier). 1 2 1

3 6 (2 × 3) Step II. Now, do cross-multiplication, i.e., multiply 3 by 1 and 1 by 2. Add the two products and write down to the left of 6. 2

1

3 56

1

2

1

3

1 5 6

Multiplication by 25

1

Step III. In the last step we multiply the left-hand figures of both multiplicand and multiplier.

( 1 × 1)

(2) 17 × 18 = ? Step I. 1 1

7 8 6 (7 × 8 = 56, write down 6 carry over 5) Step II. 1 7 1 8 0 6 (1 × 8 + 7×1+5 = 20, write down 0 and carry over 2) Step III. 1 7 1 8 3 0 6 (1 × 1 + 2 = 3, write it down) (3) 87 × 92 = ? Step I. 8 7 9 2 4 (7 × 2 = 14, write down 4 and carry over 1) Step II. 8 7 9 2 04 (8 × 2 + 9 × 7 + 1 = 80, write down 0 and carry over 8) Step III. 8 7 9 2 8004 (8 × 9 + 8 = 80) Practice questions: 1) 57 × 43 2) 51 × 42 3) 38 × 43 4) 56 × 92 5) 81 × 19 6) 23 × 99 7) 29 × 69 8) 62 × 71 9) 17 × 37 10) 97 × 89 Answers: l) 2451 2) 2142 3) 1634 4) 5152 5) 1539 6) 2277 7) 2001 8) 4402 9) 629 10) 8633 Ex 2. Solve (1) 325 × 17 = ? (2) 4359 × 23 = ? Soln: (1): Step I. 3 2 5 1 7

(3 × 1 + 2 × 1)

5 (5 × 7 = 35, put down 5 and carry over 3)

Quicker Maths

14 We can write all the steps together:

3 2 5

Step II.

1 7 2 5 (2×7+5×1+3 = 22, put down 2 and carry over 2)

3 2 5

Step III.

1 7

5 2 5 (3 × 7 + 2 × 1 + 2 = 25, put down 5 and carry over 2) Note: Repeat the cross-multiplication until all the consecutive pairs of digits exhaust. In step II, we cross-multiplied 25 and 17 and in step III, we crossmultiplied 32 and 17. Step IV. 3

2

5

1 7 5 5 2 5 (3 × 1 + 2 = 5, Put it down) (2)

Step I. 4 3 5 9 2 3 7 (9 × 3 = 27, put down 7, carry over 2)

4 3 5 9 2 3 4 2 / 4 3 2 3/ 3 3 5 2 / 5 3 9 2 / 9 3 = 10 20 22 35 27 = 100257 Or, we can write the answer directly without writing the intermediate steps. The only thing we should keep in mind is the “carrying numbers”.

4359 23 10 2 0 2 2 35 2 7

4359 23 or, 100257

Note: You should try for this direct calculation. It saves a lot of time. It is a very systematic calculation and is very easy to remember. Watch the above steps again and again until you get that systematic pattern of crossmultiplication.

Multiplication by a three-digit number Ex: 1. Solve (1) 321 × 132 = ? (2) 4562 × 345 = ? (3) 69712 × 641 = ? Soln: (1) Step I.

3 2 1

Step II. 4 3 5 9 1 3 2

2 3

5 7

(5 × 3 + 9 × 2 + 2 = 35, put down 5 and carry over 3)

2 (1 × 2 = 2)

Step II.

3 2 1

Step III. 4 3 5 9 2 3

2 5 7

1 3 2

(3×3+5×2+3 = 22, put down 2, carry over 2)

7 2 (2 × 2 + 3 × 1= 7)

Step III.

Step IV. 4 3 5 9

3 2 1

2 3

0 2 5 7

1 3 2

(4×3+3×2+2 = 20, put down 0, carry over 2)

3 7 2

Step V. 4 3 5 9 Step IV.

2 3 100257

(4 × 2 + 2 = 10, put it down)

3 2 1 1 3 2 2 3 7 2

(2 × 3 + 3 × 2 + 1 × 1 = 13, write down 3 and carry over 1)

Multiplication Step V.

15 V. 6 9 7 1 2 VI. 6 9 7 1 2 VII. 6 9 7 1 2

3 2 1

641 641 641 For each of the groups of figures, you have to crossmultiply.

1 3 2 4 2 3 7 2 (1 × 3 + 1 = 4)

or,

3 21 13 2 1 3 / 3 3 1 2 / 2 3 3 2 11/ 2 2 3 1/1 2 = 4 12 = 42372 Soln: (2):

7

13

2

5

8

3

9

1

(5 × 6 = 30, write down 0 and carry over 3)

Step II.

4 6 3 2 / 5 6 4 2 / 2 5 3

2) 646329 × 8124

4 3 2 5

0

345 4 3/ 4 4 3 5 / 5 4 4 5 3 6 / 5 5

6

Example: Solve 1) 4325 × 3216 Soln: 1) 4325 × 3216 = ? Step I.

3 2 1 6

4562

= 15 37 = 1573890 Soln: (3):

Multiplication by a four-digit number

4 3 2 5

0 3 2 1 6 0 0 (2×6+1×5+3 = 20, write down 0 and carry over 2)

6 9 712 6 41 6 6 / 4 6 6 9 / 1 6 4 9 6 7 /1 9 4 7

Step III.

6 1/ 1 7 4 1 6 2 /11 4 2 /1 2 = 44 86 88 45 23 9 2 = 44685392 Note: Did you get the clear concept of crossmultiplication and carrying-cross-multiplication? Did you mark how the digits in cross-multiplication increase, remain constant, and then decrease? Take a sharp look at question (3). In the first row of the answer, if you move from right to left, you will see that there is only one multiplication (1×2) in the first part. In the second part there are two (1×1 and 4×2), in the 3rd part three (1×7, 4×1 and 6×2), in the 4th part three (1×9, 4×7 and 6 ×1), in the 5th part again three (1×6, 4×9 and 6×7), in the 6th part two (4 × 6 and 6 × 9) and in the last part only one (6 × 6) multiplication. The participation of digits in crossmultiplication can be shown by the following diagrams. I. 6 9 7 1 2 II. 6 9 7 1 2 III. 6 9 7 1 2 IV. 6 9 7 1 2 641

641

641

641

4 3 2 5 3 2 1 6 2 0 0

Step IV.

4 3 2 5

3 2 1 6 9 2 0 0 Step V.

4 3 2 5 3 2 1 6 0 9 2 0 0

Step VI. 4 3 2 5

3 2 1 6 9 0 9 2 0 0

(4× 2 + 3 × 3 + 2 = 19, write down 9 and carry over 1)

Quicker Maths

16 Step VII. 4 3 2 5

3 2 1 6 1390 9 2 0 0

(4 × 3 + 1 = 13, write it down)

Ans: 13909200 Soln: 1) 646329 × 8124 = ? Try this question yourself and match your steps with the given diagrammatic presentation of participating digits. Step I. 6 4 6 3 2 9 8124

Step II. 6 4 6 3 2 9 8124

Step III. 6 4 6 3 2 9 8124

Step IV. 6 4 6 3 2 9 8124

Step V. 6 4 6 3 2 9 8124

Step VI. 6 4 6 3 2 9 8124

Step VII. 6 4 6 3 2 9 8124

Step VIII. 6 4 6 3 2 9 8124

Step IX. 6 4 6 3 2 9 8124

Practice Problems: Solve the following: 1) 234 × 456 2) 336 × 678 3) 872 × 431 4) 2345 × 67 5) 345672 × 456 6) 569 × 952 7) 458 × 908 8) 541 × 342 9) 666 × 444 10) 8103 × 450 11) 56321 × 672 12) 1278 × 569 13) 5745 × 562 14) 4465 × 887 15) 8862 × 341 Answers: 1) 106704 2) 227808 3) 375832 4) 157115 5) 157626432 6) 541688 7) 415864 8) 185022 9) 295704 10) 3646350 11) 37847712 12) 727182 13) 3228690 14) 3960455 15) 3021942

Checking of Multiplication Ex 1:

15 × 13 = 195 digit-sum: 6 × 4 = 6 or, 24 = 6 or, 6 = 6. Thus, our calculation is correct. Ex. 2: 69712 × 641 = 44685392 digit-sum: 7 × 2 = 5 or, 14 = 5 or, 5 =5 Therefore, our calculation is correct. Ex. 3: 321 × 132 = 42372 digit-sum: 6 × 6 = 9 or, 36 = 9 or, 9 =9 Thus, our calculation is correct. But if someone gets the answer 43227, and tries to check his calculation with the help of digit-sum rule, see what happens: 321 × 132 = 43227 digit sum: 6×6 =9 or, 36 = 9 or, 9 =9 This shows that our answer is correct, but it is not true. Thus we see that if out luck is very bad, we can approve a wrong answer.

Chapter 4

Division We now go on to the Quicker Maths of at-sight division which is based on long-established Vedic process of mathematical calculations. It is capable of immediate application to all cases and it can be described as the “crowning gem of all” for the universality of its applications. To understand the at-sight mental one-line method of division, we should take an example and its explanation.

DIVISION BY A 2-DIGIT NUMBER Ex 1. Divide 38982 by 73. Soln: Step I. Out of the divisor 73, we put down only the first digit, ie, 7 in the divisor-column and put the other digit, ie, 3 “on top of the flag”, as shown in the chart below. 7 3 38 9 8 2 The entire division will be by 7. Step II. As one digit (3) has been put on top, we allot one place at the right end of the dividend to the remainder position of the answer and mark it off from the digits by a vertical line.

7

3

38 9 8 2

Step III. As the first digit from the left of dividend (3) is less than 7, we take 38 as our first dividend. When we divide 38 by 7, we get 5 as the quotient and 3 as the remainder. We put 5 down as the first quotientdigit and just prefix the remainder 3 before the 9 of the dividend.

7

3

38 3 9 8 2 5

Step IV. Now our dividend is 39. From this we, however, deduct the product of the indexed 3 and the first quotient-digit (5), ie, 3×5 = 15. The remainder 24 is our actual net-dividend. It is then divided by 7 and gives us 3 as the second quotient-digit and 3 as the remainder, to be placed in their respective places as was done in third step.

7

3

38 39 38 2 5 3

Step V. Now our dividend is 38. From this we subtract the product of the index (3) and the 2nd quotientdigit (3), ie, 3 × 3 = 9. The remainder 29 is our next actual dividend and divide that by 7. We get 4 as the quotient and 1 as the remainder. We put them at their respective places.

7

3

38 3 9 38 12 5 3 4

Step VI. Our next dividend is 12 from which, as before, we deduct 3 × 4 ie, 12 and obtain 0 an the remainder

7

3

38 39 38 12 5 3 4 0

Thus we say : Quotient (Q) is 534 and Remainder (R) is 0. And thus finishes the whole procedure; and all of it is one-line mental arithmetic in which all the actual division is done by the single-digit divisor 7. The procedure is very simple and needs no further exposition and explanation. A few more illustrations with running comments will be found useful and helpful and are therefore given below : Ex 2: Divide 16384 by 128 (As 12 is a small number to handle with, we can treat 128 as a two-digit number). Soln:

12

8

16 43 118 64 1 2 8 0

Step I. We divide 16 by 12. Q = 1 & R = 4. Step II. 43 – 8 × 1 = 35 is our next devidend. Dividing it by 12, Q = 2, R = 11. Step III. 118 – 8 × 2 = 102 is our next dividend. Dividing it by 12, Q = 8, R = 6 Step IV. 64 – 8 × 8 = 0

Quicker Maths

18 Then our final quotient = 128 & remainder = 0 Ex 3: Divide 601325 by 76.

7

Soln :

6

60 111 63 22 7 9 1 2

25 13 (=25 - 6×2)

Step I. Here, in the first division by 7, if we put 8 down as the first quotient-digit, the remainder then left will be too small for the subtraction expected at the next step. We get -ve dividend in next step which is absurd. So, we take 7 as the quotient-digit and prefix the remainder 11 to the next dividend-digit. All the other steps are similar to the previously mentioned steps in Ex 1 & 2. Our final quotient is 7912 and remainder is 13. If we want the values in decimal, we go on dividing as per rule instead of writing down the remainder. Such as;

76

60 1 3 2 7

9 1 2

5 60 50 1 7 1

Ans = 7912.171 Note: The vertical line separating the remainder from the quotient part may be the demarcating point for decimal. Ex 4: Divide 7777777 by 38 Soln:

38 7 2

17 17 57 0 4 6

77 87 77 7 8 13 (=77 - 8×8)

Q = 204678, R = 13 You must go through all the steps of the above solution. Try to solve it. Did you find some difference? Ex 5: Divide 8997654 by 99. Try it step by step. Ex 6:(i) Divide 710.014 by 39 (to 4 places of decimals) (ii) 718.589 ÷ 23 = ? (iii) 718.589 ÷ 96 = ? Soln. (i) Since there is one flag-digit the vertical line is drawn such that one digit before the decimal comes under remainder portion.

3

9

7 41 1 8

80. 20 2 0

21 64 70 5 4

For the last section, we had 64 - 45 = 19 as our dividend, divided by 3 we choose 4 as our suitable quotient. If we take 5 as a quotient it leaves 4 as remainder (19 – 15). Now the next dividend will be 40 – 9 × 5 = –5, which is not acceptable.

The vertical line separating the remainder from quotient part may be demarcation point for decimal. Therefore, ans = 18.2054 (ii)

2

3

71 1 3 1

08. 15 2 4

18 09 3 0

8

3

Ans = 31.2430 (iii)

7

4

5

2

Ans = 7.4853 Ex 7 : Divide 7.3 by 53 Soln.

5 3 7. 23 50 6 0 40 40 1 3 7 7 3 5

Ans = 0.137735

DIVISION BY A 3-DIGIT NUMBER Ex 8 : Divide 7031985 by 823. Soln: Step I. Here the divisor is of 3 digits. All the difference which we make is to put the last two digits(23) of divisor on top. As there are two flag-digits (23), we will separate two digits (85) for remainder.

8

23

70 3 1 9 85

Step II. We divide 70 by 8 and put down 8 and 6 in their proper places.

8

23

70 63 1 9 85 8 Step III. Now, our gross dividend is 63. From that we subtract 16, the product of the tens of the flagdigits, ie 2, and the first quotient-digit, ie 8, and get the remainder 63 – 16 = 47 as the actual dividend. And, dividing it by 8, we have 5 and 7 as Q & R respectively and put them at their proper places.

8 23 70 63 71 9 85 8 5 Step IV. Now our gross dividend is 71, and we deduct the cross-products of two flag-digits 23 and the two quotient digits (8 & 5) ie 2 × 5 + 3 × 8 = 10 + 24 = 34; and our remainder is 71 – 34 = 37. We then continue to divide 37 by 8 . We get Q = 4 & R =5

Division

19

8 23 70 63 71 59 85 8 5 4 Step V. Now our gross dividend is 59. And actual dividend is equal to 59 minus cross-product of 23 and 54, ie, 59 – (2 × 4 + 3 × 5) = 59 – 23 = 36. Dividing 36 by 8, our Q = 4 and R = 4.

8 23 70 63 71 59 485 8 5 4 4 Step VI. Actual dividend = 48 – (3 × 4 + 2 × 4) = 48 – 20 = 28. Dividing it by 8, our Q = 3 & R = 4

8 23 70 63 71 59 48 45 8 5 4 4 3 Step VII. Actual dividend = 45 – (3 × 4 + 2 × 3) = 45 – 18 = 27. Dividing 27 by 8, we have Q = 3 & R = 3.

8 23 70 63 71 59 48 45 3 8 5 4 4 3 3 The vertical line separating the remainder from the quotient part may be a demarcation point for decimal. Ans = 8544.33 Our answer can be 8544.33, but if we want the quotient and remainder, the procedure is somewhat different. In that case, we do not need the last two steps, ie, the calculation upto the stage

8 23 70 63 71 59 48 5 8 5 4 4 is sufficient to answer the question. Quotient = 8544; Remainder = 485 – 10 × (Cross multiplication of 23 and 44)* – unit digit of flagged number × unit digit of quotient. = 485 –10 (4 × 2 + 4 × 3) – 3×4 = 485 – 200 –12 = 273 * Cross-multiplication of the two flag-digits and last two digits of quotient. Ex 9: Divide 1064321 by 743 ( to 4 places of decimals). Also find the remainder. Soln:

7 43 10 36 44 43 52 1 1 4 3 2

Q = 1432, Remainder = 521 – 10 (Cross-multiplication of 43 & 32) – 3×2 = 521 – 10 × 17 – 6 = 345 For decimals :

7 43 10 3 6 44 43 1 4 3 2

52 71 70 60 50 4 6 4 3

Ans = 1432.4643 Ex 10: Divide 888 by 672 ( to 4 places of decimals).

6

72

8 2 8 38 30 40 5 0 1 3 2 1 4 2

Ans = 1.3214 Note : Vertical line separating the remainder from the quotient part is the demarcation point for decimal. Can you find the quotient and remainder? Try it. Ex 11: Divide 4213 by 1234 to 4 places of decimals. Also find quotient and remainder. Soln: Although 1234 is a four-digit number, we can treat it as a 3-digit number because 12 is small enough to handle with.

12

34

42 61 3 3 Q=3, R = 613-10 (cross multiplication of 03 and 34) – 4 × 3 = 613 – 90 – 12 = 511

12

34

42 3

6 1 43 70 30 2 0 4 1 4 1 0

Ans = 3.4141 Note:Division by 4-digit or 5-digit number is not of much use. So these are not being discussed here. Now you must have seen all the possible cases which you may come across in mathematical division. Don’t escape any of the examples discussed above. Having a broad idea of at-sight mathematical division, you should solve as many questions yourself as you can.

Checking of Calculation Rule of digit-sum fails in some cases here. Ex 1: We will check the calculations of the examples one by one. 38982 ÷ 73 = 534 (Since remainder is zero) digit sum: 3 ÷ 1 = 3 or, 3 = 3 Therefore, our calculation is correct.

Quicker Maths

20 Ex 2: Check yourself. Ex 3: 601325 ÷ 76 gives quotient = 7912 and remainder = 13 We may write the above calculation as 7912 × 76 + 13 = 601325 Now, check the correctness. digit sum : 1 × 4 + 4 = 8 or 4 + 4 = 8 or, 8 = 8 Therefore, our calculation is correct. Ex 4: Check yourself. Ex 5: Check yourself. Ex 6: (i) 710.014 ÷ 39 = 18.2054... We can’t apply the digit-sum method to check our calculation because our calculation is not complete. (ii) 718.589 ÷ 23 = 31.2430 This calculation is complete because in the end we get 0 as quotient and remainder. We can apply the digit-sum rule in this case. digit-sum = 2 ÷ 5 = 4 or, 0.4 = 4 or, 4 = 4 Our calculation is correct. Note : During digit-sum (forget-nine method) we don’t take decimals into account. Ex 7: Tell whether digit-sum rule is applicable to this calculation or not. Note : Thus, we see the limitation of digit-sum rule.

Practice Problems Q.1. Divide ‘x’ by ‘y’ where a) x=135921 & y = 89

b) x=64932 & y = 99 c) x=8899 & y = 101 d) x=9995 & y =122 e) x=13596289 & y = 76 f) x= 89325 & y = 132 g) x=89 & y = 71 h) x=96 & y = 95 i) x=53 & y = 83 j) x=93 & y = 109 Q.2. Divide ‘x’ by ‘y’ where a) x= 359281 & y = 567 b) x= 8932 & y = 981 c) x= 99899 & y = 789 d) x= 1053 & y = 989 e) x= 738 & y = 895 f) x= 13569 & y = 1051 g) x= 69325 & y = 1163 h) x= 935 & y = 1259 i) x= 100002 & y = 777 j) x= 12345 & y = 567 Answers: 1. a) 1527.2022 c) 88.10891 e) 178898.539 g) 1.2535 i) 0.6385 2. a) 633.6525 c) 126.6147 e) 0.8245 g) 59.6087 i) 128.7027

b) 655.87878 d) 81.926229 f) 676.7045 h) 1.0105 j) 0.8532 b) 9.1049 d) 1.0647 f) 12.9105 h) 0.7426 j) 21.7724

Chapter 5

Divisibility We now take up the interesting question as to how one can determine whether a certain given number, however large it may be, is divisible by a certain given divisor. There is no defined general rule for checking the divisibility. For different divisors, the rules differ at large. We will discuss the rule for divisors from 2 to 19.

Divisibility by 2 Rule: Any number, the last digit of which is either even or zero, is divisible by 2. For example: 12, 86, 130, 568926 and 5983450 are divisible by 2 but 13, 133 and 596351 are not divisible by 2.

Divisibility by 3 Rule: If the sum of the digits of a number is divisible by 3, the number is also divisible by 3. For example: 1) 123: 1 + 2 + 3 = 6 is divisible by 3; hence 123 is also divisible by 3. 2) 5673: 5 + 6 + 7 + 3 = 21; therefore divisible by 3. 3) 89612: 8 + 9 + 6 + 1 + 2 = 26 = 2 + 6 = 8 is not divisible by 3. Therefore, the number is not divisible by 3.

Divisibility by 4 Rule: If the last two digits of a number is divisible by 4, the number is divisible by 4. The number having two or more zeros at the end is also divisible by 4. For example: 1) 526428: 28 is divisible by 4. Therefore, the number is divisible by 4. 2) 5300: There are two zeros at the end, so it is divisible by 4. 3) 134000: As there are more than two zeros, the number is divisible by 4. 4) 134522: As the last two-digit number (22) is not divisible by 4, the number is not divisible by 4. Note: The same rule is applicable to check the divisibility by 25. That is, a number is divisible by 25 if its last two digits are either zeros or divisible by 25.

Divisibility by 5 Rule: If a number ends in 5 or 0, the number is divisible by 5. For example: 1) 1345: As its last digit is 5, it is divisible by 5. 2) 1340: As its last digit is 0, it is divisible by 5. 3) 1343: As its last digit is neither 5 nor 0, it is not divisible by 5.

Divisibility by 6 Rule: If a number is divisible by both 3 and 2, the number is also divisible by 6. So, for a number to be divisible by 6, 1) the number should end with an even digit or 0 and 2) the sum of its digits should be divisible by 3. For example: 1) 63924 : The first condition is fulfilled as the last digit (4) is an even number and also (6+3+9+2+4=)24 is divisible by 3; therefore, the number is divisible by 6. 2) 154 : The first condition is fulfilled but not the second; therefore, the number is not divisible by 6. 3) 261 : The first condition is not fulfilled; therefore, we need not to check for the second condition.

Special Cases The rules for divisibility by 7, 13, 17, 19 ... are very much unique and are found very rarely. Before going on for the rule, we should know some terms like “one-more” osculator and negative osculator. “One-more” osculator means the number needs one more to be a multiple of 10. For example: osculator for 19 needs 1 to become 20 (=2 × 10), thus osculator for 19 is 2 (taken from 2 × 10 = 20). Similarly osculator for 49 is 5 (taken from 5 × 10 = 50). Negative osculator means the number should be reduced by one to be a multiple of 10. For example:

Quicker Maths

22 Negative osculator for 21 is 2 (taken from 2 × 10 = 20). Similarly, negative osculator for 51 is 5 (taken from

5 × 10 = 50). Note: (1) What is the osculator for 7? Now, we look for that multiple of 7 which is either less or more by 1 than a multiple of 10. For example 7 × 3 = 21, as 21 is one more than 2 × 10; our negative osculator is 2 for 7. And 7 × 7 = 49 or 49 is one less than 5 × 10; our ‘one-more’ osculator is 5 for 7. Similarly, osculators for 13, 17 and 19 are: For 13: 13 × 3 = 39, “one more” osculator is 4 (from

4 × 10) For 17: 17 × 3 = 51, negative osculator is 5 (from

5 × 10) For 19: 19 × 1. “one-more” osculator is 2 (from 2 × 10) (2) Can you define osculators for 29, 39, 21, 31, 27 and 23. (3) Can you get any osculator for an even number or a number ending with ‘5’? (No. But why?)

Divisibility by 7 First of all we recall the osculator for 7. Once again, for your convenience, as 7 × 3 = 21 (one more than 2 × 10), our negative osculator is 2. This oscuator ‘2’ is our key-digit. This and only this digit is used to check the divisibility of any number by 7. See how it works: Ex 1: Is 112 divisible by 7? Soln: Step 1. 11 2: 11 – 2 × 2 = 7 As 7 is divisible by 7, the number 112 is also divisible by 7. Ex 2: Is 2961 divisible by 7? Soln: Step I: 296 1: 296 – 1 × 2 = 294 Step II: 29 4: 29 – 4 × 2 = 21 As 21 is divisible by 7, the number is also divisible by 7. Ex 3: Is 55277838 is divisible by 7? Soln: 5527783 8: 5527783 – 8 × 2 = 5527767 552776 7: 552776 – 7 × 2 = 552762 55276 2: 55276 – 2 × 2 = 55272 5527 2: 5527 – 2 × 2 = 5523 552 3: 552 – 3 × 2 = 546 54 6: 54 – 6 × 2 = 42

As 42 is divisible by 7, the number is also divisible by 7. Note: (1) In all the examples, each of the numbers obtained after the equal sign (=) is also divisible by 7. Whenever you find a number which looks divisible by 7, you may stop there and conclude the result without any hesitation. (2) The above calculations can be done in one line or even mentally. Try to do so.

Divisibility by 8 Rule: If the last three digits of a number is divisible by 8, the number is also divisible by 8. Also, if the last three digits of a number are zeros, the number is divisible by 8. Ex. 1. 1256: As 256 is divisible by 8, the number is also divisible by 8. Ex. 2. 135923120: As 120 is divisible by 8, the number is also divisible by 8. Ex. 3. 139287000: As the number has three zeros at the end, the number is divisible by 8. Note: The same rule is applicable to check the divisibility by 125.

Divisibility by 9 Rule: If the sum of all the digits of a number is divisible by 9, the number is also divisible by 9. Ex. 1. 39681: 3 + 9 + 6 + 8 + 1=27 is divisible by 9, hence the number is also divisible by 9. Ex. 2. 456138: 4 + 5 + 6 + 1 + 3 + 8= 27 is divisible by 9, hence the number is also divisible by 9.

Divisibility by 10 Rule: Any number which ends with zero is divisible by 10. There is no need to discuss this rule.

Divisibility by 11 Rule: If the sums of digits at odd and even places are equal or differ by a number divisible by 11, then the number is also divisible by 11. Ex. 1. 3245682: S1 = 3 + 4 + 6 + 2= 15 and

S2 = 2 + 5 + 8 = 15 As S1 = S2 , the number is divisible by 11. 1. Ex. 2. 283712: S1 = 2 + 3 + 1= 6 and S2 = 8 + 7 + 2 = 17. As S1 and S2 differ by 11 (divisible by 11), the number is also divisible by 11.

Divisibility Ex. 3. 84927291658: S1 = 8 + 9 + 7 + 9 + 6 + 8 = 47 and S2 = 4 + 2 + 2 + 1 + 5 = 14 As (S1 – S 2 ) 33 is divisible by 11, the number is also divisible by 11.

Divisibility by 12 Rule: Any number which is divisible by both 4 and 3, is also divisible by 12. To check the divisibility by 12, we 1) first divide the last two-digit number by 4. If it is not divisible by 4, the number is not divisible by 12. If it is divisible by 4 then 2) check whether the number is divisible by 3 or not. Ex. 1. 135792: 92 is divisible by 4 and also (l + 3 + 5 + 7 + 9 + 2=)27 is divisible by 3; hence the number is divisible by 12. Remark: Recall the method for calculation of digit-sum. What did you do earlier (in the 1st chapter)? “Forget nine”. Do the same here. For example: digit-sum of 135792 __ 1 plus 3 plus 5 is 9, forget it. 7 plus 2 is 9, forget it. And finally we get nothing. That means all the “forget nine” adds to a number which is multiple of 9. Thus, the number is divisible by 9.

Divisibility by 13 Oscuator for 13 is 4 (See note). But this time, our osculator is not negative (as in case of 7). It is ‘onemore’ osculator. So, the working principle will be different now. This can be seen in the following examples. Ex 1: Is 143 divisible by 13? Soln: 14 3: 14 + 3 × 4 = 26 Since 26 is divisible by 13, the number 143 is also divisible by 13. or, This working principle may further be simplified as: Step I : 1 4 3 16 [4 × 3 (from 143) + 4 (from 143)] Step II : 1 4 3 26/16 [4 × 6 (from 16) +1 (from 16) +1 (from 143) = 26] As 26 is divisible by 13, the number is also divisible by 13. Note: The working of second method is also very systematic. At the same time, it is more acceptable because it has less writing work.

23 Ex 2: Check the divisibility of 24167 by 13. 2 4 1 6 7 26/6/20/34 [4 × 7 (from 24167) + 6 (from 24167) = 34] [4 × 4 (from 3 4) +3 (from 34) +1 (from 24167) = 20] [4 × 0 (from 20) +2 (from 20) +4 (from 24167) = 6] [4 × 6 (from 6) +2 (from 24167) = 26] Since 26 is divisible by 13 the number is also divisible by 13. Remark: Have you understood the working principle? If your answer is no, we suggest you to go through each step carefully. This is very simple and systematic calculation. Ex 3: Check the divisibility of 6944808 by 13. Soln: 6 9 4 4 8 0 8 39/18/12/41/19/32 4 × 8 + 0 = 32 4 × 2 + 3 + 8 = 19 4 × 9 + 1 + 4 = 41 4 × 1 + 4 + 4 = 12 4 × 2 + 1 + 9 = 18 4 × 8 + 1 + 6 = 39 Since 39 is divisible by 13, the given number is divisible by 13. Note: (1) This method is applicable for “one-more” osculator only. So we can’t use this method in the case of 7. (2) This is a one-line method and you don’t need to write the calculations during exams. These are given merely to make you understand well.

Divisibility by 14 Any number which is divisible by both 2 and 7, in also divisible by 14. That is, the number’s last digit should be even and at the same time the number should be divisible by 7.

Divisibility by 15 Any number which is divisible by both 3 and 5 is also divisible by 15.

Divisibility by 16 Any number whose last 4 digit number is divisible by 16 is also divisible by 16.

Quicker Maths

24

Divisibility by 17

Divisibility by 18

Negative osculator for 17 is 5 (see note). The working for this is the same as in the case of 7. Ex. 1: Check the divisibility of 1904 by 17. Soln: 190 4: 190 – 5 × 4 = 170 Since 170 is divisible by 17, the given number is also divisible by 17. Note: Students are suggested not to go upto the last calculation. Whenever you find the number divisible by the given number on right side of your calculation stop further calculation and conclude the result. Ex. 2: 957508: 95750 8: 95750 – 5 × 8 = 95710 9571 0: 9571 – 5 × 0 = 9571 957 l: 957 – 5 × l = 952 95 2: 95 – 5 × 2 = 85 Since 85 is divisible by 17, the given number is divisible by 17. Ex. 3: 8971563: 897156 3: 897156 – 5 × 3 = 897141 89714 1: 89714 – 5 × 1 = 89709 8970 9: 8970 – 5 × 9 = 8925 892 5: 892 – 5 × 5 = 867 86 7: 86 – 5 × 7 = 51 Since 51 is divisible by 17, the given number is also divisible by 17.

Rule: Any number which is divisible by 9 and has its last digit (unit-digit) even or zero, is divisible by 18. Ex. 1. 926568: Digit-sum is a multiple of nine (ie, divisible by 9) and unit-digit (8) is even, hence the number is divisible by 18. Ex. 2. 273690: Digit-sum is a multiple of nine and the number ends in zero, so the number is divisible by 18. Note: During the calculation of digit-sum, follow the method of “forget nine”. If you get zero at the end of your calculation, it means the digit-sum is divisible by 9.

Divisibility by 19 If you recall, the ‘one-more’ osculator for 19 is 2. The method is similar to that of 13, which is well known to you. Let us take an example. Ex. 1: 149264 Soln: 1 4 9 2 6 4 19/9/12/11/14 Thus, our number is divisible by 19. Note: You must have understood the working principle (see the case of 13).

Chapter 6

Squaring Squaring of a number is largely used in mathematical calculations. There are so many rules for special cases. But we will discuss a general rule for squaring which is capable of universal application. The method of squaring is intimately connected with a procedure known as the “Duplex Combination” process. We now go on to a brief study of this procedure.

Duplex Combination Process The first one is by squaring; and the second one is by cross-multiplication. In the present context, it is used in both senses (a2 and 2ab). In the case of a single central digit, the square is meant; and in the case of an even number of digits equidistant from the two ends, double the cross-product is meant. A few examples will elucidate the procedure. Ex. 1: For 2, Duplex (D) = 22 = 4 Ex. 2: For 8, D = 82 = 64 Ex. 3: For 34, D = 2 × (3 × 4) = 24 Ex. 4: For 79, D = 2 × (7 × 9) = 126 Ex. 5: For 103, D = 2(1 × 3) + 02 = 6 Ex. 6: For 346, D = 2(3 × 6) + 42 = 52 Ex. 7: For 096, D = 2(0 × 6) + 92 = 81 Ex. 8: For 1342, D = 2(1 × 2) +2(3 × 4) = 28 Ex. 9: For 5156, D = 2(5 × 6) + 2(1 × 5) = 70 Ex.10: For 23564, D = 2(2 × 4) +2(3 × 6) + 52 = 77 Ex.11: For 123456, D = 2(1 × 6) +2 (2 × 5) +2 (3 × 4) = 56 Now, we see the method of squaring in the following examples. Ex. 1. 2072 = ? Soln: 2072 = D for 2 / D for 20 / D for 207 / D for 07 / D for 7 = 22/ 2(2 × 0) / 2(2 × 7) + 02/ 2(0 × 7) / 72 =4 / 0 / 28 / 0 / 49 =4 / 0 / 8 / 0 / 49 2 = 4 / 0 + 2 / 8 / 0 + 4 / 9 = 42849 If you have understood the duplex method and its use in squaring, you may get the answer in a line. For example: 2072 = 4228449.

Explanations. 1. Duplex of 7 is 72 = 49. Put the unit digit (9) of duplex in answer line and carry over the other (4). 2. 2 × 0 × 7 + 4(carried) = 4; write it down at 2nd position. 3. 2 × 2 ×7 + 02 = 28; write down 8 and carry over 2. 4. 2 × 2 × 0 + 2(carried) = 2; write it down. 5. 22 = 4; write it down. Note: (1) If there are n digits in a number, the square will have either 2n or 2n-l digits. (2) Participation of digits follows the same systematic pattern as in multiplication. Ex. 2: (897)2 = 8016420613049 = 804609 Explanations: 1. 72 = 49; write down 9 and carry over 4. 2. 2 × 9 × 7 + 4 (carried) = 130; write down 0 and carry over 13. 3. 2 × 8 × 7 + 92 + 13 = 206; write down 6 and carry over 20. 4. 2 × 8 × 9 + 20 = 164; write down 4 and carry over 16. 5. 82 + 16 = 80; write it down. Ex. 3: (1 4 3 2)2 = 210253026124 = 2050624 Explanations: 1. 22 = 4; write it down. 2. 2 × (3 × 2) = 12; write down 2 and carry over 1. 3. 2 × (4 × 2) + 32 + l = 26; write down 6 and carry over 2. 4. 2(1 × 2) + 2(4 × 3) + 2 = 30; write down 0 and carry over 3. 5. 2(1 × 3) + 42 +3 = 25; write down 5 and carry over 2. 6. 2(1 × 4) + 2 = 10; write down 0 and carry over 1. 7. 12 +1 = 2; write it down. Ex 4: (73214)2 = 53464032682917916 = 5360289796 Ex 5: (5432819)2 = 29455155115102162122128661472681 = 29515522286761

26 Practice problem: Q: Find the squares of the following numbers: 1) 835 2) 8432 3) 45321 4) 530026 5) 73010932 Answers: 1) 697225 2) 71098624 3) 2053993041 4) 280927560676 5) 5330596191508624 Note: To find the square of a fractional (decimal) number, we square the number without looking at decimal. After that we count the number of digits after the decimal in the original value. In the squared value, we place the decimal after double the number of digits after decimal in the original value. For example: (12.46)2 = 151532455l36 = 155.2516 Some special cases derived with help of Duplex Combination Process 1. Square of numbers from 51 to 59. We take a general representative of the numbers (from 51 to 59), say 5A. Now, (5A)2 = 52 / 2 × 5 × A / A2 = 25 / 10 × A / A2 We have; 10 × A = A0 [like 10 × 4 = 40, 10 × 6 = 60, etc] (5A)2 = 25 / A0 / A2 = (25 + A) / A2 ; where A2 should be written as a two-digit number Now, we see that our duplex combination process reduces to a simplier form. Using the above equation: Ex. 1: (51)2 = 25 + 1 / (1)2 = 26 / 01 = 2601 Ex. 2: (52)2 = 25 + 2 / (2)2 = 27 / 04 = 2704 Ex. 3: (54)2 = 25 + 4 / (4)2 = 29 / 16 = 2916 Ex. 4: (59)2 = 25 + 9 / (9)2 = 34 / 81 = 3481 2. Square of a number with unit digit as 5. We take a general representative of such number, say A5. Now, (A5)2 = A2/2 × A × 5/52 = A2/10 × A / 25

Quicker Maths = A2 / A0 / 25 [ 10 × A = A0] = A2 + A / 25 = A (A + 1) / 25 Using the above equation: Ex. 1: (15)2 = 1 × (1 + 1) / 52 = 2 / 25 = 225 Ex. 2: (25)2 = 2 × (2 + 1) / 52 = 6 / 25 = 625 Ex. 3: (85)2 = 8 × (8 + 1) / 52 = 72 / 25 = 7225 Ex. 4: (115)2 = 11 × (12) / 52 = 132 / 25 = 13225 Ex. 5: (225)2 = 22 × (23) / 52 = 506 / 25 = 50625 3. Square of a number wich is nearer to 10x We use the algebraic formula x2 = (x2 – y2) + y2 = (x + y) (x – y) + y2 Ex. 1: (98)2 = (98 + 2) (98 – 2) + 22 = 9600 + 4 = 9604 Ex. 2: (103)2 = (103 – 3) (103 + 3) + 32 = 10600 + 9 = 10609 Ex. 3: (993)2 = (993 + 7) (993 – 7) + 72 = 986000 + 49 = 986049 Ex. 4: (1008)2 = (1008 – 8) (1008 + 8) + 82 = 1016000 + 64 = 1016064 To check the calculation We use the digit-sum method for checking calculations in squaring. For example: In Ex 1: (207)2 = 42849 digit-sum: (0)2 = (0)2 Hence, our calculation is correct. In Ex 2: (897)2 = 804609 digit-sum: (6)2 = 18 or, 36 =18 or, 0 = 0 Thus, our calculation is correct. In Ex 3: (1432)2 = 2050624 digit-sum: 12 = 1 or, 1 = 1 Thus, our calculation is correct. Note: 1. Follow the “forget-nine” rule during the calculation of digit-sum. 2. Check all the calculations mentally. 3. Check the correctness of calculations in other examples without using pen.

Chapter 7

Cube Cubes of large numbers are rarely used. During our mathematical calculations, we sometimes need the cube value of two-digit numbers. So, an easy rule for calculating the cubes of 2-digit numbers is being given. In its process the cube values of the “first ten natural numbers”, ie, 1 to 10, are used. Readers are suggested to remember the cubes of only these “first ten natural numbers.” 13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216, 73 = 343, 83 = 512, 93 = 729 and 103 = 1000. To calculate the cube value of two-digit numbers we proceed like this: Step I: The first thing we have to do is to put down the cube of the tens-digit in a row of 4 figures. The other three numbers in the row of answer should be written in a geometrical ratio in the exact proportion which is there between the digits of the given number. Step II: The second step is to put down, under the second and third numbers, just two times of second and third number. Then add up the two rows. For example: Ex 1. 123 = ? Soln. Step I: We see that the tens-digit in the number is 1, so we write the cube of 1. And also as the ratio between 1 and 2 is 1 : 2, the next digits will be double the previous one. So, the first row is 1 2 4 8 Step II: In the above row our 2nd and 3rd digits (from right) are 4 and 2 respectively. So, we write down 8 and 4 below 4 and 2 respectively. Then add up the two rows. 1 2 4 8 4 8 1 7 12 8 = 1728 Ex. 2: 113 = ? (Solve it yourself.) Ex. 3: 163 = ? Soln: 1 6 36 216 12 72 4 30 9 6 = 4096 12 21 Explanations: 13 (from 16) = 1. So, 1 is our first digit in the first row. Digits of 16 are in the ratio 1 : 6, hence our other digits should be 1 × 6 = 6, 6 × 6 = 36, 36 × 6 = 216. In the second row, double the 2nd and 3rd number

is written. In the third row, we have to write down only one digit below each column (except under the last column which may have more than one digit). So, after putting down the units-digit, we carry over the rest to add up with the left-hand column. Here, (i) Write down 6 of 216 and carry over 21. (ii) 36 + 72 + 21 (carried) = 129, write down 9 and carry over 12. (iii) 6 + 12 + 12 (carried) = 30, write down 0 and carry over 3. (iV) 1 + 3 (carried) = 4, write down 4. Ex 4: 183 = ? Soln:

1

8

64

512

5

16 48

128 24 3

512

= 5832

(i) Write down 2 and carry over 51 of 512. (ii) 64 +128 + 51 = 243, write down 3 and carry over 24. (iii) 8 + 16 + 24 = 48, write down 8 and carry over 4. (iv) 1 + 4 = 5 write it down. Ex 5: 173 = ? (Solve it yourself) Ex 6: 193 = ? (Solve it yourself) Ex 7: 213 = ? Soln: 8 4 2 1 [8 = 23, 8 ÷ 2 = 4, 4 ÷ 2 = 2, 2 ÷ 2 = 1, since ratio is 2:1] 8 4 [4 × 2 = 8, 2 × 2 = 4, double is written below] 9 12 6 1 = 9261 Do you mark the difference? If no, go through the following explanations. Step I: (i) 23 = 8 is the first figure (from left) in the first row. (ii) Ratio between the two digits is 2 : 1, ie, the number should be halved subsequently. Therefore, the next three numbers in the first row should be 4, 2 & 1.

Quicker Maths

28 Step II:It should be clear to all of you because it has nothing new. Ex 8: 233 = ? Soln: 8 12 18 27 24 36 12 41 12167 56 27 Explanations: Step I: (i) 23 = 8 ---------- the first figure (from left) in the first row. 3 (ii) 2 : 3 the next numbers should be of 2 the previous ones. So,

we have 8 ×

3 3 3 = 12, 12 × =18, 18 × = 27. 2 2 2

Ex 9: 333 = ? Soln:

27

27 54

27 54

27

35

89

83

27

= 35937

Explanations: Step i) 33 = 27 ------ the first figure in first row. ii) 3 : 3 = 1 : 1 the subsequent numbers should be the same. Ex 10: 343 = ? Soln:

27

36 72

48 96

64

39

12 3

15 0

64

= 39304

(ii) Ratio is 3 : 4, ie, the next numbers should be of their previous ones. Here, 27 ×

24 5

27

4 3

4 = 36, 3

4 4 36 × = 48, 48 × = 64. 3 3 3 Ex 11: 93 = ? Soln: 729 243 81 27 486 162 753

Soln:

729

567 1134

441 882

343

912

1836

1357

34 3

= 912673

Explanations: (i) 93 = 729 ---------- first figure (from left) in the first row. 7 (ii) Ratio = 9 : 7 Next numbers should be of 9 7 the previous ones. Therefore, 729 × = 567, 9 7 7 567 × = 441, 441 × = 343. 9 9 Explanation of the above method: (It leads us to more Quicker Method.) Any two-digit number can be represented as 10x + y. Now,

10x y 3 103 x 3 3102 x 2 y 310xy 2 y 3

Explanations: (i) 33 = 27 --------- the first figure (from left) in the first row.

804

Explanations: (i) 93 = 729 --------- the first figure (from left) in the first row, (ii) 9 : 3 3 : 1 the subsequent figures should be 1 of their previous ones. 3 Ex. l2: 973 = ?

= 1000x 3 100 3x 2 y 10 3xy 2 y 3 The above expansion gives us four parts. The leftmost part (1000x3) has three zeroes at the end. Similarly, the second part has two zeroes and the third part has one zero. This implies that the firstpart leaves three places for the other parts. Similarly, second part leaves two places for the other parts and the third part leaves one place for the last part. You can better understand by an example: Suppose we have to find (23)3. Now ,

233 20 33 203 3202 3 32032 33 = 2 3 1000 322 3 100 32 32 10 33 = 8000 + 36 × 100 + 54 × 10 + 27 The answer is in the form

= 804357

8 – – – 36 – – 54 – 27

Cube

29

523 140

8 (36) (54) (27) 12

1

6

7

Note: In place of 27 we have to write only a single digit, ie 7, and carry forward 2 to the ten’s position. Now at the ten’s position we have 54 + 2 = 56. So, we write 6 at the ten’s position and carry forward 5 to the hundred’s position, and so on. The summary of the above explanation is as follows:

a b 3 233

a3

3a 2 b

3ab 2

b3

23 3(2)2 (3) 3(2)(3)2 (3)3

= 8

36

54

27

= 12

1

6

7

Now, we reach at the stage where we can write the answer directly if a b a 3 3a 2 b 3ab 2 b 3 is at our fingertip. We start from the rightmost position and move towards the leftmost positions. Take another example: 3

213

1

6

Step I:

3

1

3 2 1 6 2

Step II:

2

1 1

6

1

Step III: 3 2 2 1 12 Put 2, carry over 1

9

2

6

1

Step IV: 23 = 8; 8 + 1(carried)= 9 Combining all the steps into a single step:

343

39

3

0

4

253 15

6

2

5

6

0

8

Practice Problems. Q. Find the cubes of the following 1) 17 2) 26 4) 32 5) 41 7) 49 8) 51 10) 55 11) 57 13) 67 14) 69 16) 77 17) 88 19) 95 20) 99 Answers: 1) 4913 4) 32768 7) 117649 10) 166375 13) 300763 16) 456533 18) 857375

2) 17576 5) 68921 8) 132651 11) 185193 14) 328509 17) 681472 19) 970299

numbers. 3) 27 6) 43 9) 53 12) 64 15) 73 18) 92

3) 19683 6) 79507 9) 148877 12) 262144 15) 389017 18) 778688

Note: Don't use this method for getting the cubes of 20, 30, 40, --- Do you know the other quick method?

Checking the correctness (with the help of digitsum) Ex. 1: 123 1728 digit sum:

(3) 3 0 (7 2 9, forget it. 1 8 9, forget it.) or,

0 = 0, thus the cube value is correct.

Ex. 2: 16 4096 3

digit-sum:

73 1 or, (49) 7 1 or, 4×7 =1 or, 28 = 1 or, 1 = 1, Thus cube value is correct. Practice-Problem: Check all the calculations (from Ex 2 to Ex 12) done in this chapter.

30

Quicker Maths

Chapter 8

HCF and LCM Factor: One number is said to be a factor (Gunankhand) of another when it divides the other exactly. Thus, 6 and 7 are factors of 42. Common Factor: A common factor of two or more numbers is a number that divides each of them exactly. Thus, 3 is a common factor of 9, 18,21 and 33. Highest Common Factor (HCF): HCF of two or more numbers is the greatest number that divides each of them exactly. Thus, 6 is the HCF of 18 and 24. Because there is no number greater than 6 that divides both 18 and 24. Note: The terms Highest Common Divisor and Greatest Common Measure are often used in the sense of Highest Common Factor (HCF).

To find the HCF of two or more numbers Method I: Method of Prime Factors Rule: Break the given numbers into prime factors and then find the product of all the prime factors common to all the numbers. The product will be the required HCF. Ex. 1. Find the HCF of 42 and 70. Soln: 42 = 2 × 3 × 7 70 = 2 × 5 × 7 HCF = 2 × 7 = 14 Ex. 2. Find the HCF of 1365, 1560 and 1755. Soln: 1365 = 3 × 5 × 7 × 13 1560 = 2 × 2 × 2 × 3 × 5 × 13 1755 = 3 × 3 × 3 × 5 × 13 HCF = 3 × 5 × 13 = 195 Note: (1) In finding the HCF, we need not break all the numbers into their prime factois. We may find the prime factors of one of the numbers. Then the product of those prime factors which divide each of the remaining numbers exactly will be the required HCF. In Ex. (1), the prime factors of 42 are 2 × 3 ×7. Of these three factors, only 2 and 7 divide 70 exactly. Hence, the required HCF = 2 × 7 = 14 In Ex. (2) the prime factors of 1365 are 3 × 5 × 7 × 13. Of these four factors, only 3,5 and

13 divide the other two numbers 1560 and 1755 exactly. Hence, the required HCF = 3 × 5 × 13 = 195 (2) We must remember that the quotient obtained by dividing the numbers by their HCF are prime to each other. In Ex. (1) 42 ÷ 14 = 3 70 ÷ 14 = 5. We see that 3 is prime to 5, i.e., 3 can’t divide 5 exactly. In Ex. (2), 1365 ÷ 195 = 7, 1560 ÷ 195 = 8 and 1755 ÷ 195 = 9, We see that 7, 8 and 9 are prime to one another, i.e. none divides the other.

Method II: Method of Division Rule: Divide the greater number by the smaller number, divide the divisor by the remainder, divide the remainder by the next remainder, and so on until no remainder is. left. The last divisor is the required HCF. Ex. 1. 42) 70 (1 42 28) 42 (1 28 14) 28 (2 28 0 HCF =14 Note: The above rule for finding the HCF of numbers is based on the following two principles: (i) Any number which divides a certain number also divides any multiple of that number; for example, 6 divides 18 therefore, 6 divides any multiple of 18. (ii) Any number which divides each of the two numbers also divides their sum, their difference and the sum and difference of any multiples of that numbers. Thus 5, being a common factor of 25 and 15, is also a factor of (25 + 15), and (25 – 15). Again, 5 is also a factor of (25 × a + 15 × b) and of (25 × a – 15 × b), where a and b are integers.

Quicker Maths

32 In accordance with these principles, HCF of 42 and 70 = HCF of 28 and 42 [ 28 = 70 – 42 ] = HCF of 14 and 28 [14 = 42 – 28] = HCF of 14 and 14 [ 14 = 28 – 14 ] HCF of 42 and 70 = 14 HCF of 13281 and 15844 = HCF of 2563 and 13281 [2563 = 15844 –13281] = HCF of 466 and 2563 [466 = 13281 – 5 × 2563] = HCF of 233 and 466 [233 = 2563 – 5 × 466 ] = HCF of 233 and 233 [233 = 466 – 233] HCF of 13281 and 15844 = 233 The above discussed method is very much interesting. It gives results very quickly. But one should have a good understanding of this method.

4199) 4693 (1 4199 494 494 is divisible by 2 but 4199 is not. We, therefore, divide 494 by 2 and proceed with 247 and 4199 (by rule iii). 247) 4199 (17 247 1729 1729 0 The HCF of 4199 and 4693 is 247. Hence, the HCF of the original numbers is 247 × 9 = 2223. Note: If the HCF of two numbers be unity, the numbers must be prime to each other.

To find the HCF of more than two numbers

HCF of smaller numbers

Rule: Find the HCF of any two of the numbers and then find the HCF of this HCF and the third number and so on. The last HCF will be the required HCF. Ex. 1. Find the HCF of 1365, 1560 and 1755. Soln: 1365) 1560 (1 1365 195) 1365 (7 1365 0 Therefore, 195 is the HCF of 1365 and 1560. Again, 195 ) 1755 ( 9 1755 0 the required HCF = 195 Method III: The work of finding the HCF may sometimes be simplified by the following devices: (i) Any obvious factor which is common to both numbers may be removed before the rule is applied. Care should however be taken to multiply this factor into the HCF of the quotients. (ii) If one of the numbers has a prime factor not contained in the other, it may be rejected. (iii) At any stage of the work, any factor of the divisor not contained in the dividend may be rejected. This is because any factor which divides only one of the two cannot be a portion of the required HCF. Ex. Find the HCF of 42237 and 75582. Soln: 42237 = 9 × 4693 75582 = 2 × 9 × 4199 We may reject 2 which is not a common factor (by rule (i). But 9 is a common factor. We, therefore, set it aside (by rule ii) and find the HCF of 4199 and 4693.

If the numbers are not very large, we can follow the following steps to get the HCF very quickly. For example, (i) Find the HCF of 8, 20, 28 and 44. Soln: Step I: As we know that HCF is the highest common factor of all the numbers, it cannot be larger than the smallest number. So, take the smallest number, ie 8. Step II: Divide the other numbers by 8. As it does not divide 20, we reject 8 as our HCF. Step III: Take the second highest factor of 8, ie 8 ÷ 2 = 4. Check the divisibility of the other numbers by 4. As it divides all the other numbers, our HCF is 4. (ii) Find the HCF of 21, 15 and 36. Soln: Step I: Take the smallest number 15. As it does not divide the other numbers it is rejected as our HCF. Step II: Take the second highest factor of 15, ie 15 ÷ 3 = 5. As it does not divide the other numbers, it is rejected as our HCF. Step III: Take the next highest factor, ie 15 ÷ 5 = 3. As it divides the other numbers also, our HCF is 3. (iii) Find the HCF of 72, 126 and 198. Soln: Step I: Take the smallest number, ie 72. As it does not divide any other number, it is not our HCF. Step II: Take the second highest factor of 72, ie 72 ÷ 2 = 36. It is rejected as it does not divide other numbers.

HCF and LCM Step III: Take the next fact, ie 72 ÷ 3 = 24. It is also rejected, as it does not divide 126. Step IV: Take the next factor ie 72 ÷ 4 = 18. As it divides all the other numbers, our HCF is 18. (iv) Find the HCF of 24, 60, 84 and 108. Soln: The smallest number 24 is rejected as HCF. The next largest factor of 24, ie 12 is the required HCF as it divides all the other numbers.

33

Thus, the fractions are

HCF of 6, 60, 12 6 HCF LCM of 1, 17, 17 17 Note: We see that each of the numbers is perfectly divisible by

HCF of decimals Rule: First make (if necessary) the same number of decimal places in all the given numbers; then find their HCF as if they are integers and mark off in the result as many decimal places as there are in each of the numbers. Ex. 1: Find the HCF of 16.5, 0.45 and 15. Soln: The given numbers are equivalent to 16.50, 0.45 and 15.00 Step I: First we find the HCF of 1650, 45 and 1500. Which comes to 15. Step II: The required HCF = 0.15 Ex. 2: Find the HCF of 1.7, 0.51 and 0.153. Soln: Step I: First we find the HCF of 1700, 510 and 153. Which comes to 17. Step II: The required HCF = 0.017

HCF of vulgar fractions Def: The HCF of two or more fractions is the highest fraction which is exactly divisible by each of the fractions. Rule: First express the given fractions in their lowest terms: HCF of numerators Then, HCF = LCM of denominators Note: The HCF of a number of fractions is always a fraction (but this is not true with LCM).

54 9 36 , 3 and 9 17 51 54 6 9 60 36 12 ,3 Soln: Here, and 9 1 17 17 51 17 Ex. 1: Find the HCF of

6 . 17

Ex. 2: Find the greatest length that is contained an exact

1 3 number of times in 3 m and 8 m. 2 4

To find the HCF of two or more concrete quantities First, the quantities should be reduced to the same unit. Ex. Find the greatest weight which can be contained exactly in 1 kg 235 gm and 3 kg 430 gm. Soln: 1 kg 235 gm = 1235 gm 3 kg 430 gm = 3430 gm The greatest weight required is the HCF of 1235 and 3430, which will be found to be 5 gm.

6 60 12 , and 1 17 17

Soln:

3

1 7 3 35 and 8 2 2 4 4

The greatest length will be the HCF of

7 35 and . 2 4

the required length

HCF of 7 and 35 7 3 1 m LCM of 2 and 4 4 4

Miscellaneous Examples on HCF Ex. 1: What is the greatest number that will divide 2400 and 1810 and leave remainders 6 and 4 respectively? Soln: Since on dividing 2400 a remainder 6 is left, the required number must divide (2400 – 6) or 2394 exactly. Similarly, it must divide (1810 – 4) or 1806 exactly. Hence, the greatest required number should be the HCF of 2394 and 1806, ie., 42. Ex. 2. What is the greatest number that will divide 38, 45 and 52 and leave remainders as 2, 3 and 4 respectively? Soln: The required greatest number will be the HCF of (38 – 2), (45 – 3) and (52 – 4) or 36, 42 and 48. Ans = 6 Ex. 3: Find the greatest number which will divide 410, 751 and 1030 so as to leave remainder 7 in each case? Soln: The required greatest number = HCF of (410 – 7), (751 – 7) and (1030 – 7). Ans = 31 Ex. 4: Find the greatest number which is such that when 76,151 and 226 are divided by it, the remainders are all alike. Also find the common remainder.

34 Soln:

Let k be the remainder, then the numbers (76 – k), (151 – k) and (226 – k) are exactly divisible by the required number. Now, we know that if two numbers are divisible by a certain number, then their difference is also divisible by that number. Hence, the numbers (151 – k) – (76 – k), (226 – k) – (151 – k) and (226 – k) – (76 – k) or 75, 75 and 150 are divisible by the required number. Therefore, the required number = HCF of 75, 75 and 150 = 75 And the remainder will be found after dividing 76 by 75, as 1. Ex. 5: The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find that number of three digits. Soln: The required number must be a factor of (11284 – 7655) or 3629. Now, 3629 =19 × 191 191 is the required number.. Ex 6: The product of two numbers is 7168 and their HCF is 16; find the numbers. Soln: The numbers must be multiples of their HCF. So, let the numbers be 16a and 16b where a and b are two numbers prime to each other. 16a × 16b = 7168 or, ab = 28 Now, the pairs of numbers whose product is 28 are (28), (1); (14), (2); (7, 4). 14 and 2 which are not prime to each other should be rejected. Hence, the required numbers are 28 × 16, 1 × 16; 7 × 16, 4 × 16 or, 448, 16; 112, 64 Common multiple: A common multiple of two or more numbers is a number which is exactly divisible by each of them. Thus, 30 is a common multiple of 2, 3, 5, 6, 10 and 15. Least common multiple (LCM): The LCM of two or more given numbers is the least number which is exactly divisible by each of them. Thus, 15 is a common multiple of 3 and 5. 30 is a common multiple of 3 and 5. 45 is a common multiple of 3 and 5. But 15 is the least common multiple (LCM) of 3 and 5.

Quicker Maths

To find the LCM of two or more given numbers Method I: Method of Prime Factors Rule: Resolve the given numbers into their prime factors and then find the product of the highest power of all the factors that occur in the given numbers. This product will be the LCM. Ex. 1: Find the LCM of 8,12, 15, and 21. Soln: 8 = 2 × 2 × 2 = 23 12 = 2 × 2 × 3 = 22 × 3 15 = 3 × 5 21 = 3 × 7 Here, the prime factors that occur in the given numbers are 2, 3, 5 and 7 and their highest powers are respectively 23 , 3, 5 and 7. Hence, the required LCM = 23 × 3 × 5 × 7 = 840 Ex. 2: Find the LCM of 18, 24, 60 and 150. Soln: 18 = 2 × 3 × 3 = 2 × 32 24 = 2 × 2 × 2 × 3 = 23 × 3 60 = 2 × 2 × 3 × 5 = 22 × 3 × 5 150 = 2 × 3 × 5 × 5 = 2 × 3 × 52 Here, the prime factors that occur in the given numbers are 2, 3 and 5, and their highest powers are 23 , 32 and 52 respectively . Hence, the required LCM = 23 × 32 × 52 = 1800 Note: The LCM of two numbers which are prime to each other is their product. Thus, the LCM of 15 and 17 is 15 × 17 = 255 Method II: The LCM of several small numbers can be easily found by the following rule: Write down the given numbers in a line separating them by commas. Divide by any one of the prime numbers 2,3,5,7, etc., which will exactly divide at least any two of the given numbers. Set down the quotients and the undivided numbers in a line below the first. Repeat the process until you get a line of numbers which are prime to one another. The product of all divisors and the numbers in the last line will be the required LCM. Note: To simplify the work, we may cancel, at any stage of the process, any one of the numbers which is a factor of any other number in the same line.

HCF and LCM

35

Ex. 1: Find the LCM of 12, 15, 90, 108, 135, 150. Soln:

2 12, 15, 90, 108, 135, 150 ....(1) 3 45, 54, 135, 75 ....(2) 3 18, 45, 25 ....(3) 5 6, 15, 25 ....(4) 6, 3, 5 ....(5) the required LCM = 2 × 3 × 3 × 5 × 6 × 5 = 2700 In line (1), 12 and 15 are the factors of 108 and 90 respectively, therefore, 12 and 15 are struck off. In line (2), 45 is a factor of 135, therefore 45 is struck off. In line (5), 3 is a factor of 6, therefore 3 is struck off. Note: The product of two numbers is equal to the product of their HCF and LCM. For example: The LCM and HCF of 12 and 15 are 60 and 3 respectively. Multiplication of two numbers =12 × 15 = 180; HCF × LCM = 3 × 60 =180 Thus, we see that the product of two numbers is equal to the product of their LCM and HCF.

Rule: First make (if necessary) the same number of decimal places in all the given numbers; then find their LCM as if they were integers, and mark in the result as many decimal places as there are in each of the numbers. Ex. Find the LCM of 0.6, 9.6 and 0.36. Soln: The given numbers are equivalent to 0.60, 9.60 and 0.36. Now, find the LCM of 60, 960 and 36. Which is equal to 2880. the required LCM = 28.80

LCM of fractions The LCM of two or more fractions is the least fraction or integer which is exactly divisible by each of them. Rule: First express the fractions in their lowest terms, then

LCM of numerator HCF of denominator

Ex. 1: Find the LCM of a)

1 5 , 2 8 1 2

c) 4 , 3, 10

a) The required LCM

b)

b)

1 2

108 17 54 , 1 , 375 25 55

LCM of 1 and 5 5 1 2 HCF of 2 and 8 2 2

108 36 17 42 ,1 375 125 25 25

Thus, the fractions are

36 42 54 , and 125 25 55

the required LCM =

LCM of 36, 42 and 54 756 1 151 HCF of 125, 25, 55 5 5

1 9 1 21 c) 4 , 10 2 2 2 2 Thus, the fractions are

9 3 21 , and 2 1 2

the required LCM

LCM of decimals

LCM

Soln:

LCM of 9, 3 and 21 63 63 HCF of 2, 1, 2 1

Note: In Ex. 1 (c), we see that the LCM of fractions is an integer. Thus, we may conclude that LCM of fractions may be a fraction or an integer.

Miscellaneous Examples on LCM Ex. 1: The LCM of two numbers is 2079 and their HCF is 27. If one of the numbers is 189, find the other. Soln: The required number

LCM HCF 2079 27 297 = First number 189 Ex. 2: Find the least number which, when divided by 18, 24, 30 and 42, will leave in each case the same remainder 1. Soln: Clearly, the required number must be greater than the LCM of 18, 24, 30 and 42 by 1. Now, 18 = 2 × 32 24 = 23 × 3 30 = 2 × 3 × 5 42 = 2 × 3 × 7 LCM = 32 × 23 × 5 × 7 = 2520 the required number = 2520 + 1 = 2521 Ex 3. What is the least number which, when divided by 52, leaves 33 as the remainder, and when divided by 78 leaves 59, and when divided by 117 leaves 98 as the respective remainders?

36 Soln:

Since 52 – 33 = 19, 78 – 59 = 19, 117 – 98 = 19 We see that the remainder in each case is less than the divisor by 19. Hence, if 19 is added to the required number, it becomes exactly divisible by 52,78 and 117. Therefore, the required number is 19 less than the LCM of 52, 78 and 117. The LCM of 52,78 and 117 = 468 the required number = 468 – 19 = 449 Ex 4: Find the greatest number of six digits which, on being divided by 6, 7, 8, 9 and 10, leaves 4, 5, 6, 7 and 8 as remainders respectively. Soln: The LCM of 6, 7, 8, 9 and 10 = 2520 The greatest number of 6 digits = 999999 Dividing 999999 by 2520, we get 2079 as remainder. Hence, the 6-digit number divisible by 2520 is (999999 – 2079), or 997920. Since 6 – 4 = 2, 7 – 5 = 2, 8 – 6 = 2, 9 – 7 = 2, 10 – 8 = 2, the remainder in each case is less than the divisor by 2. the required number = 997920 – 2 = 997918 Ex 5: Find the greatest number less than 900, which is divisible by 8,12 and 28. Soln : The least number divisible by 8,12 and 28 is 168. Clearly, any multiple of 168 will be exactly divisible by each of the numbers 8, 12 and 28. But since the required number is not to exceed 900, it is 168 × 5 = 840. Ex 6: Find the least number which, upon being divided by 2, 3, 4, 5,6 leaves in each case a remainder of 1, but when divided by 7 leaves no remainder. Soln: The LCM of 2, 3, 4, 5, 6 = 60 the required number must be = 60k +1, where k is a positive integer. = (7 × 8 + 4)k + 1 = (7 × 8k) + (4k + 1) Now, this number is to be divisible by 7. Whatever may be the value of k, the portion (7 × 8k) is always divisible by 7. Hence, we must choose that least value of k which will make 4k + 1 divisible by 7. Putting k = 1, 2, 3, 4, 5 etc. in succession, we find that k should be 5.

Quicker Maths the required number = 60k + 1 = 60 × 5 + 1 = 301 Note: The above example could also be worded as follows. A person had a number of toys to distribute among children. At first, he gave 2 toys to each child, then 3, then 4, then 5, then 6, but was always left with one. On trying 7 he had none left. What is the smallest number of toys that he could have had? Ex. 7: What least number must be subtracted from 1936 so that the remainder when divided by 9, 10, 15 will leave in each case the same remainder 7? Soln: The LCM of 9, 10 and 15 = 90 On dividing 1936 by 90, the remainder = 46 But 7 is also a part of this remainder . the required number = 46 – 7 = 39 Ex. 8: What greatest number can be subtracted from 10,000 so that the remainder may be divisible by 32, 36, 48 and 54? Soln: LCM of 32, 36, 48, 54 = 864 the required greatest number = 10,000 – 864 = 9,136 Ex. 9: What is the least multiple of 7, which when divided by 2, 3, 4, 5 and 6, leaves the remainders 1, 2, 3, 4, 5 respectively? Soln: LCM of 2, 3,4, 5, 6 = 60 Other numbers divisible by 2, 3, 4, 5, 6 are 60k, where k is a positive integer. Since 2 – 1 = 1, 3 – 2 = 1, 4 – 3 = 1, 5 – 4 = 1 and 6 – 5 = 1, the remainder in each case is less than the divisor by 1, the required number = 60k – 1 = (7 × 8k) + (4k × 1) Now, this number is to be divisible by 7. Whatever may be the value of k the portion 7 x 8k is always divisible by 7. Hence, we must choose the least value of k which will make (4k –1) divisible by 7. Putting k equal to 1, 2, 3, etc. in succession, we find that k should be 2. the required number = 60k –1=60 × 2 – 1 = 119

HCF and LCM

37

EXERCISES 1. What is the greatest number that will divide 2930 and 3250 and will leave as remainders 7 and 11 respectively? 2. What is the least number by which 825 must be multiplied in order to produce a multiple of 715? 3. The LCM of two numbers is 2310 and their HCF is 30. If one of the numbers is 7 × 30, find the other number. 4. Three bells commence tolling together and they toll after 0.25, 0.1 and 0.125 seconds. After what interval will they again toll together? 5. What is the smallest sum of money which contains 2.50, 20, 1.20 and 7.50? 6. What is the greatest number which will divide 410, 751 and 1030 so as to leave the remainder 7 in each case? 7. What is the HCF and LCM of

4 5 7 , and ? 5 6 15

8. Three men start together to travel the same way around a circular track of 11 km. Their speeds are 4, 5.5 and 8 km per hour respectively. When will they meet at the starting point? 9. Find the smallest whole number which is exactly

1 2

1 3

1 4

divisible by 1 , 1 , 2 , 3

1 1 and 4 2 5

10. How many times is the HCF of 48, 36, 72 and 24 contained in their LCM? 11. Find the least square number which is exactly divisible by 4, 5, 6, 15 and 18. 12. Find the least number which, when divided by 8, 12 and 16, leaves 3 as the remainder in each case; but when divided by 7 leaves no remainder. 13. Find the greatest number that will divide 55, 127 and 175 so as to leave the same remainder m each case. 14. Find the least multiple of 11 which, when divided by 8,12 and 16, leaves 3 as remainder. 15. What least number should be added to 3500 to make it exactly divisible by 42, 49, 56 and 63? 16. Find the least number which, when divided by 72, 80 and 88, leaves the remainders 52, 60 and 68 respectively. 17. Find the greatest number of 4 digits which, when divided by 2, 3, 4, 5, 6 and 7, should leave remainder 1 in each case.

18. Find the greatest possible length which can be used to measure exactly the lengths 7m, 3m 85cm, 12m 95cm. 19. Find the least number of square tiles required to pave the ceiling of a hall 15m 17cm long and 9m 2cm broad. 20. What is the largest number which divides 77, 147 and 252 to leave the same remainder in each case? 21. The traffic lights at three different road crossings change after every 48 sec., 72 sec., and 108 sec. respectively. If they all change simultane-ously at 8:20:00 hrs, then at what time will they again change simulta-neously? 22. The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is devided by 2, the quotient is 44. What is the other number? 23. The product of two numbers is 2160 and their HCF is 12. Find the possible pairs of numbers. 24. Find the greatest number of 4 digits and the least number of 5 digits that have 144 as their HCF. 25. Find the least number from which 12, 18, 32 or 40 may be subtracted, each an exact number of times. 26. Find the least number that, being increased by 8, is divisible by 32, 36 and 40. 27. The sum of two numbers is 528, and their HCF is 33. How many pairs of such numbers can be formed? 28. In a school, 391 boys and 323 girls have been divided into the largest possible equal classes, so that each class of boys numbers the same as each class of girls. What is the number of classes ? 29. Is it possible to divide 1000 into two parts such that their HCF may be 15? 30. Show that 2205 and 4862 are prime to each other. 31. WTiat least number must be subtracted from 1294 so that the remainder, when divided by 9, 11, 13 will leave in each case the same remainder 6? 32. Find the sum of the numbers between 300 and 400 such that when they are divided by 6, 9 and 12 (a) it leaves no remainder; and (b) it leaves remainder as 4 in each case. 33. Three friends J, K and L jog around a circular stadium and complete one round in 12, 18 and 20 seconds respectively. In how many minutes will all the three meet again at the starting point?

Quicker Maths

38

Solutions (Hints) 1. The greatest such number will be the HCF of (2930 – 7) and (3250 –11), i.e. 79. 2. 825 = 3 × 5 × 5 × 11; 715 = 5 × 11 × 13 Any multiple of 715 must have factors of 5, 11 and 13. So, 825 should be multiplied by the factor(s) of 715, which is (are) not present in 825. the required least number = 13 3. 4.

5.

6. 7.

2310 30 330 7 30 They will toll together after an interval of time equal to the LCM of 0.25 sec, 0.1 sec and 0.125 sec. LCM of 0.25, 0.1 and 0.125 = (LCM of 250, 100 and 125) × 0.001 = 500 × 0.001 = 0.5 sec. LCM of 2.5, 20, 1.2 and 7.5 = (LCM of 25, 200,12 and 75) × 0.1 = 600 × 0.1 = 60 The required number willl be the HCF of (410 – 7), (751 – 7), (1030 – 7), i.e. 31. HCF of Numerators 1 HCF = = LCM of Denominators 30 The required number

LCM =

LCM of Numerators 140 = = 140 HCF of Denominators 1

8. Time taken by them to complete one revolution

11 11 11 11 2 11 , and hrs respectively , and 4 5.5 8 4 1 8

LCM of

11 2 11 , and 4 1 8

LCM of 11, 2, 11 22 22 hrs. HCF of 4, 1, 8 1

they will meet after 22 hrs. 9. The required smallest number = LCM of the given numbers 10. HCF of 48, 36, 72, 24 = 12; LCM of 48, 36, 72, 24 = 144 LCM = 12 x HCF 11. LCM of 4, 5, 6, 15, 18 = 180, which is exactly divisible by the given numbers. 180 = 2 × 2 × 3 × 3 × 5 = 22 × 32 × 5 Therefore, if 180 is multiplied by 5 (180 × 5 = 900) then the number will be a perfect square as well as divisible by 4, 5, 6, 15 and 18. 12. The reast number which, when divided by 8, 12 and 16, leaves 3 as remainder = (LCM of 8, 12, 16)+ 3 = 48 + 3 = 51

Other such numbers are 48 × 2 + 3 = 99, 48 × 3 + 3 = 147,.... the required number which is divisible by 7 is 147. Note: This is a hit-and-trial method. Can you get the answer by the defined method? (see Ex. 6). 13. Let x be the remainder, then the numbers (55 – x), (127 – x) and (175 – x) are exactly divisible by the required number. Now, we know that if two numbers are divisible by a certain number, then their difference is also divisible by the number. Hence the numbers (127 – x) – (55 – x), (175 – x) – (127– x) and (175 – x) – (55 – x) or, 72, 48 and 120 are divisible by the required number. HCF of 48, 72 and 120 = 24, therefore the required number = 24. Note: Find the HCF of the positive differences of numbers. It will serve your purpose quickly. 14. LCM of 8, 12 and 16 = 48. Such numbers are (48 × 1 + 3) = 51, (48 × 2 + 3) = 99, which is divisible by 11. the required number = 9. Note: This is a hit-and-trial method. Try to solve by the detailed method. 15. LCM of 42, 49, 56, 63 = 3528; therefore, the required least number = 3528 – 3500 = 28 16. 72 – 52 = 20, 80 – 60 = 20, 88 – 68 = 20. We see that in each case, the remainder is less than the divisor by 20. The LCM of 72, 80 and 88 = 7920, therefore, the required number = 7920 – 20 = 7900 17. The greatest number of 4 digits = 9999. LCM of 2, 3, 4, 5, 6, 7 = 420 On dividing 9999 by 420, we get 339 as remainder. the greatest number of 4 digits which is divisible by 2, 3, 4, 5, 6 and 7 = 9999 – 339 = 9660 the required number = 9660 + 1 = 9661 18. The required length = HCF of 7m, 3.85m and 12.95m = (HCF of 700, 385, 1295) × .01m = 35 × .01m = 0.35m = 35 cm 19. Side of each tile = HCF of 1517 and 902 = 41 cm. Area of each tile = 41 × 41 cm2 1517 902 814 the number of tiles 41 41 20. Solve as in Ex. 13. The required number = HCF of (147 – 77), (252 – 147) and (252 – 77) = HCF of 70, 105, 175 = 35 21. LCM of 48, 72, 108 = 432 The traffic lights will change simultaneously after 432 seconds or 7 min = in 12 secs.

HCF and LCM they will change simultaneously at 8 : 27 : 12 hrs. 22. The first number = 2 × 44 = 88 The second number HCF LCM 44 264 132 88 88 FICF = 12. Then let the numbers be 12x and 12y. Now l2x × 12y = 2160 xy = 15 Possible values of x and y are (1, 15); (3, 5); (5, 3); (15, 1) the possible pairs of numbers (12, 180) and (36, 60) The required numbers shcld be multiples of 144. We have the greatest number of 4 digits = 9999. On dividing 9999 by 144, we get 63 as the remainder. the required greatest number of 4 digits = 9999 – 63 = 9936 Again, we have the least number of 5 digits = 10000 On dividing 10,000 by 144, we get 64 as the remainder. the required least number of 5 digits = 10,000 + (144 – 64) = 10,080 The required number = LCM of 12, 18, 32, 40 =1440 LCM of 32, 36 and 40 = 1440, therefore, the required number = 1440-8 = 1432 Let the numbers be 33a and 33b. Now, 33a + 33b = 528 or, 33 (a + b) = 528 a + b = 16 The possible values of a and b are (1, 15); (2, 14); (3, 13); (4, 12); (5, 11); (6, 10); (7, 9); and (8, 8). Of these the pairs of numbers that are prime to each other are (1, 15); (3, 13); (5, 11); and (7, 9). the possible pairs of numbers are (33, 495); (99, 429); (165, 363); (231,297)

23.

24.

25. 26. 27.

39 28. Number of classes = HCF of 391 and 323 = 17 29. If two numbers are divisible by a certain number, then their sum is also divisible by that number. According to this rule: if 15 is the HCF of two parts of 1000, then 1000 must be divisible by 15. But it is not so. Therefore, it is not possible to divide 1000 into two parts such that their HCF may be 15. 30. If the numbers are prime to each other, then their HCF should be unity. Conversely, if their HCF is unity, the numbers are prime to each other. In this case, the HCF is 1, so they are prime to each other. 31. LCM of 9, 11 and 13 = 1287 Therefore, the number which, after being divided by 9, 11 and 13, leaves in each case the same remainder 6 = 1287 + 6 = 1293 the required least number = 1294 – 1293 = 1. 32. The LCM of 6, 9 and 12 = 36 (a) Multiples of 36 which lie between 300 and 400 are 324, 360 and 396. the required sum = 324 + 360 + 396= 1080 (b) Here, the remainder is 4 in each case. So, the numbers are (324 + 4 =) 328 and (360 + 4 =) 364. (The no. 396 + 4 = 400 does not lie between 300 and 400 so it is not acceptable.) the required sum = 328 + 364 = 692. 33. J, K and L will meet again at the starting point after LCM of 12, 18 and 20. LCM of 12, 18 and 20 = 180 seconds = 3 min Note: Why LCM? Because to meet again at the starting point all of them should take a certain number of complete rounds. We can see LCM (180 sec) is exactly divisible by 12, 18 or 20 seconds. So, J completes (180 ÷ 12 =) 15 rounds, K completes (180 ÷ 18 =) 10 rounds and L completes (180 ÷ 20 =) 9 rounds when they meet at starting point.

40

Quicker Maths

Chapter 9

Fractions If any unit be divided into any number of equal parts, one or more of these parts is called a fraction of the unit. The fraction one-fifth, two-fifths, three-fourths are written

1 2 3 , and respectively.. 5 5 4 The lower number, which indicates the number of equal parts into which the units is divided, is called denominator. The upper number, which indicates the number of parts taken to form the fraction, is called the numerator. The numerator and the denominator of a fraction are called its terms. Note: 1. A fraction is unity when its numerator and denominator are equal. 2. A fraction is equal to zero when its numerator alone is zero. The denominator of a fraction is always assured to be non-zero. 3. A fraction is also called a rational number. 4. The value of a fraction is not altered by multiplying or dividing the numerator and the denominator by the same number. as

2 25 24 5 55 5 4 5. When the numerator and the denominator of a fraction have no common factor, the fraction is said to be in its lowest terms. Ex.:

15 3 5 20 4 5 The numerator and the denominator have a Ex.:

other i.e. they have no common factor. 6. If the numerator and the denominator are large numbers, or if their common factors cannot be guessed easily, we may find their HCF. After dividing the terms by their HCF, the fraction is reduced to its lowest term.

385056 ; HCF of 385056 and 715104 715104 = 55008 Ex.:

Now,

385056 385056 55008 7 = 715104 715104 55008 13

7 is in its lowest term because the terms 13 have no common factor. 7. An integer can be expressed as a fraction with any denominator we want. For example: if we want to express 23 as a fraction whose denominator is 17, the process will be as follows: Here,

23

23 17 391 17 17

Proper fraction: A proper fraction is one whose numerator is less than the denominator.

3 17 21 , , are proper fractions. The 4 19 42 value of a proper fraction is always less than 1. For example:

common factor 5, so

Improper fraction: A fraction whose numerator is equal to or greater than the denominator is called an improper fraction.

3 3 , which has no common factor. Hence is 4 4

17 12 18 , , are improper fractions. 12 7 5 The value of an improper fraction is always more than or equal to 1. In the above examples,

15 in its lowest terms. 20 When a fraction is reduced in its lowest terms, its numerator and denominator are prime to each

17 5 12 5 18 3 1 , 1 , 3 12 12 7 7 5 5 Thus, we see that an improper fraction is made up of a whole number and a proper fraction. When an improper

15 is not in its lowest 20 terms. If we cancel out 5 by dividing both the numerator and the denominator by 5, we get

the fraction

For example:

Quicker Maths

42 fraction is changed to consist of a whole number and a proper fraction, it is called a mixed number.

5 5 3 In the above examples 1 , 1 and 3 are mixed 12 7 5 numbers. Fractions in which the denominators are powers of 10 are called decimal fractions. e.g.

3 7 , , 10 10

3 9 , , etc. 100 100 Fractions in which the denominators are the same are called like fractions.

13 19 8 20 1 5 17 , , , ; , , etc. For example: 17 17 17 17 12 12 12 In this case, the fraction having the greatest numerator is the greatest. 20 19 13 8 1 5 17 and 17 17 17 17 12 12 12 Fractions in which the denominator are different are called unlike fractions. So,

For example:

13 15 379 , , are unlike fractions. 17 8 1000

Solved Examples: Ex. 1: Find the sum of a) b)

1 2 3 , and 2 3 4

125 50 48 1 , , and 1 100 36 45 2

Soln: a) All the fractions are in their lowest terms. Now, LCM of 2, 3, 4 = 12 Then, 1 2 3 (12 2) 1 (12 3) 2 (12 4) 3 2 3 4 12

6 8 9 23 11 1 12 12 12 Note: Students should not write the step =

(12 2) 1 (12 3) 2 (12 4) 3 12 It should be done mentally. b) First reduce the given fractions into their lowest terms. 125 5 50 25 48 16 ; ; 100 4 36 18 45 15

Now, change the fractions into mixed numbers.

5 1 25 7 16 1 1 ; 1 ; 1 4 4 18 18 15 15 Thus,

the

given

expression

becomes:

1 7 1 1 1 1 1 1 4 18 15 2 Now, add all the whole numbers together and all the fraction together. Thus, 1 7 1 1 1 7 1 1 1 1 1 1 (1 1 1 1) 4 18 15 2 4 18 15 2

4

45 70 12 90 180

217 37 37 4 1 5 180 180 180 Ex. 2: Solve : 4

a)

7 11 13 1 – – 9 12 16 8

b) 3

10 7 9 9 5 – 2 – 4 11 15 22 10

5 17 c) 3 81 9 16 5 d) 10 91 6 4 e) 50 14 7 f)

15 3 4 20 4 5

g)

6 3 7

h)

6 4 7

Soln: a) LCM of 9, 12, 16, 8 = 144

7 11 13 1 16 7 – 11 12 13 9 – 18 – – 9 12 16 8 144

112 – 132 117 – 18 229 – 150 79 144 144 144

Fractions b) 3

10 7 9 9 5 – 2 – 4 11 15 22 10

9 9 10 7 (3 5 – 2 – 4) – – 11 15 22 10

300 154 – 135 – 297 2 330 22 1 1 2 2 2 330 15 15 5 17 c) 3 81 9 16 Change the mixed fraction into an improper fraction. Then the expression becomes: 32 17 81 306 9 16 5 65 65 1 5 91 d) 10 91 6 6 6 91 42 4 410 410 1 205 9 14 4 e) 58 14 7 7 7 14 49 49 Or, if the integral part of the mixed number be greater than the divisor, we proceed as follows:

43 Soln: "To divide a fraction by a fraction, multiply the fraction by the reciprocal of the divisor." a) First change the mixed number into an improper fraction. Then multiply the fraction by the reciprocal of the divisor.

90 108 90 31 155 17 1 23 31 23 108 138 138 4 1 85 3 5 b) 9 11 9 3 9 34 6 Compound Fractions: A fraction of a fraction is called a compound fraction. Thus,

1 3 1 3 3 of 2 7 2 7 14 Combined Operations In simplifying fractions involving various signs and brackets the following points should be remembered. (i) The operations of multiplication and division should be performed before those of addition and subtraction. (ii) Each of the signs × or ÷ should be applied only to the number which immediately follows it.

4 4 18 1 9 9 58 14 56 2 14 4 4 4 7 7 7 14 49 49

f)

15 3 4 9 20 4 5 20

6 3 ; when the numerator is perfectly divisible 7 by the divisor, divide it without changing the ÷ sign into the × sign. 63 2 Ans. = 7 7 6 h) 4 7 In this case, the numerator 6 is not perfectly divisible by 4. Hence, 6 6 1 3 4 7 7 4 14 Ex. 3: Simplify: g)

a) 3

21 15 3 23 31

4 1 b) 9 11 9 3

1 3 of is a compound fraction. And 2 7

Ex. 1.

4 7 5 4 7 24 56 5 12 24 5 12 5 25

Ex. 2.

4 7 5 4 12 5 2 5 12 24 5 7 24 7

4 7 5 4 12 24 1152 5 12 24 5 7 5 175 (iii) The operations within brackets are to be carried out first. (iv) The rule of ‘BODMAS’ is applied for combined operations. Complex Fractions: A complex fraction is one in which the numerator or denominator or both are fractions. For example: Ex. 3.

5 7, 8

8 , 5 7

8 9, 7 9

4 15 Ex. 1: Simplify (i) 2 5

1 2 2 3 are complex fractions. 3 2 – 4 5 1 2 2 3 (ii) 3 2 – 4 9

Quicker Maths

44 4 15 4 2 4 5 2 Soln: (i) 2 15 5 15 2 3 5 1 2 3 4 7 2 3 6 6 7 36 42 2 4 (ii) 3 2 27 – 8 19 6 19 19 19 – 4 9 36 36

1 1 5 24 10 1 5 25 – 1 – 24 25 – 24 1 Soln: 1 1 5 2 1 25 5 5 Continued Fractions: Fractions that contain an additional fraction in the (numerator or the) denominator are called continued fraction. Fractions of the form (i) 4

or, multiply the numerator and denominator by the LCM of the denominators 2, 3, 4, 9 namely 36. Thus,

1 2 1 2 36 36 18 24 42 4 2 32 3 2 3 2 3 2 19 – 36 – 36 27 – 8 19 4 9 4 9 Note: The second method works quickly, so we suggest that you adopt this method. Ex. 2: Simplify:

3– 7 8 5– 3

2

3

4 3 – 5 7 6 5 21 5 7 10 5 6 5 2

1

or (ii)

1

5

4–

2 5

Rule: To simplify a continued fraction, begin at the bottom and work upwards. Ex. 1: Simplify:

1 2

1

1 1 1 of 555 5 5 5 1 1 – 5 – Ex. 3: Simplify: 5 5 5 1 of 1 1 5 2 5 5 5 10

1

3

1

2

1 5 2 19

1 1 3 5 4

1 8

1 6 81 10

5

2

1 6

1

1

1 19 43 43 19

Ex. 2: Simplify: 5

Soln: 5

1 4

1

Soln: The fraction

3–

2 5 49 9 7 2 – – 1 3 7 10 2 2 2

2 3

are called continued fractions.

2

1–

5 2 5 7 21 3 – 3 5 7 10 5

3

1

1

5

1 1 3 3 –2 3– 1 6 2 2 –5 of 3 of 3 5 47 3 7 1 – 2 4– 21 7 2

2 3 10 5 3 – 7 5 7 6 3 2 2 – Soln: 7 5 7 10 5 47 – 2 3 2 21

1

5

1 10

1 10 6 81

1 81 81 5 5 496 496 496 81

1 4 3 5

1 2

1 19 5

Fractions

45

Note: No shortcut method has been derived for solving these questions. But you can solve these questions more quickly by (i) more mental calculations; and (ii) skipping the steps.

Miscellaneous Solved Examples on Fractions Ex. 1: One-quarter of one-seventh of a land is sold for 30,000. What is the value of an eight thirty-fifths of land?

1 1 1 Soln: One-quarter of one-seventh 4 7 28 Now,

1 of a land costs = 30,000 28

8 30, 000 28 8 35 of the land will cost 35 = 1,92,000 1 1 of its contents, 5 12 of the remainder was found to be 7.40. What sum did the purse contain at first?

Ex. 2: After taking out of a purse

Soln: After taking out remains with

1 of its contents, the purse 5

4 of contents. 5

1 4 1 of = 7.40 or, = 7.40 12 5 15 1 = 1111 Ex. 3: A sum of money when increased by its seventh part amounts to 40. Find the sum. Now,

S 8S = 40 = 40 S = 135 7 7 Ex. 4: A train starts full of passengers. At the first station, it drops one-third of these and takes in 96 more. At the next, it drops half of the new total and takes in 12 more. On reaching the next station, there are found to be 248 left. With how many passengers did the train start? Soln: Let the train start with x passengers. After dropping one-third and taking in 96 x 2x 96 passengers, the train has x – 96 3 3 Soln: S

passengers

2x 288 passengers 3

At the second station, the number of passengers 2x 288 12 6

2x 288 12 248 6 or, 2 x 288 1416 x = 564 Ex. 5: Determine the missing figures (denoted by stars) in the following equations, the fractions being given in their lowest terms. 3 2 (i) 6 * 30 * 3 1 1 ** –* 1 (ii) 4 6 * 17 6* Soln: (i) Since (6 + a fraction) is contained (4 + a fraction) times in 30 (because 6 × 4 is just less than 30), the integral portion of the second mixed number must be 4. Then, 3 2 6 4 30 * 3 Now,

3 30 3 90 45 3 6 or 6 * 14 14 7 7 Hence, the missing digits are 7 and 4. 1 1 ** (ii) 4 6 * – * 17 1 6 * The denominator in the first mixed number and denominator in the third mixed number are in sixties, which must be multiple of 17. Thus, the equation becomes: 1 1 ** 4 – * 1 68 17 68 1 ** 1 or, (4 – *) – 1 68 17 68

Since

1 1 1 , will borrow 1 from 4. 68 17 68

** 69 1 – 1 Then 3 – * 68 68 17 or, (3 – 2)

65 ** 65 65 1 1 1 68 68 68 68

Hence, the equation becomes: 4

1 1 65 – 2 1 68 17 68

Quicker Maths

46

EXERCISES 1. Which one of the following is the smallest fraction?

2.

3.

4.

5.

6 13 15 19 5 , , , , 11 18 22 36 6 What smallest fraction should be added to 2 7 9 1 3 6 4 5 7 to make the sum a whole 3 12 36 12 number? What must be subtracted from the sum of 7 5 13 and 4 to have a remainder equal to their 66 66 difference? Find the smallest fraction which, when added to 2 15 7 3 gives a whole number.. 5 21 10 8 Find out the missing figures (denoted by stars) in the following equations, the fractions being given in their lowest terms: 3 3 * i) * 2 14 7 * 14

1 1 1 1 144 3 1 8 * 9 143 v) 2 5 4 5 15

5 hour 6 service time every 90 days, while after overhauling, 5 it requires hour service time every 120 days. What 6 fraction of the pre-overhauling service time is saved in the latter case? 7. If the numerator of a fraction is increased by 150% and the denominator of the fraction is increased by 5 300%, the resultant fraction becomes . What is the 18 original fraction? 6. A motorcycle, before overhauling, requires

8.

* 5 * ii) 7 – * 3 3 11 *3

3 4 th of the girls and th of the boys of a primary 8 9 school participated in the annual sports. If the number of participating students is 155 out of which 92 are boys, what is the total number of students in the primary school?

2 3 of what he received to his mother. Vijay’s mother gave 5 of the money she received from Vijay to the grocer.. 8 Vijay's mother is now left with 600. How much money did Abhay have initially?

9. Abhay gave 30% of his money to Vijay. Vijay gave

9 2 16 iii) 8 * 7 ** 27 17

1 2 5 2 iv) 2 – 3 1 – 0 2 3 6 *

Solutions (Hints) 1. Out of the first two fractions, we see that 6 6 × 18 < 11 × 13, so, is smaller.. 11 6 15 and Now, from , we see that 6 22 11 15 , 11 22

6 is smaller.. 11 6 19 , we see that 19 11 6 36 , Now, from and 11 36 19 so is smaller.. 36 so

Now, from so

19 5 , and , we see that 19 6 5 36, 36 6

19 is smaller.. 36

19 we conclude that 36 is the smallest. Note: The above method does not need accurate calculations. You can decide which of the multiplications is greater by your keen observation only.

2.

2 7 9 1 3 6 4 5 7 3 12 36 12

Fractions

47

Add only the fractional parts:

2 7 9 1 3 12 36 12

24 21 9 3 57 19 7 1 36 36 12 12 Thus, to make the expression a whole number we should

add 1 –

7 5 12 12

3. Difference 13

7 5 5 7 – 4 (13 – 4) – 66 66 66 66

9

2 1 9 66 33

7 5 12 2 4 17 17 66 66 66 11 the required answer Sum 13

2 1 6 1 5 –9 17 – 9 8 11 33 33 33 33 Direct formula: The required answer = 2 × smaller value 17

24 4.

5 5 8 66 33

2 15 7 3 3 5 21 10 8 40 3 37 the required fraction = 1 – 40 40

3 3 13 5. i) 5 2 14 7 4 14 2 5 7 ii) 7 – 4 3 3 11 33 iii) 8

9 2 16 1 7 17 27 17

1 2 5 2 iv) 2 – 3 1 – 0 2 3 6 3 1 3 6. LCM of 90 and 120 = 360 So, in 360 days, the pre-overhauling service time V) * 7

5 360 10 hrs and after overhauling, the service 6 90 3

5 360 5 hrs. 6 120 2 10 5 5 Time saved – hrs 3 2 6 time

5 6 5 3 1 The required answer 10 6 10 4 3 7. Let the original fraction be So,

x . y

x 2.5 5 = y4 18 x 5 4 4 = = y 18 2.5 9

Note: Increase by 150% means the original numerator becomes (100 + 150=) 250% or 2.5 times. And similarly the original denominator becomes 4 times. Now, we can find a direct formula as given below: 5 2.5 5 4 4 Original fraction = 18 4 18 2.5 9 8. Total students who participated in annual sports =155 Boys = 92 Girls = 155 – 92 = 63

Total number of boys in school =

92 9 207 4

63 8 168 3 Total number of students = 207+ 168 = 375 9. Suppose Abhay has initially x.

Total number of girls =

30 3x Then Vijay got = x = 100 10 3x 2 x = 10 3 5 Vijay’s mother gave money to the grocer

Vijay’s mother got

x5 x = 5 8 8 Money left with Vijay’s mother

=

=

x x 8x 5x 3x = 5 8 40 40

Now,

3x = 600 40

x=

600 40 = 8000 3

Quicker Maths

48 Logical Approach: In such questions, we should think in reverse. Like 5 3 the mother is left with 1 8 th of the money,, 8 which is equal to 600. So, she must have

1 1 8 600 5 = 600 3 / 8 = 600 3 = 1600. 1 8 1 3 Similarly, Vijay must have 1600 2 / 3 1600 2

= 2400

1 100 and Abhay must have 2400 30% = 2400 30

= 8000. Combining all the above steps, we get the following quicker method: Quicker Method: 1 Abhay's initial money = 600 1 5 / 8 8 3 10 = 600 3 2 3 = 8000

1 1 2 / 3 30%

Chapter 10

Decimal Fractions Decimal Fraction: Fractions in which the denominators are powers of 10 are called decimal fractions. 1 7 9 , , , etc. 10 10 1000 Reading a decimal: In reading a decimal, the digits are named in order. Thus 0.457 is read as zero point (or decimal) four, five, seven. Decimal places: The number of figures which follow the decimal point is called the number of decimal places. Thus, 1.432 has three decimal places and 7.82 has two decimal places.

For example:

Converting a decimal into a vulgar fraction Rule: Write down the given number without the decimal point, for the numerator, and for the denominator write 1 followed by as many zeros as there are figures after the decimal point. Ex. 1: Reduce the following decimal values into the vulgar fractions. a) 0.63 b) 0.0032 c) 3.013 Soln: a) 0.63

63 100 0032 32 10000 10000

b) 0.0032

013 13 3 1000 1000 Note: Adding zeros to the extreme right of a decimal fraction does not change its value. For example: 0.9 = 0.90 = 0.9000 = 0.90000000 (Why is this so?) If numerator and denominator of a fraction contain the same number of decimal places, then we may remove the decimal sign. 0.53 053 53 Ex. 2: i) 3.21 321 321 9.83051 983051 ii) 18.53342 1853342

c) 3.013 3

iii) Change

1.53 into vulgar fraction. 2.4321

1.53 1.5300 15300 2.4321 2.4321 24321 Thus, to remove the decimal, we equate the number of figures in the decimal part of the numerator and the denominator (by putting zeros) and then remove the decimals. Note: An integer may be expressed as a decimal by putting zeros in the decimal part. Ex: 17 = 17.0 = 17.00 Addition and Subtraction of Decimals Rule: Write down the numbers under one another, placing the decimal points in one column. The numbers can now be added or subtracted in the usual way. Ex. 3: Add together 5.032, 0.8, 150.03 and 40. Soln: 5.032 Soln:

0.8 150.03 40.00 195.862

Ex. 4: Subtract 19.052 from 24.5. Soln: Here, we write two zeros to the right of 24.5 and then subtract as in the case of integers. 24.500 19.052 5.448

Multiplication of decimals I: To multiply by 10, 100, 1000, etc. Rule: Move the decimal point by as many places to the right as many as there are zeros in the multiplier. Ex.: (i) 39.052 × 100 = 3905.2 (ii) 42.63 × 1000 = 42630 II. To multiply by a whole number Rule: Multiply as in the case of integers, and in the product mark as many decimal places as there are in the multiplicand, prefixing zeros if necessary. Ex.: (i) 0.9 × 12 = ? Soln: Step I: 9 × 12 = 108 Step II: As there is one decimal place in multiplicand, the product should also have only one decimal place. So, the answer = 10.8

Quicker Maths

50 Ex.: (ii) 0.009 × 12 = ? Soln: Step I: 9 × 12 = 108 Step II: As there are three decimal places in the multiplicand, the product shuld also have three decimal places. So, the answer = 0.108 Ex.: (iii) 0.00009 × 12 = ? Soln: Step I: 9 × 12 = 108 Step II: As there are five decimal places in the multiplicand, the product should also have five decimal places. So, we prefix two zeros to get the answer. Ans = 0.00108 III: To multiply a decimal by a decimal Rule: Multiply as in integers, and in the product mark as many decimal places as there are in the case of the multiplier and the multiplicand together, prefixing zeros, if necessary. Ex.: (i) 0.61 × 0.07 = ? Soln: Step I: 61 × 7 = 427 Step II: As there are (2 + 2 =)4 decimal places in the multiplier and the multiplicand together, the product should also have 4 decimal places. But there are only three digits in the product; so we prefix one zero to the product before placing the decimal. So, the answer = 0.0427 Ex.: (ii) 0.2345 × 0.24 = ? Soln: Step I: 2345 × 24 = 56280 Step II: There should be (4 + 2 =)6 decimal places in the product. Thus, answer = 0.056280 = 0.05628

Division of decimals I: When the divisor is 10, 100, 1000, etc. Rule: To divide a decimal by 10, 100, 1000 etc., move the decimal point 1, 2, 3 etc. places to the left respectively. Thus, (i) 463.8 ÷ 10 = 46.38 (ii) 4.309 ÷ 100 = 0.04309 (iii) 0.003 ÷ 10000 = 0.0000003 (iv) 234.789 ÷ 1000000 = 0.000234789 (v) 5.08 ÷ 100000000 = 0.0000000508 II: When the divisor is a decimal fraction Rule: Move the decimal points as many places to the right in both the divisor and dividend as will make the divisor a whole number, annexing zeros to the dividend, if necessary. Divide as in simple divison and when you take a figure from the decimal part (if any) of the dividend so altered, set down the decimal point in the quotient. Ex.

32.5 325000 5078.125 (i) 0.0064 64

0.0323 3230 190 0.00017 17

(ii)

Recurring Decimals A decimal in which a figure or set of figures is repeated continually is called a recurring or periodic or circulating decimal. The repeated figures or set of figures is called the period of the decimal. For example: (1) (2)

1 0.333..... 3 1 0.142857142857142857...... 7

13 0.29545454...... 44 Recurring is expressed by putting a bar (or dots) on the set of repeating numbers. So, in the above examples:

(3)

(1) 0.333 ... = 0.3 (or , 0.3 ) (2) 0.142857 142857.......

0.142857 (or, 0.142857) (3) 0.295454... 0.2954 (or , 0.295 4 ) Pure Recurring Decimal: A decimal fraction in which all the figures after the decimal point are repeated, is called a pure recurring decimal. For example, 0.142857 is a pure recurring decimal. Mixed Recurring Decimal: A decimal fraction in which some figures do not recur is called a mixed recurring decimal. For example: 0.2954 is a mixed recurring decimal. In example (1), the period is 3, in (2) it is 142857 and in (3) it is 54. Note: 1. If the denominator of a vulgar fraction in its lowest terms be wholly made up of powers of 2 and 5, either alone or multiplied together, the fraction is convertible into a terminating decimal. Ex.:

(i)

12 12 0.48 25 52

(ii)

19 19 0.38 50 52 2

(iii)

13 13 2 0.65 20 2 5

(iv)

3 3 0.375 8 23

Decimal Fractions

51

2. A pure recurring decimal is produced by a vulgar fraction in its lowest terms, whose denominator is neither divisible by 2 nor by 5. Ex.:

2 (i) 0.6 3

3 0.27 (ii) 11

4 5 0.571428 (iv) 0.5 7 9 3. A mixed recurring decimal is produced by a vulgar fraction in its lowest terms, whose denominator contains powers of 2 or 5 in addition to other factors.

(iii)

Ex.:

By note 2, a pure recurring decimal will be produced. 1 1 18 9 2 By note 3, a mixed recurring decimal will be produced.

e)

7 7 45 9 5 By note 3, a mixed recurring decimal will be produced.

f)

(i)

1 1 0.16 6 23

1 1 g) 80 (2)4 5

(ii)

1 1 0.06 15 5 3

By note 1, no recurring decimal will be produced.

(iii)

To convert a recurring decimal fraction into a vulgar fraction

7 7 0.093 75 52 3

8 8 0.53 15 5 3 Question: Which of the following vulgar fractions will produce recurring decimal fractions?

(iv)

a)

12 50

b)

12 75

c)

3 18

d)

8 14

e)

1 18

f)

7 45

1 80 Soln: a) Reduce the fraction to its lowest terms, so

g)

Case I: Pure Recurring Decimals Rule: A pure recurring decimal is equal to a vulgar fraction which has for its numerator the period of the decimal, and for its denominator the number which has for its digits as many nines as there are digits in the period. Ex.: Express the following recurring decimals into vulgar decimals. a) 0.5

b) 0.45

c) 0.532

Soln: a) We have 0.5 0.555..... -------(1) Multiplying both sides by 10, we get

10 0.5 5.55... ----- (2) Subtracting (1) from (2), we get

9 0.5 5

5 0.5 9

12 6 6 2 50 25 (5)

Directly from the rule, we also get the same result.

By note 1, the above vulgar fraction will not produce recurring decimal.

0.5 has its period 5, so the numerator is 5. And as there is only one digit in the period, the denominator

12 4 4 b) 75 25 (5)2

will have one nine. Thus, the vulgar fraction

By note 1, the above vulgar fraction will not produce recurring decimal. 3 1 1 18 6 3 2 By note 3, a mixed recurring decimal will be produced.

c)

d)

8 4 14 7

5 9

b) 0.45 By the rule, numerator is the period (45) and the denominator is 99 because there are two digits in the period. Ans

45 99

c) 0 .532 Numerator = period = 532

Quicker Maths

52 Denominator = as many nines as the number of digits in the denominator 532 999 Try to solve Ex. (b) & (c) by detailed method. Ans

Case II: Mixed Recurring Decimals Rule: A mixed recurring decimal is equal to a vulgar fraction which has for its numerator the difference between the number formed by all the digits to the end of first period, and that formed by the digits which do not recur; and for its denominator the number formed by as many nines as there are recurring digits, followed by as many zeros as there are non-recurring digits. Ex.: Express the following as vulgar fractions. a) 0.18

b) 0.43213

Soln: a) 0.18 0.18888

c) 5.00983 --------- (1)

10 0.18 1.8888

--------- (2)

and 100 × 0.18 18.8888 -------- (3) Subtracting (2) from (3), we have,

90 0.1 8 18 – 1 17 0.18 90

Also, by the rule, numerator = 18 – 1 (all-digitnumber – non-recurring-number) and denominator = one nine (as there is one recurring digit) followed by one zero (as there is one non-recurring digit) = 90

Vulgar fraction =

18 – 1 17 90 90

b) 0.43213 ; Numerator = 43213 – 43 = 43170 Denominator = 3 nines (as there are three recurring digits) followed by 2 zeros (as there are two nonrecurring digits) = 99900 43170 4317 Vulgar fraction = 99900 9990 00983 – 00 983 5 c) 5.00983 5 99900 99900 Solve Ex. (b) and (c) by the detailed method.

Addition and Subtraction of Recurring Decimals Addition and subtraction of recurring decimals can be understood well with the help of given examples. Example: Add and subtract 3.76 and 1.4576 .

3.7 676767 67 1.4 576576 57 (+)5.2 253344 24

5.2253344

(-)2.3 100191

2.3100191

10

Explanation: Step I: We separate the expansion of recurring decimals into three parts. In the left-side part, there is integral value with non-recurring decimal digits. In the above case, 1.4576 has 1 as integral part and 4 is as non-recurring decimal digit. So, 1.4 has been separated from 1.4576576576...... Although the first value does not have any nonrecurring decimal digit (i.e. in 3.76 , both the digits 7 and 6 are recurring), we put as many digits in the left-side part as there are nonrecurring digits in the second value. Thus, we have put 3.7 in the left-side part. Step II:In the middle part, the number of digits is equal to the LCM of the number of recurring digits in two given values. In this case, the first value has 2 recurring digits and the second value has 3 recurring digits, so the middle part has 6 digits (as LCM of 2 and 3 = 6). Step III: In the right-side part take two digits. Now, add or subtract as you do with simple addition and subtraction. Step IV: Now, leave the right-side part and put a bar on the middle part of the resultant. The left-side part will remain as it is in the resultant. You can see the same in the above example.

Verification: 76 76 3 and 99 99 4576 – 4 4572 1.4576 1 1 9990 9990 76 4572 3.76 1.4576 3 1 99 9990 3.76 3

4 4

76 9990 4572 99 99 9990

76 1110 4572 11 11 9990 134652 24762 4 4 1 109890 109890

5

24762 109890

-------- (1)

Decimal Fractions Now, 5.2253344 5

53 2253344 – 2 2253342 5 9999990 9999990

24762 91 24762 5 5 ------ (2) 109890 91 109890 From (1) and (2); it is verified that our addition is correct. Now, for subtraction: 76 4572 3.76 – 1.4576 3 – 1 99 9990 2

76 9990 – 4572 99 99 9990

2

76 1110 – 4572 11 11 9990

2

34068 11 9990

Now, 2.3100191 2

Ex. 2: Find 324 .786 – 10.193 Soln: 324.78

Soln: 324.78

10.19

678 333

67 33

334.98

012

00

10.193 10

786 999

193 – 19 174 10 900 900

786 174 324.786 10.193 324 10 999 900 786 900 174 999 334 999 900 78600 174 111 97914 334 334 99900 99900

334

98012 – 98 334.98012 99900

345

34

786 900 – 174 999 999 900 786 100 – 174 111 314 99900 59286 59345 – 59 314 314 314.59345 99900 99900

Ex. 3: Find 17.83 0.007 310.0202 Soln:

17.838 0.007 310.020 327.866

38 77 22 38

38 77 22 37

Thus, the answer = 327.86638 Ex. 4: Find 17.1086 – 7.9849 Soln: 17.108

Thus, the answer = 334.98012 Verification: 324.786 324

314.59

314

91 34068 34068 2 = 2 ------(2) 91 11 9990 11 9990 From (1) and (2); it is verified that our subtraction is correct.

Ex. 1: Find 324.786 10.193

67 33

Thus, the answer = 314 .59345 Verification: 786 174 324.786 – 10.193 324 – 10 999 900

------ (1)

3100191 3 3100188 2 9999990 9999990

10.19

678 333

7.984

68 99

9.123

68

68 99 69

Thus, the answer = 9.12368

Multiplication of Recurring Decimals Case A: While multiplying a recurring decimal by a multiple of 10, the set of repeating digits is not altered. For example: 1.

4.03 10 4.030303... 10 40.30303... 40.303

2.

124.427 1000 124.427427... 1000 124427.427

3.

0.006379 10000 0.006379379... 10000

63.79379... 63.79379 Case B: While multiplying a recurring decimal by a number which is not a multiple of 10, first of all, the recurring decimal is changed into the vulgar fraction and then the calculation is done. For example:

Quicker Maths

54 63 7 84 11 7 11 11 84 99 11 11

1.

7.63 11 7

2.

13.345 15 13

345 – 34 311 15 13 15 900 900

11700 311 12011 12000 11 11 15 200 900 60 60 60

200 0.1833... 200 0.183 200.183

3.

27 1.22 5.5262 1.6

27 1

22 – 2 5262 – 52 6 5 90 9900 9

2 521 11 5471 5471 18 1 5 18 9 990 9 990 45 121.5777... 121.57

Division of Recurring Decimals Case A: When the divisor is a multiple of 10. 1.

0.06 1000 0.060606... 1000

0.000060606... 0.00006

2. 16.45379 10 1.645379 Case B: When the divisor is not a multiple of 10. Ex. 1: 17.26 2 17

26 – 2 24 1 4 1 2 17 17 90 90 2 15 2

259 240 19 19 8 30 30 30

57 63 – 6 8 8 0.63 8.63 90 90 Another way of calculation is 8

17.26 2 8

19 6.33... 8 8 0.633... 30 10

8 0.63 8.63 Second Method: 2) 17.266... (8.633... 16 12 12 06 6 06 Thus, the answer = 8.63

Ex. 2: 36.343 7 36.34343 ... 7 7) 36.34343... (5.19 35 13 7 64 63 13 The process of division repeats; so, our quotient becomes 5.1919.... i.e. 5.19 .

Approximation and Contraction Rule: Increase the last figure by 1 if the succeeding figure be 5 or greater than 5. For Example: (i) The approximate value of 0.3689 up to three decimal places is 0.369. (ii) The approximate value of 0.3684 up to three decimal places is 0.368. (iii) The approximate value of 0.3685 up to three decimal places is 0.369. (iv) The approximate value of 0.3689 up to two decimal places is 0.37. (v) The approximate value of 0.3468 up to one decimal place is 0.3. Note: Decimals which are correct at one, two, three ... places are said to be correct to the nearest tenth, hundredth, thousandth, ... respectively. Significant figures: The following examples are given to explain the meaning of the term significant figures. (a) The population of a certain place is 189000 correct to the nearest thousand. Here, the assumed unit is one thousand, and the population is stated to be 189 such units correct to the nearest unit. The figure 189, which gives the number of units, is said to be significant while the three zeros, which indicate magnitude of the unit, are said to be non-significant. (b) The distance between two places is 1400 km correct to the nearest hundred km. The figure 14 is significant and the two zeros are nonsignificant. (c) The distance between two places is 1400 km correct to the nearest km. The figure 1400 is significant but there is no non-significant figure. (d) The length of a string is 0.07 cm correct to the 2nd decimal place. This means 7 is the significant figure but the zero at the beginning is non-significant.

Decimal Fractions

55

Note:

Zeros at the beginning of a decimal are always non-significant. Thus, significant figures are those which in any approximate result express the number of units correct to the nearest such unit.

Contracted Addition Rule: Set down the decimals under one another. Then add in the usual way, taking care that the last figure retained be increased by 1 if the succeeding figure be 5 or greater than 5. Ex: Find the sum of 320.4321, 29.042934, 0.0085279 and 0.3412 correct to 3 decimal places. Soln:

Remark: If you are asked to get the answer correct to three decimal places, then use not more than 5 (two more) decimal places during calculation.

Contracted Subtraction Rule: Write down the subtrahend under the minued in the usual way, retaining two places of decimals more than what is required. Ex: Find 160.342195 – 32.0048326 correct to four decimal places. Soln: As we are asked to get the solution correct to four decimal places, we will use not more than 6 decimal places during our calculation.

160.342195 32.004832

320.4321 29.04293

128.337363

0.00852 0.3412

Answer = 128.3374

349.82475

Answer = 349.825

EXERCISES 1. Express the following decimals as fractions in their lowest terms:a) 0.0375 b) 0.00625 c) 1.008125 2. Simplify: a) 0.25 + 0.036 + 0.0075 b) 34.07 + 0.007 + 0.07 c) 30.9 + 3.09 + 0.309 + 0.039 d) 35 – 7.892 + 0.005 – 10.345 e) 0.6 + 0.66 + 0.066 – 6.606 + 66.06 3. Remove decimals a) 0.35 × 106 b) 0.275 × 10–7 c) 0.0034 × 10,000 d) 0.132 500 e) 5.302 × 513 4. Divide: a) 28.9 17 b) 0.457263 ÷ 18 c) 64 ÷ 800 d) 64 ÷ 0.008 e) 2.375 ÷ 0.0005 f) 0.1 ÷ 0.0005 g) 0.1 ÷ 5000

5. Find the quotient to three places of decimals:a) 0.5 ÷ 0.71 b) 4.321 ÷ 0.77 c) 5.002 ÷ 0.00078 6. Simplify the following: a) 12 ÷ 0.09 of 0.3 × 2 b)

0.0025 1.4 0.0175

c)

9.5 0.085 0.0017 0.19

d)

3 of 0.176 11

7. What should come in place of question mark (?) ? a) 3 × 0.3 × 0.03 × 0.003 × 300 = ? b) 0.25 0.0025 0.025 of 2.5 ? c) 0.00033 ÷ 0.11 of 30 × 100 = ? d) 0.8 × ? = 0.00004 e)

3420 ? 7 19 0.01

f)

17.28 ? 0.2 400 3.6

Quicker Maths

56 d) 2.432

0.324 0.081 4.624 1.5625 0.0289 72.9 64

g)

e) 10 .036

20 8 0.05 16 40 – ? i) ?% of 10.8 = 32.4 j) 3.79 × 31 + 3.79 × 37 + 3.79 × 32 = ? k) 321 × 11.54 – 203 × 11.54 – 105 × 11.54 = ?

1 0.0276 , then what is the value of 36.18

h)

10. If

l) (1.27)3 3(1.23)2 1.27 3(1.27)2 1.23 (1.23)3 ?

1 ? 0.0003618 11. If 13324 ÷ 145 = 91.9, then what is the value of 133.24 ÷ 9.19?

m) (2.3) 3 – 3 (2.3) 2 (0.3) 3( 2.3)(0.09) – (0.3) 3 ? 8. State in each case whether the equivalent decimal is terminating or non-terminating: a)

1 6

b)

8 625

c)

17 90

d)

104 111

33 e) 165 9. Express the following recurring decimals in their vulgar fractions:-

12. If

3 5 5 2.24 , then what is the value of 2 5 – 0.48 ?

13. If

15 3.88 , then what is the value of

14. If

2916 54 , then what is the value of

29.16 0.2916 0.002916 0.00002916 ? 15. What decimal of an hour is a second? y–x 16. If 1.5x = 0.05y, then what is the value of y x ?

a) 0.3

17. (0.6 0.7 0.8 0.3) 9000 ?

b) 0.037

18. 0.3467 (0.45 0.5) 11 ?

c) 0.09

Answers 375 3 1. a) 0.0375 10, 000 80 625 1 b) 0.00625 1, 00,000 160 8125 13 c) 1.008125 110, 00,000 11600 2. a) 0.2935 b) 34.147 c) 34.338 d) 16.768 e) 60.78 3. a) 35 10 4 b) 275 10 –10 c) 34

5 ? 3

d) 66 e) 2719926 10 3 4. a) 1.7 b) 0.0254035 c) 0.08 d)8000 e) 4750 f) 200 g) 0.00002 5. a) 0.704 b) 5.612 c) 6412.821 6. a) 888.88 b) 0.2 c) 2500 d) 0.048

Decimal Fractions

57

7. a) 0.0243 b) 6.25 c) 0.01 d) 0.00005 e) 0.257 f) 0.0024

d) 2.432 2

e) 10.036 10

0.324 0.081 4.624 1.5625 0.0289 72.9 64

g)

(0.0276) 105 2760 This question should not be solved by detailed method. If you observe the following points you will be able to answer within seconds.

324 81 4624 10 15625 289 729 64 10 –9 –9

18 9 68 3 0.024 125 17 27 8 125 h) 38.725 i) 300 j) Given expression = 3.79 (31+37+32) = 3.79(100) = 379 k) Given expression = 11.54 (321 – 203 – 105) = 11.54 × 13 = 150.02 l) Given expression = (1.27 + 1.23)3 = (2.5)3 = 15.625 m) Given expression = (2.3 – 0.3)3 = 23 = 8 8. If the denominator of a vulgar fraction in its lowest terms be wholly made up of powers of 2 and 5,either alone or multiplied together, the fraction is convertible into a terminating decimal. Following the same rule:

b)

1 is non-terminating since its denominator has a 6 factor (3) other than 2 and 5.

8 8 4 is terminating since its denominator 625 (5) is made up of powers of 5 only.

c)

d)

17 17 is non-terminating since its 90 2 3 3 5 denominator has factors other than 2 and 5. 104 104 is non-terminating. 111 3 37

33 1 is terminating. e) 165 5

9. a) 0.3

3 1 9 3

b) 0.037 c) 0.09

37 999

9 1 99 11

36 – 3 33 11 10 10 900 900 300

1 1 (10)5 10. 0.0003618 36.18 (10) –5 36.18

a)

432 – 4 428 214 2 2 990 990 495

1 1 from the decimal in 0.0003618 36.18 denominator is moved 5 positions left. ii) So, to get the answer from 0.0276 the decimal should be moved 5 places right. Note: Without going into details, we can get the

i) To get

answer. You should know that

1 is larger 0.0003618

1 . Therefore the required answer should 36.18 also be larger than 0.0276. So, the required larger value should be obtained by moving the decimal to the right. 11. We have, 13324 ÷ 145 = 91.9 or, 13324 ÷ 91.9 = 145. Then we have to find 133.24 ÷ 9.19 = ? To get the answer divide the whole number of dividend by the whole number of divisor and observe the number of digits in the whole number of quotient. In this case, when 133 is divided by 9, the quotient (whole number) will have two digits. So, our answer should have a 2-digit whole number. required answer = 14.5

than

12.

13.

3 5 2 5 – 0.48

6.72 1.68 4

5 5 3 15 3.88 1.2933 3 3 3 3 3

14. If 2916 54 , then the required expression = 5.4 + 0.54 + 0.054 + 0.0054 = 5.9994 15.

1 0.00028 (approx) 60 60

Quicker Maths

58 16. 1.5x 0.05y

x 0.05 or, y 1.5 By the rule of componendo-dividendo, we have,

y – x 1.5 – 0.05 0.935 y x 1.5 0.05

6 7 8 3 17. The given expression 9000 9 9 9 9

24 9000 24, 000 9

3467 – 3 45 5 18. The given expression 11 9990 99 9

3464 225 93642 31214 15607 . 9990 81 9990 3 9990 4995

Chapter 11

Elementary Algebra Algebraic Expressions: A number, including literal numbers, along with the signs of fundamental operations is called an algebraic expression. They may be monomials, having only one term as +6x, –3y etc. They could also be binomials, i.e. having two terms as p2 +2r, –x2 –2x etc. They may even be polynomials, i.e. having more than two terms. Addition and Subtraction of Polynomials: The sum or difference of coefficients of like terms is performed. For example, let the two polynomials be 3x2 + 2x – 3, 5x3 – 2x2 – x. The sum of these two expressions is 5x3 + (3x2 – 2x2) + (2x + x) – 3 = 5x3 + x2 + 3x – 3. The difference between these two is the subtraction of smaller expression from the greater expression i.e. 5x3 – 2x2 + x – (3x2 + 2x – 3) = 5x3 – 2x2 + x – 3x2 – 2x +3 = 5x3+ (–2x2 – 3x2) + (x – 2x) + 3 = 5x3 – 5x2 – x + 3. Ex. 1: Find the sum of

15a 2 3ab 6b2 , a 2 5ab 11b2 , 7a 2 18ab 13b2 and 26a 2 16ab 7b 2 Soln: (15a 2 3ab 6b 2 ) (a 2 5ab 11b 2 )

( 7a 2 18ab 13b 2 ) (26a 2 16ab 7b2 ) = (15a 2 a 2 7a 2 26a 2 )

(3ab 5ab 18ab 16ab)

(6b 2 11b 2 13b2 7b2 ) = 5a 2 36ab 15b 2

Ex. 2: If P a 4 a 3 a 2 6 , Q a 2 2a 3 2 3a and

R 8 3a 2a 2 a 3 Find the value of P + Q + R Soln: P + Q + R = (a 4 a 3 a 2 6)

(a 2 2a 3 2 3a) (8 3a 2a 2 a 3 ) = a 4 (a 3 2a 3 a 3 ) (a 2 a 2 2a 2 )

(3a 3a) (–6 2 8) = a4 + 0 + 0 + 0 + 0 = a 4

Remainder Theorem: This theorem represents the relationship between the divisor of the first degree in the form (x–a) and the remainder r (x). If an integral function of x is divided by x – a, until the remainder does not contain x, then the remainder is the same as the original expression with ‘a’ put in place of ‘x’. In other words, if f(x) is divided by x – a, the remainder is f(a); e.g. when f (x) x 3 2x 2 3x 4 is divided by x – 2, the remainder is f(2).

f 2 2 3 22 2 32 4 2 Note: (1) If f(x) is divided by x + a, the remainder is f(–a) (2) If f(x) is divided by ax + b, the remainder is b f a (3) If f(x) is divided by ax – b, the remainder is

b f a (4) The method of finding the remainder is as follows: I: Put the divisor equal to zero. Like, if the divisor b a II: The remainder will be a function of the value of

is ax + b then ax b 0 x

b x, i.e., f . a Similarly, you can find the other remainders. Ex. 1: Without using the division process, find the remainder when x3 + 4x2 + 6x – 2 is divided by (x + 5). Soln: Step I: Put divisor equal to zero and find the value of x. x+5=0 x=–5 Step II: The remainder will be f(–5). f 5 5 4 5 6 5 2 3

2

= –125 + 100 – 30 – 2 = – 57.

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60 Ex. 2: Find the remainder when 27x 3 9x 2 3x 8 is divided by 3x + 2. Soln: Step I: 3x + 2 = 0 x =

2 3

2 Step II: Remainder is f 3

2 3

2 3

3

2 3

2

2 3

f 27 9 3 8 = – 8 – 4 – 2 – 8 = – 22

Ex. 3: If the expression Px 3 3x 2 3 and 2x 3 5x P when divided by x – 4 leave the same remainder, find the value of P. Soln: The remainder are:

R1 f (4) P(4)3 3(4)2 3 64P 45 R 2 f 4 2 4 5 4 P P 108 3

Since, R 1 R 2 , we have 64P + 45 = P + 108 or, 63P = 63 P=1 Ex. 4: Find the values of p and q when px 3 x 2 2x q is exactly divisible by (x – 1) and (x + 1). Soln: When the expression is exactly divisible by any divisor, the remainder will be zero. Now, the remainder, when the divisor is x – 1, is f(1) = p + 1 – 2 – q = 0 p – q = 1 .............. (1) and the remainder, when the divisor is x + 1, is

f 1 p 13 12 2 1 q 0 –p + 1 + 2 – q = 0 p + q = 3 .............. (2) Solving (1) & (2), we have, p = 2, q = 1 Factorisation of Polynomials: The factor theorem based on the remainder theorem is useful in the factorisation of polynomials. Factor Theorem: Let f(x) be a polynomial and a be a real number. Then the following two results hold: (i) If f(a) = 0 then (x – a) is a factor of f(x). (ii) If (x – a) is a factor of f(x) then f(a) = 0. Ex. 1: Let f x x 3 12x 2 44x 48 Find out whether (x – 2) and (x – 3) are factors of f (x).

Soln: (a) To check whether (x – 2) is a factor of f (x) we find f (a) i.e., f(2). If this becomes zero then (x – 2) is a factor of f(x) according to the factor theorem (i) of division algorithm. f (2) = 23 – 12 × 22 + 44 × 2 – 48 = 0 Hence, by the factor theorem, (x – 2) is a factor of f (x). (b) To check whether (x – 3) is a factor of f (x), we repeat the above process with f (3). f (3) = 33 – 12 × 32 + 44 × 3 – 48 = 3 f(a) = f(3) 0. Hence (x – 3) is not a factor of f (x). Ex. 2: Find out whether (3x – 1) is a factor of 27x 3 – 9x 2 – 6x 2 by the above rule. Soln: We have, 3x – 1 = 0 x

1 3

1 If (3x – 1) is a factor of f(x) then f should be 3 equal to zero. 3

2

1 1 1 1 Here, f 27 9 6 2 3 3 3 3 = 1–1–2+2=0 (3x – 1) is a factor of the above expression. Note: (1) We can see how the factor theorem has been derived from the remainder theorem: “When remainder is zero after dividing an expression.” (2) If f(a) = 0, then x – a is a factor of f(x) (3) If f(–a) = 0, then x + a is a factor of f(x) (4) If f(x) = 0, when x = a and x = b then f(x) is exactly divisible by (x – a) (x – b) i.e., (x – a) and (x – b) both are the factors of f(x). (5) If an integral function of two or more variables is equal to zero when two of these variables are supposed to be equal, then the function is exactly divisible by the difference of these variables; e.g. (b – c), (c – a) and (a – b) are the factors of

a ( b – c ) 3 b (c – a ) 3 c (a – b ) 3 Proof: Put b = c in the given expression. a(c – c)3 c(c a)3 c(a c)3 0 c(c – a)3 c(c a)3 0 (b – c) is a factor of the given expression. Similarly, it can be shown that c – a and a – b are the factors of the given expression.

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Conditions of Divisibility 1. x n a n is exactly divisible by (x + a) only when n is odd. e.g. a 5 b5 is exactly divisible by a + b. 2. x n a n is not exactly divisible by (x + a) when n is even. e.g. a 8 b8 is not exactly divisible by a + b. 3. x n a n is never divisible by (x – a). e.g. a 7 b 7 or a10 b10 is not divisible by a – b. 4. x n a n is exactly divisible by x + a when n is even. e.g. x 6 a 6 is exactly divisible by x + a 5. x n a n is exactly divisible by x – a (whether n is odd or even). e.g. x 9 a 9 and x10 a10 are exactly divisible by x – a. Proof: All the above statements can be proved. We are going to prove only statements (1) and (5). You should try to prove the remaining three yourself. (1) x n a n is exactly divisible by x + a. Put x + 4 = 0 x = –a Then f(–a) = (–a)n + an = 0 This is possible only when n is odd.

Soln: (26 )2n – (62 )2n (64)2n – (36) 2n Since 2n is an even integer, the above expression must be divisible by (64 + 36) {By condition (4)}. Hence, 64 + 36 = 100 is the factor of the above expression. Hence, the last two digits of its expansion must be 00. Ex. 6: For all integral values of n, the expression 7 2n – 33n is a multiple of 1) 22 2) 12 3) 10 4) 11 5) 24 2n

3n

2 n

3 n

n

n

Soln: 1; 7 – 3 (7 ) – (3 ) ( 49) – ( 27 ) By the condition (5), 49 – 27 = 22 is a factor of the expression 7 2n – 33n . Ex. 7: What should be subtracted from 3 2 27x 9x 6x 5 to make it exactly divisible by (3x – 1)? Soln: We will find the remainder by the rule of remainder. We have, 3x – 1 = 0 x

(5) x n a n is exactly divisible by x – a. Put x – a = 0 x = a

1 3

Thus, the remainder is 3

2

1 1 1 1 f 27 – 9 – 6 5 3 3 3 3 =1–1–2–5=–7 If we reduce the given expression by the remainder (–7), the expression will be exactly divisible by the given divisor. Hence, our required value is –7.

n n Then f(a) = a a 0 This is true in all cases.

Ex. 3: The expression 5 2 n 2 3n has a factor 1) 3 2) 7 3) 10 4) 17 5) None of these Soln: 4; 52n 23n (52 )n (23 ) n (25)n (8)n As we don’t know about n, by the condition (5), (25 – 8) is a factor, i.e. 17 is a factor of 5

is a factor, the last digit in the expansion should be 0. Ex. 5: Find the last two digits of the expansion of 212n – 64n when n is any positive integer..

2n

3n

2 .

n

Ex. 4: The last digit in the expansion of (41) – 1 when n is any +ve integer is 1) 2 2) 1 3) 0 4) –1 5) None of these Soln: 3; The last digit in the (41)n for any value of n is 1. Thus, the last digit in (41)n – 1 is 0. or, (41 – 1) is a factor or exact divisor of (41)n – (1) n for any +ve integer of n. Now, when 41 – 1 = 40

Theorem for Zero of a Polynomial Let f ( x ) a 0 x n a1x n –1 a 2 x n – 2 ..... a n be a polynomial with integral co-efficients. If an integer k is a zero polynomial then k is a factor of a n . Solved Example: Find all integral zeros of the polynomial,

f ( y) y 3 – 2 y 2 y 4 . Solution: Suppose (k) is an integral zero of the polynomial f(y). Then by the above theorem, k is a factor of a n i.e., 4. Hence, possible values of k are 1, –1, 2, –2, 4 and –4. Now, test each of them to see whether it is zero of the polynomial or not.

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62 (i) f (1) 13 – 2 12 1 4 4. Since f (1) 0 , so 1 is not a zero of f(y). (ii) f(–1) = ( 1)3 (2)( 1) 2 ( 1) 4 0 . Since f(–1) = 0, therefore –1 is the zero of f(y). (iii) Similarly, 2, –2, 4 and –4 are not the zeros of f(y). Thus, the only integral zero of f(y) is –1. If there are other zeros of the f(y), they are not integers. Quadratic Equations: An equation in which the highest power of the variable is 2 is called a quadratic equation. For example, the equation of the type ax 2 bx c 0 denotes a quadratic equation. The product of multiplication of two linear polynomials also gives a quadratic polynomial. Let the two linear polynomials be (lx + m) and (px + q), where 1 0, p 0. Then the product of these polynomials is the quadratic polynomial, lpx2 + (lq + mp)x + mq. This is written in the standard form as ax2 + bx + c. Then a = lp, b = (lq + mp) and c = mq. Factorisation of Quadratic Equations: The quadratic polynomial ax2 + bx + c can be factorized only if there exist two numbers r and s such that (i) r = lq, s = mp (ii) r + s = b = co-efficient of x = lq + mp (iii) r × s = ac = l.p.m.q = co-efficient of x2 × constant Solved Example: Factorize 2 x 2 11x 5. Solution: (i) Here a = 2, b = 11 and c = 5 (ii) Find two numbers r and s such that r + s = b = 11 and r × s = a × c = 2 × 5 = 10 So, the numbers are 10 and 1. (iii) Now, break up the middle term 11x of the given polynomial as 10x + 1x

2x 2 11x 5 2x 2 10x lx 5 = 2x(x + 5) + l(x + 5) = (2x + 1) (x + 5) Conditions for Factorisation of a Quadratic Equation: All quadratic expressions cannot be factorized. To test whether it can be factorized or not follow the points given below:

1.

a 2 b2 (a b)(a b)

2.

(a b)2 a 2 2ab b 2

3.

(a – b)2 a 2 2ab b 2

4.

a 3 b3 (a b)(a 2 ab b 2 )

5.

a 3 b3 (a b)(a 2 ab b2 )

6.

a 3 b3 (a b)3 3ab(a b)

a 3 b3 (a b)3 3ab(a b) 8. (a b)2 (a b)2 4ab 9. (a b) 2 (a b)2 4ab 10. a 3 b3 c3 3abc (a b c)(a 2 b2 c2 – ab bc ca) 7.

1 (a b c)[(a – b) 2 (b – c) 2 (c – a) 2 ] 2 11. (a b c)3 a 3 b3 c3 3(b c)(c a)(a b) 12. a 2 b2 (a b)2 2ab

13. a 2 b 2 (a – b)2 2ab 14. If a b c 0 then the value of a 3 b3 c3 is 3abc. 15. x 2 x(a b) ab (x a)(x b) 16. a 2 (b c) b2 (c a) c2 (a b) 3abc (a b c)(ab bc ca) 17. ab(a b) bc(c b) ca(c a) 2abc (a b)(b c)(c a) 18. a 2 (b c) b 2 (c a) c2 (a b) (a b)(b c)(c a) 19. (a b c)2 a 2 b2 c2 2(ab bc ca) Theory of Indices: The expression of a m means the product of m factors, each equal to ‘a’ . When m is a positive integer, m is called the exponent or index or the power of ‘a’. Any quantity raised to the power zero is always equal to 1. For example,

a 0 1(where a 0) 0

(1) If b 2 4ac 0 , then the quadratic equation can be factorized.

1 10 1, ( 1) 0 1, 1, (3)0 1, (1000) 0 1 etc. 2 Whenever any index of the power is taken from the numerator to denominator or from the denominator to numerator, the sign of the power changes. For example,

(2) If b 2 4ac 0 , then the quadratic equation cannot be factorized. Some important formulae that are used in basic operations and in finding the factors of an expression are summarized below:

1 am a m , n a m a n a m n m a a a Law of Indices: While solving the problems of exponents, the following laws are useful: am

1

m

,

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63

(1) x m x n x m n n (2) x m x mn (3) x mn x (mn ) (4) (xy)m x m ym

xm xm 1 x m n , if m > n and n n m ; if n > m n x x x Note: The difference between (2) & (3) can be seen in the following examples: (2) (33 )2 332 729 2 (3) 33 39 36 33 = 729 × 27 = 19683 In all the above laws, m and n are positive integers. In algebra, the root problems are solved by laws of indices, (5)

for example 4

2

a means a

1

2

,

4

a means a

1

4

,a

3

4

Step IV: Divide the second divisor by the second remainder. Step V: Continue the process till the remainder become zero. Step VI: The divisor which does not leave remainder is the HCF of the two numbers and thus, the last divisor is required (HCF) of given number. L.C.M.: The minimum multiplied quantities out of two or more algebraic expressions is called LCM. If only two expressions are given, then at first we find their HCF by the factor method or by the division method and apply the following formula to find their LCM. LCM of two expressions =

means

a . 3

H.C.F.: The highest common factor of two or more algebraic expressions can be determined by the following methods: 1. By the factor method 2. By the division method Factor Method: In case of the factor method, at first, the factors of the given expressions are found separately. Then the maximum number of common factors are taken out and multiplied, the product of which becomes the H.C.F. For example, if we are asked to find the H.C.F. of 8(x 2 5x 6) and 12(x 2 9) we shall first find the factors of the given expressions separately. Thus, 8(x 2 – 5x 6) 4 2(x 3)(x 2) and 12(x 2 9) 4 3(x 3)(x 3) Then the maximum number of common elements of the two will give the HCF of the expressions. Therefore, HCF = 4 (x – 3) = 4x –12 Division Method: To find highest common factor by using division method in discussed here: Finding highest common factor (HCF) by prime factorization for larger number is not very convenient. The method of log division is more useful for large numbers. We use the repeated division method for finding HCF of two or more number. By following steps: Step I: Divide the larger number by smaller one. Step II: Then remainder is treated as divisor and the divisor or dividend. Step III: Divide the first divisor by the first remainder.

Product of the two expressions Their HCF

Thus, LCM of 8(x 2 5x 6) and 12(x 2 9) will be equal to

8(x 2 5x 6) 12(x 2 9) (by the formula) 4(x 3) 24(x 2 5x 6)(x 3) = 24(x 3 5x 2 6x 3x 2 15x 18)

24(x 3 2x 2 9x 18) Solving Quadratic Equations: A quadratic equation when solved will always give two values of the variable. These values of the variable are called the roots of the equation. Any quadratic equation can either be solved by the factor method or by a formula. (1) By the Factor Method: When we solve a quadratic equation by the factor method, we first find the factors of the given equation making the right-hand side equal to zero and then by equating the factors to zero, we get the values of the variable. For example, in solving the quadratic equation x2 – 5x + 6 = 0 we first find the factors of x2 – 5x + 6 which are (x – 3) and (x – 2). Then we say that (x – 3) (x – 2) = 0 After that, we put each bracket equal to zero and find the values of x. i.e. when x – 3 = 0, x = 3 and when x –2 = 0, x = 2 Therefore, solution is x = 3 or x = 2. (2) By Formula: When we want to solve a quadratic equation ax 2 bx c 0, we apply the following formula

b b2 4ac 2a Taking + and – signs separately, we get two values of x. The quantity b2 – 4ac is called the discriminant. x

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64 The two values of x obtained from a quadratic equation are called the roots of the equation. These roots are denoted by and . It is seen that the sum of the roots b of a quadratic equation ax 2 bx c 0 is equal to , a b c i.e., and product of the root is equal to a a c i.e., . a For a quadratic equation ax2 – bx + c = 0. (i) The roots will be equal if b2 = 4ac. (ii) The roots will be unequal and real if b2 > 4ac. (iii) The roots will be unequal and unreal if b2 < 4ac. Formulating a quadratic equation from given Roots: Whenever we are given the roots of a quadratic equation x2 – x (sum of the roots) + product of the roots = 0. For example, if the roots are given as 2 and 3, then the quadratic equation will be as follows: x 2 (3 2)x 3 2 0 x 2 5x 6 0 Note: (i) A quadratic equation ax 2 bx c 0 will have reciprocal roots, if a = c. (ii) When a quadratic equation ax 2 bx c 0 has one root equal to zero, then c = 0. (iii) When the roots of the quadratic equation ax 2 bx c are negative reciprocals of each other, then c = – a. (iv) When both the roots are equal to zero, b = 0 and c = 0. (v) When one root is infinite, then a = 0 and when both the roots are infinite, then a = 0 and b = 0. (vi) When the roots are equal in magnitude but opposite in sign, then b = 0. (vii)If two quadratic equations ax2 + bx + c = 0 and a 1x 2 b1x c1 0 and have a common root (i.e., one root common), then (bc 1 – b 1c) (ab1 – a1b) = (ca 1 c1a ) 2 . (viii) If they have both the roots common, then a b c . a1 b1 c1 (ix) The square root of any negative number is an

imaginary number (e.g.

–2, –4, –25,

13 , etc), we do not 17 have to deal with the problems regarding –0.04, 20.2, 34 ,

imaginary numbers. So a simple introduction is sufficient i 2 1 , so, i 1 So, 4 ( 1) 4 1 4 2i , which is an imaginary number. Solved examples Ex. 1: Find the discriminant of the following quadratic equations. Tell the nature of the roots of the equations. Verify them. (i) x2 – 4x + 3 = 0 (ii) 3x2 + 4x – 3 = 0 (iii) –5x2 + 7x = 0 (iv) 2x2 – 6x + 7 = 0 Soln: (i) In x2 – 4x + 3 = 0, a = 1, b = –4, c = 3 2 2 D b 4ac (4) 4 1 3 16 12

4 0 and also a perfect square. So, the roots will be distinct rational numbers.

b D (4) 4 4 2 2a 2 1 2 2 1 3 and 1. (These are distinct rational numbers.) (ii) In 3x2 + 4x – 3 = 0, a = 3, b = 4, c = –3 2 2 D b 4ac (4) 4 3 (3) 16 36 52 0 and is not a perfect square. So, the roots will be distinct irrational numbers. Now, the roots are Now, the roots are

b D 4 52 4 2 13 2a 23 23 2 13 2 13 2 13 and 3 3 3 (iii) Here, a = –5, b = 7, c = 0. =

2 D b 4ac becomes b 2 7 2 49 0 and also a perfect square. So, the roots will be distinct rational numbers.

Now, the roots are

b D 7 7 2 7 7 2a 2 ( 5) 10

0 14 and i.e. 0 and 1.4. 10 10 As c = 0, we can simply solve this equation as –5x2 + 7x = 0 or, x (–5x + 7) = 0 either x = 0 or –5x + 7 = 0, i.e., 5x = 7 7 i.e. x 1.4 5 Finally, we get x = 0 and 1.4.

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65

(iv) Here a = 2, b = –6, c = 7

D b 4ac (6) 4 2 7 36 56 2

2

20 0; so the roots will be imaginary.. Now, the roots are

b D (6) 20 2a 22

6 2 5i 3 5i 3 5 3 5 i and i 2 2 2 2 2 2 2 Ex. 2: Form a quadratic equation whose roots are (i) 3 and –5 (ii) 2 and 7, and verify them. Soln: (i) The quadratic equation will be x2 – sum of the roots × x + products of the roots = 0 =

or, x 2 {3 ( 5)} x 3 (5) 0 or, x 2 (2)x (15) 0 or, x2 + 2x – 15 = 0 Verification: D = 4 + 60 = 64

The roots are

2 64 2 8 1 4 i.e. 2 1 2

3 and (–)5.

27 3 6 4 4 Ex. 4: Find the roots of the equations (i) 2x2 + 3x – 5 = 0 (ii) x2 – 8x + 7 = 0 Note: Whenever we get a + b + c = 0, one of the roots will always be 1. Soln: (i) Here a + b + c = 2 + 3 + (–5) = 0, so, 1 is one of the roots of the equation 2x2 + 3x – 5 = 0 The other root = sum of the roots – one of the 3 5 roots = 1 2 2 5 So, the required roots are 1 and 2 (ii) Try yourself.

i.e.

Ex. 5: A motorcycle travels 20 km an hour faster than a cycle over a journey of 600 km. The cycle takes 15 hours more than the motorcycle. Find their speeds. Soln: Let the speed of the cycle be x kmph then that of the motorcycle = x + 20.

distance 600 speed x

(ii) x 2 ( 2 7) x 2 7 0

Time taken by the cycle =

or, x 2 9x 14 0 Verification: D = 81 – 56 = 25

and the time taken by the motorcycle

The roots are

– (–9) 25 9 5 14 and 2 1 2 2

4 7 and 2. 2 Ex. 3: (i) If one of the roots of the equation x2 – 19x + 88 = 0 is 8, find the other root. (ii) If one of the roots of the equation 4x2 – 27x + 18 = 0 is 6,find the other root.

Soln: (i) We have the product of the roots =

600 15 hours less than the time taken by x 20 the cycle.

c 88 88 a 1

88 The other root = one root i.e. 8 11

19 b 19 a 1 the other root = 19 – 8 = 111 (ii) The required other root = sum of the roots Or, the sum of the roots =

b 27 one of the roots (= 6) a 4

600 600 15 x 20 x

i.e.

40 40 40 x 1 x 20 x x or, 40x = 40(x + 20) – x(x + 20) or, 40x = 40x + 800 – x2 – 20x or, x2 + 20x – 800 = 0

or,

x

20 (20)2 4 1(800) 2 1

20 400 3200 2

20 60 10 30 20 or 40 2 As x, the speed of the cycle, cannot be negative, so x = –40 is not acceptable. x, the speed of the cycle = 20 kmph and the speed of the motorcycle = x + 20 = 40 kmph

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66

It is the real number line. As we move right, the value becomes greater. So, 2 < 3, –3 < –2, (on the real number line as –2 is in the right side of –3, –2 is greater than –3) Also –2 < 0, –2 < 1, –1.5 < –0.5, –1.999 > –2, and so on.

Ex. 6: Solve the equation 32x 1 3x 3x 3 32 Soln: 32 x 3 3x 3x 33 32 or, 3(3x)2 – 3x = 27 × 3x – 9 or, 3m2 – m = 27m – 9 Where m = 3x or, 3m2 – 28m + 9 = 0 28 (28) 2 4 3 9 28 676 23 23 28 26 14 13 1 9, 23 3 3 When m = 9, then 3x = 9 = 32 x = 2

m

and when m =

1 1 x 1 then 3 3 3 3

x = –1 Ex. 7: For what value of m can the equation –9x2 + 12x – m = 0 be a perfect square of a linear expression? Soln: A linear expression is of the form ax + b = 0; (a 0), a and b are constants. A quadratic equation whose roots are and is given by (x – ) (x – ) = 0 If = then the equation becomes (x – )2 = 0. For both the roots to be equal, we have D = 0. So, the given equation –9x2 + 12x – m = 0 can be a perfect square of a linear expression if D = 0 or, b2 – 4ac = (12)2 – 4 × (–9) (–m) = 0 or, 144 – 36m = 0 144 m = 36 4 Verification: If we put m = 4 in the equation –9x2 + 12x – m = 0, the equation becomes –9x2 + 12x – 4 = 0 or, 9x2 – 12x + 4 = 0 or, (3x – 2)2 = 0 Clearly, 3x – 2 is a linear expression of the form ax + b. Here a = 3 and b = –2.

Quadratic Expression An expression of the form ax2 + bx + c, ( a 0 ) where a, b, c are real numbers is called a quadratic expression in x. The corresponding equation of the expression ax2 + bx + c is ax2 + bx + c = 0 Before discussing the sign scheme for the quadratic expression, let us study the concept of real number line. -ve side

+ve side

– -3 -2 -1 0 (left)

smaller no.

1 2 3 greater no.

(right)

Sign scheme for the quadratic expression Note: 1. The sign sheme for the quadratic expression is always meant for the real values of x. We cannot compare any two imaginary numbers. So to say that ri > 2i or 4i < wi is absolutely incorrect. Let and be the roots of the corresponding quadratic equation (i.e. ax2 + bx + c = 0) of the quadratic expression ax2 + bx + c (= y, suppose). Then, we have, for all the real values of x: Case I: If and are real and equal or both imaginary i.e., D 0 then y i.e., ax2 + bx + c will have the same sign as that of a, the coefficient of x2. That is, if D 0 and a is +ve, y will always be +ve, and if a is -ve y will always be -ve. Case II: If and are real and unequal i.e., if D > 0, the sign scheme for y i.e. ax2 + bx + c is as follow:

Ex. 1: Find the sign scheme for the quadratic expression x2 – 4x + 7. Soln: The corresponding equation is x2 – 4x + 7 = 0

D (4)2 4 1 7 16 – 28 –12 0 i.e. roots are imaginary. Here, a = 1 > 0. Therefore, for all the real values of x, the given expression x2 – 4x + 7 is always positive. Note: 1.For all the real values of x (x may be 0.2, 32.07, 3 , –4, – 32.07, 2 5 , etc) x2 – 4x + 7 > 0. 8 You can put x = any real number and verify that x2 – 4x + 7 > 0 Let us suppose x = –4.2, then x2 – 4x + 7 = (4.2)2 – 4 × (–4.2) + 7 = 17.64 + 16.8 + 7 > 0 When x = 0.07 then x2 – 4x + 7 = (0.07)2 – 4 × 0.07 + 7 = 0.07(0.07 – 4) + 7 = 0.07 × (3.93) + 7 = 7 – 0.2751 > 0 When x = 2 – 5 then x2 – 4x + 7 = (2 – 5 )2 – 4(2 – 5 ) + 7

Elementary Algebra

67 So, the sign scheme for the given expression –x2 + 3x + 28 is as follows:

=4+5– 4 5 –8+ 4 5 +7=1+7>0 2.Here, D = –12 < 0 i.e., the roots are imaginary.

b D i.e. The roots are 2a –(–4) 12 4 2 3i 2 3i 2 1 2 So, if we put x = 2 3 i or 2 3 i , the given expression will become zero. The expression cannot be +ve or -ve. When x = 2 3i , x2 – 4x + 7 = ( 2 3 i ) 2 4( 2 3 i ) 7

4 3i 2 4 3 i 8 4 3 i 7 = 4 + 3(–1) – 8 + 7 (we have i 2 1 ) = 4 – 3 – 1= 0 Similarly, when x 2 – 3 i, x 2 4 x 7 0 3. The sign scheme is not valid for imaginary values of x. If we put x = 4i in the given expression, we get x2 – 4x + 7 = (4i)2 – 4(4i) + 7 2

16i 16i 7 = 16 ( 1) 16i 7 = –16 –16i + 7 = –9 –16i which is imaginary and it cannot be compared with any real or imaginary number. So the purpose of the sign scheme becomes meaningless except for the roots ( 2 3 i which are imaginary) Ex. 2: For what values of x, 9x2 + 42x + 49 > 0? Soln: The corresponding equation is 9x2 + 42x + 49 = 0

D ( 42) 2 4 9 49 1764 – 1764 0 Here, the coefficient of x2 = a = 9 > 0 So, for all real values of x, 9x2 + 4x + 49 > 0 i.e. the given expression is always positive. Ex. 3: Find the sign scheme for –x2 + 3x + 28. Soln: The corresponding equation is –x2 + 3x + 28 = 0 D (3) 2 4 ( 1) 28 9 112 121 0 The roots of the equation are 3 121 3 11 8 14 , = – 4, 7 2 ( 1) 2 2 2 Here, the coefficient of x2 is –1 < 0.

(+ve) – 7(–ve)

(–ve)–4

That is, Case (i): if –4 < x < 7, –x2 + 3x + 28 > 0 Case (ii): if x < –4 or x > 7, –x2 + 3x + 28 < 0 Case (iii): if x = –4 or 7, –x2 + 3x + 28 = 0 Note: You can verify the above cases (i) and (ii) by putting such values of x as may be easier for calculation. If we put x = 0, –4 < 0 < 7 we see, –x2 + 3x + 28 = 0 +0 + 28 = 28. If we put x = –5 < –4, –x2 + 3x + 28 = –25 – 15 + 28 = –12 < 0 If we put x = 10 > 7, –x2 + 3x + 28 = –100 + 30 + 28 = –42 < 0 Ex. 4: For what range of the values of x, is 2x2 – 4x – 7 non-positive? Soln: The corresponding equation is 2x2 – 4x – 7 = 0 D ( 4)2 4 2 ( 7) 16 56 72 0 Here, the coefficient of x2 is 2 > 0 The roots

are

–(–4) 72 4 6 2 2 3 2 22 22 2 So, the sign scheme for the given expression 2x2 – 4x + 7 is as follows: (–ve) –

(– ve) for x =

23 2 2

23 2 (– ve) 2

23 2 , the given expression will be zero 2

2–3 2 23 2 x the given 2 2 expression will be negative. For other values, the expression will be +ve. and for

2–3 2 23 2 x the given 2 2 expression is non-positive i.e., the expression is either zero or negative. Note: If you want to check the sign scheme, you simply Thus, for

take an approximate value of For example, we take

2 and then proceed.

2 = 1.4 (approximately)

Quicker Maths

68

Now, the roots are

2 3 1.4 1 3 0.7 2

(ii) is more suitable than option (i). Though option (iii) is correct, it is not as close as option (ii). Options (iv) and (v) are incorrect. Hence (ii) is the answer.

1 2.1 3.1, 1.1 As the roots have been found by approximation, so while checking for the sign scheme, you should not take such values of x which are nearer to the roots.

Summary of the above discussions

23 2 23 2 1 Putting x = 1 2 2

or a1x 2 b1x c1 0 lies in the two parts of the sign scheme

The solution of the inequality ax 2 bx c 0

2

2 x 4 x 7 2 4 7 9 0 Putting x = 5, 2x2 – 4x – 7 = 2 × 25 – 20 – 7 = 23 > 0 Putting x = –2, 2x2 – 4x – 7 = 2 × 4 + 8 – 7 = 9 > 0 Thus, we find that the sign scheme is correct. Ex. 5: The inequality of b2 + 8b 9b + 14 is correct for (i) b 5, b 5 (ii) b 5, b 4 (iii) b 6, b 6

(iv) b 4, b 4

(v) b 6, b 4 Soln: The given equality is b + 8b 9b + 14 . i.e. b2 + 8b – 9b – 14 0 or, b2 – b – 14 0. For the equation b2 – b – 14 = 0, we have 2

D (–1)2 4 1 (–14) 1 56 57 0

Here, the coefficient of b 2 1 0 and the roots are

( 1) 57 1 7.6 2 1 2

[ 57 7.6 (approximately)] = 4.3 and –3.3 So, the sign scheme for the expression b2 – b – 14 is as follows: (+ve) 4.3 (-ve)

Thus for b 3.3 or, b 4.3, b 2 – b 14 0 To decide the correct option, we draw the points of the option on the real number line of the sign scheme (+ve)

(+ve)

(–ve)

– .... -7 -6 -5 -4 -3.3 -3

4

The solution (ie, the value of x) will lie either between its roots (ie, x ) or outside its roots (ie x or

x ). Now, the question is, in which case will the value of x lie between its roots and in which case outside its roots? Remember the following two points and forget all the above discussion. For the above two inequalities. where a and a1 are +ve, (i) ax 2 bx c 0 x (ii) a1x 2 b1x c1 0 x or x Note: To remember the above points, mark that when the inequality is less than zero (ie 0), the value of x is in the smaller range (ie x )and when the inequality is more than zero (ie 0) the value of x is in the larger range (ie x or x ) Now we shall discuss the above examples again. Take Ex. 4: For 2x 2 4x 7 0, find the range of x. The corresponding equation is 2x 2 4x 7 0 and 23 2 2 Here we see the inequality is less than zero so the range of x will lie between its roots. the roots are

– (-ve) -3.3

4.3

5

6

7 .....

Now, we tally each of the options one by one. Clearly, option (i) is correct. Option (ii) is also correct. Between these two options, we observe that option

So,

23 2 23 2 x 2 2

Take Ex 5: b2 8b 9b 14

b2 b 14 0 b = –3.3, 4.3 Since the inequality is greater than zero, the value of b will lie in the larger range b 3.3 or b 4.3 So, the closest answer is (ii), ie b 5, b 4

Elementary Algebra

69

Take Ex 3: x 2 3x 28 The roots of the corresponding equation are –4 and 7. Case I: When –x2 + 3x + 28 0 Change the -ve sign of x2 ie multiply by –1. Then, x 2 3x 28 0 Now, as the inequality is greater than zero, the value of x will lie in the larger range ie x 4, x 7 Case II: When –x2 + 3x + 28 0 Then, x 2 3x 28 0 As the inequality is less than zero, the value will lie in the smaller range, ie 4 x 7 . Note: In the above case, don't follow the original sign of the inequality. First make the coefficient of x2 positive and accordingly change the sign of the inequality. Ex. 6: If x + y is constant, prove that xy is maximum when x = y. 2

2

x y x y Soln: xy = , 2 2 We know that the square of any real no. 0 2

2

x y x y So, 0 and 0 2 2 As x + y is constant, xy is the maximum when 2

x y 0 2

xy = 0 or, x – y = 0 or, x = y 2 Ex. 7: Which of the following values of P satisfies the inequality = P (P – 3) < 4P – 12? 1) P > 14 or P < 13 2) 24 P 71 3) P > 13; P < 51 4) 3 < P < 4 5) P = 4, P = +3 Soln: P(P – 3) < 4 (P – 3); P(P – 3) – 4(P – 3) < 0; (P – 3) (P – 4) < 0 This means that when (P – 3) > 0 then (P – 4) < 0 ... (i) or, when (P – 3) < 0 then (P – 4) > 0 ... (ii) From (i) P > 3 and P < 4 3

QUICKER MATHS

Miraculous for Banks, LIC, GIC, UTI, SSC, CPO, Management, Railways and other competitive exams Stimulating for general use

M. TYRA Revised and Enlarged Edition

BSC PUBLISHING CO. PVT. LTD. C-37, GANESH NAGAR, PANDAV NAGAR COMPLEX DELHI-110092 Phone: 011-22484910, 22484911, 22484912, 22484913 e-mail: [email protected]; Website: www.bsccareer.com

BSC PUBLISHING CO. PVT. LTD. © Author First Edition — 1995 Second Edition — 1999 Third Edition — 2000 Fourth Edition — 2013 Fifth Edition — 2018

ISBN: 978-81-904589-2-4 DISCLAIMER Every effort has been made to avoid errors or omissions in the publication. In spite of this, some errors might have crept in. The publisher shall not be responsible for the same. Any mistake, error or discrepancy noted may be brought to our notice which shall be taken care of in the next edition. It is notified that neither the publisher nor the author or sellers will be responsible for any damage or loss of action of any one, of any kind, in any manner, therefrom. All rights reserved. No part of this publication may be reproduced in any form without the prior permission of the publisher.

CONFIRM THE ORIGINALITY OF THE BOOK Published by

BSC PUBLISHING CO. PVT. LTD. C-37, Ganesh Nagar, Pandav Nagar Complex Delhi-110092

Ch. 560

Dedicated to Sri Jagdeo Singh (Dadaji)

.

Contents Chapters 1. How to Prepare for Maths 2. Addition

Page No. 1 5

3. Multiplication

11

4. Division

17

5. Divisibility

21

6. Squaring

25

7. Cube

27

8. HCF and LCM

31

9. Fractions

41

10. Decimal Fractions

49

11. Elementary Algebra

59

12. Problems on Comparison of Quantities

89

13. Surds

97

14. Number System

101

15. Binary System

125

16. Permutation & Combination

129

17. Probability

143

18. Ratio and Proportion

157

19. Partnership

183

20. Percentage

191

22. Average

215

22. Problems Based on Ages

223

23. Profit and Loss

233

24. Simple Interest

265

25. Compound Interest

277

26. Alligation

293

27. Time and Work

313

28. Work and Wages

333

29. Pipes and Cisterns

337

30. Time and Distance

343

31. Trains

359

32. Streams

373

33. Elementary Mensuration-I

381

34. Elementary Mensuration-II

401

35. Series

415

36. Data Sufficiency

433

37. Data Analysis

467

38. Caselets

533

39. Trigonometry

545

Foreword When did you get up this morning? At 6.35 am. Did you reach your destination on time? Yes, I arrived 10 minutes earlier even though I got up five minutes late. I usually walk but today I took a bus. Why not a taxi? Isn’t it faster? Faster, yes. But, in my case the speed of the bus was sufficient. Besides, I compared the fares of the two means of transport and concluded that taking a taxi would mean incurring unnecessary expenditure. Wow! what a clever guy! Saves money, saves time. Does he know some magic? Is he a student of occult sciences? No, dear. He is an ordinary mortal like any of us. But he exudes an uncommon confidence, thanks to his ability to compute at a magical pace. In other words, Quicker Maths is an asset at every step of one's life. By Quicker Maths, the book means speed and accuracy at both simple numerical operations and complex problems. And it is heartening that the book takes into account all kinds of problems that one may encounter in the ordinary run of life. There are several books one comes across which take care of problems asked in examinations. But they are conventional and provide solutions with the help of detail method. This book provides you both the detail as well as the quicker method. And it is the latter that makes the book irresistible. That the book provides you both the methods is a pointer to the fact that you are not being led into a faith where you have to blindly follow what the guru says. The conclusions have been rationally drawn. The book, therefore, serves as a guide—the true function of a guru—and once you get well-versed with the book, you will feel empowered enough to evolve formulas of your own. I think the real magic lies there! At the same time, a careful study of the book endows you with the magical ability to arrive at an answer within seconds. There may be some semi-educated persons who sneer at this value of the book. Can you beat a computer?— a contemptuous question is asked. Absurd question. For one, the globe is yet to get sufficiently equipped with computers. Two, even when we enter a full-fledged hi-tech age, let us hope it is not at the cost of our minds. Let not computers and calculators become the proverbial Frankenstein's monster. The mind is a healthy organ and computing a healthy function. Computers and calculators were devised by the brain to aid it, not to consume it. That the mathematical ability of the brain be intact is a concern of every individual. In most of the examinations even calculators are not allowed, let alone computers. And it is here that this magical book proves immensely useful. If you are a reader of this book, you can definitely feel more confident—miles ahead of others. There is no doubt that there has been a lot of labour involved in the book. For all those students who are gearing up to drive their Mathematics Marutis at an amazing 200 kmph, the book definitely provides the requisite infrastructure. And what is more, the methods are accident-free with proper cautions at necessary places. Here is a book that will help exam-takers glide and enthusiastic students enjoy the ride. In an age when speed is being maniacally pursued, a careful study of the book will serve as a powerful accelerator. At the same time, its simple language makes it easily accessible across the linguistic barriers. Besides, the fact that you vividly see the Quicker Method makes things very interesting. And so the book provides you speed not at the cost of joy. It is not merely a mechanical device, but has an organic charm. One concludes: fast driving is fun. Chetananand Singh Editor, Banking Services Chronicle

Author’s Preface We, at Banking Services Chronicle (BSC), analyse students' problems. If a student is not able to perform well in an exam, our research group members try to penetrate the student's psyche and get at the roots of the problems. In the course of our discussions we found that the mathematics section often proves to be the Achilles' heel for most of the students. Letters from our students clearly indicated that their problem was not that they could not solve the questions. No, the questions asked in general competitions are in fact so easy that most of the students would secure a cent per cent score, if it were not for the time barrier. The problem then is: INABILITY TO SOLVE the question IN TIME. Unfortunately, there was hardly any book available to the student which could take care of the time aspect. And this prompted the BSC members to action. We decided to offer a comprehensive book with our attention targeted at the twin advantage factors: speed and accuracy. Sources were hunted for: Vedic Mathematics to computer programmings. Our aim was to get everything beneficial from wherever possible. The most-encountered questions were categorised. And Quicker Methods were intelligently arrived at and diligently verified. How does the book help save your time? Probably all of you learnt by heart the multiplication-tables as children. And you have also been told that multiplication is the quicker method for a specific type of addition. Similarly, there exists a quicker method for almost every type of problem, provided you are well-versed with some key determinants and formulas. For the benefit of understanding we have also given the detail method and how we arrive at the Quicker Method. However, for practical purposes you need not delve too much into the theory. Concentrate on the working formulas instead. For the benefit of non-mathematics students, the book takes care to explain the oft-used terms in an ordinary language. So that even if you are vaguely familiar with numbers, the book will prove beneficial for it is self-explanatory. The mathematics students are relatively in a comfortable position. They do not have to make an effort to understand the concepts. But even in their case, there are certain aspects of questions asked in the competitive exams which have been left by them untouched since school days. So, a revision is desirable. In the case of every student, however, the unique selling proposition of the book lies in its ability to increase the student's problem-solving speed. Due caution has been observed to proceed methodically. Gradual progress has been made from simple to complex examples. There are theorems and solved examples followed by exercises. A systematic, chapter-by-chapter study will definitely result in a marked improvement of the student's mathematical speed. The students are requested to send their responses to the book and suggestions for further improvement. And, finally, I would extend my thanks to all those who have played a role in making the book available to the reader. I specially thank Mr Madhukar Pandey for having played a key role in promoting the endeavour, Mr Chetananand Singh for the meticulous editing of the book and Mr Niranjan Bharti for having carefully verified the results. Mr Niranjan Singh's all-round assistance cannot be forgotten. Friends kept on encouraging me at every step. The inspiration I received from Mr Sanjay, Mr Deven Bharti, Mr Nagendra Kumar Sinha, Mr Sandeep Varma, Mr Manoj Kumar, Mr Vijay Kumar, Mr Rajeev Raman, Mr Anil Kumar and Mr JK Singh, to name a few, has been invaluable. And thanks to Mr Pradeep Gupta for printing.

Preface to the Second Edition It is a great pleasure to note that Magical Book on Quicker Maths continues to be popular among the students who are looking for better results in this world of cut-throat competition. This book has brought the new concept of timesaving quicker method in mathematics. So many other publications have tried to publish similar books but none could reach even close to it. The reason is very simple. It is the first and the original book of its kind. Others can only be duplicate and not the original. Some people can even print the duplicate of the same book. It will prove dangerous to our publication as well as to our readers. So, we suggest our readers to confirm the originality of this book before buying. The confirmation is very simple. You can find a three-dimensional HOLOGRAM on the cover page of this book. This edition has been extensively revised. Mistakes in its first edition have been corrected. Some new chapters like Permutation-Combination, Probability, Binary System, Quadratic Expression etc. have been introduced. Some old chapters have been rewritten. Hope you will now find this book more comprehensive and more useful.

Preface to the Third Edition The pattern of question paper as well as the standard of questions have changed over the past couple of years. Besides, Permutation-Combination and Probability, and questions from trigonometry—in the form of Height and Distance—have also been introduced. In chapters like Data Analysis, Data Sufficiency and Series, new types of questions are being asked. With the above context in mind, a few new chapters have been introduced and a few old ones enlarged. Important Previous Exam questions have been added to almost all the chapters. But they have been added in larger numbers in the chapters specially mentioned above. An introductory chapter has been added on “How to Prepare for Maths”. I suggest going through this chapter before setting any targets. A revision in the cover price was long due. The first edition (1995), which cost Rs 200, had only 612 pages. The price remained the same even in the second edition (1999) in spite of the number of pages being increased to 749. But the third edition (2000) has gone into 807 voluminous pages. So, the price is being increased to Rs 280. Kindly bear with us.

Preface to the Fourth Edition Another edition of this book had long been overdue. No matter how good a book — well, that has been the verdict of generations of readers — there is always scope for improvement. And this edition is an endeavour in this direction. The chapter on “Division” has been introduced once again. Besides, simplicity has been the hallmark of this book for decades. I have tried to further simplify the methods wherever I could. Hope you will benefit more from this book after the incorporation of these changes.

Preface to the Fifth Edition Questions in competitive exams have changed quite a lot over the past few years. Being the pioneer in Mathsbased competitive exams, it was incumbent upon us to guide the students in the changed environment. Hence the Fifth Edition of the book that has been so dear to generations of readers. We have tried to add questions based on the latest patterns to the chapters of this book. Besides, two new chapters have been introduced: Comparison of Quantities; and Caselets. Some other chapters have been specially enhanced. In Mensuration and Percentage, we have added an Exercise each after Solved Examples. In Data Interpretation (DI), an Exercise has been added with the latest questions. This includes DI questions based on missing data as well as those based on Arithmetic. With so many additions the content became voluminous. In order to address this issue, we have changed the format of the book. Hope the book in its new format and with its revised content serves you adequately.

Chapter 1

How to Prepare for MATHS (Using this book for Competitive Exams) 1. Importance of Maths paper (PO) Quantitative Aptitude is a compulsory paper. You can't neglect. So make sure you are ready to improve your mathematical skills. Each question values 1.2 marks whereas each question of Reasoning values only 1.066 marks in PO exam. So, if you devote relatively more time on this paper you get more marks. Also, the answers of Maths questions are more confirmed than answers of Reasoning questions, which are often confusing. Most of you feel it is a more time-consuming paper, but if you follow our guidelines, you can save your valuable time in examination hall. Other exams: There are very few competitive exams without Maths paper. SSC exams have different types of Maths paper. The mains exam of SSC contains Subjective Question paper. Keeping this in mind, I have also given the detail method of each short-cut or Quicker Method given in this book. Each theorem, which gives you a direct formula also contains proof of the theorem, which is nothing but a general form (denoting numerical values by letters say X, Y, Z etc) of detail method.

2. Preparation for this paper (A) How to start your preparation Maths is a very interesting subject. If you don't find it interest-ing, it simply means you havn't tried to understand it. Let me assure you it is very simple and 100% logical. There is nothing to be assumed and nothing to be confused about. So, nothing to worry if you come forward with firm determination to learn maths. The most basic things in Maths are: (a) Addition - Subtraction (b) Multiplication - Division All these four things are most useful. At least one of these four things is certainly used in any type of mathematical question. So, if we do our basic calculations faster we save our valuable time in each question. To calculate faster, I suggest the following tips:

(i) Remember the TABLE upto 20 (at least): You should know that tables have been prepared to make calculations faster. You can see the use of table in the following example: Evaluate: 16 × 18 If you don’t remember the table of either 16 or 18 you will proceed like this: 16 18 128 16 288 But if you know the table of 16, your calculations would be: 16 × 18 = 16(10 + 8) = 16 × 10 + 16 × 8 = 160 + 128 = 288 Or, if you know the table of 18; your calculation would be 16 × 18 = 18(10 + 8) = 18 × 10 + 18 × 6 = 180 + 108 = 288 If you can, you remember the table upto 30 or 40. It will be precious for you. Note: (1) You should try the above two methods on some more examples to realise the beauty of tables. Try to evaluate: 19 × 13; l7 × 24; 18 × 32 (18 × 30 + 18 × 2 = 540 + 36 = 576); 19 × 47; 27 × 38; 33 × 37 etc. (2) All the above calculations should be done mentally. Try it. (ii) LEARN the one-line Addition or Subtraction method from this book In the first chapter we have given some methods of faster addition and subtraction. Suppose you are given to calculate: 789621 – 32169 + 4520 – 367910 = .... If you don’t follow this book you will do like:

Quicker Maths

2 789621 +4520 794141

32169 +367910 400079

794141 –400079 394062 The above method takes three steps, i.e. (i) add the two +ve values; (ii) add the two -ve values; (iii) subtract the second addition from the first addition. But you can see the one-step method given in the chapter. Have mastery over this method. It takes less writing as well as calculating time. (iii) Learn the one-line Multiplication or Division method from this book Method of faster multiplication is given in the second chapter. I think it is the most important chapter of this book. Multiplication is used in almost all the questions, so if your multiplication is faster you can save at least 35% of your usual time. You should learn to use the faster one-line method. It needs some practice to use this method frequently. The following example will show you how this method saves your valuable time. Ex. Multiply: 549 × 36 If you don't follow the one-line method of multiplication, you will calculate like: 549 36 3294 1647 19764 If this method takes 30 seconds I assures: you that one-line method given in this book will take at the most 15 seconds. Try it. One-line method of calculation for Division is also very much useful. You should learn and try it if you find it interesting. But, as division is less used, some of you may avoid this chapter.

(iv) Learn the Rule of Fractions In the chapter Ratio and Proportion on Page No. 269, I have discussed this rule. It is the faster form of unitary method. It is nothing but simplified form of Rule of Three and Rule of Proportional Division. No doubt, it works faster and is used in almost all the mathematical questions where unitary method (Aikik Niyam) is used. See the following example: Ex. If 8 men can reap 80 hectares in 24 days, how many hectares can 36 men reap in 30 days? Soln: I don't know how much time you will take to answer the question but if we follow the rule of fraction our calculation would be:

80

36 30 450 hectares. 8 24

In this book this method is used very frequently. It is only when you go through the various chapters of this book that you will find how wonderful the method is. It saves at least half of your usual time. I think this method should be adopted by all of you at any cost. First learn it and then use it wherever you can. So, these four points are necessary for your strong and firm start. And only strong and firm start is the key to sure success.

(B) Clear the Fundamentals behind each chapter There are 39 chapters in this book. Each chapter has some important basic fundas. Those fundas should be clear to you. Any doubt with the basics will hamper your further steps. Now the question arises - what are those basic fundas? Naturally, for each chapter, there are different fundas. I will discuss one chapter and its basic fundas. You can understand and find the same with different chapters. My chapter is PARTNERSHIP (on page 309)

Partnership

Simple (Different investments for the same period) Ratio of profit = P : Q.

Compound [Different investments (P & Q); Diff periods (t 1 & t 2) Ratio of profit = P × t1: Q × t 2

How to Prepare for MATHS

3

It is natural that your fundas of ratio should be clear before going through this chapter. Now, you can understand what I actually mean by the fundas. In a similar way, you can collect all the fundas and basic formulae at one place.

sources: Guides, Books, Magazines etc. The most standard and reliable sets are available in the magazine Banking Services Chronicle. Also, with our Correspondence Course we give at least 60 sets of Maths papers separately and 60 sets of Maths with full-length Practice Sets.

(C) More and More Examples

(C) Now start your practice:

You are suggested to go through as many examples as possible. Each question given in examples has some uniqueness. Mark it and keep it in mind. To collect more examples of different types you may consult different books available in the market.

From the beginning to the end, the complete session of practice should be divided into five parts. Part (i): Take your first test with previous paper without taking time into consideration. Try to solve all questions. Note down the total time and score in your performance diary. Also note down the questions which took more than one minute. Now you have to find out the reason of your low performance, if it is so. Naturally, you would find the following reasons: Some questions were difficult and time consuming. Some questions were unsolvable for you. You lost your concentration. You lost your patience. You did more writing job. You could not use Quicker Methods. Try to find out the solutions to all the above problems. If any of the questions was difficult for you, it means your initial preparation was not good. But don't worry. Go through that chapter again and clear your basic concepts. Because the standard of a question is always within your reach. If you have passed your 10th exam with maths, you can solve all the questions. You should take at least 10-12 tests in this part. Part (ii): This time the paper may be either previous or model (sample). Fix your alloted period (say 50 minutes for PO). And solve as many questions as possible within that period. Once you have completed your test, count the number of correct questions. Note down the number of questions solved by you and the no. of correct solutions in your performance diary. Now, you can find the reasons for your low scoring. If the reasons are the same as in Part (i), you try to resolve the problem again. Take at least 5-6 tests in this part. After analysing your performance and problems you should be ready for your third part of test. Part (iii): This time you try to solve all the questions within a time period fixed by you in advance. This period should be less than that for the tests in part (ii). If you couldn't do it, try it again on another test paper. Part (iii) should not be

(D) Use of Quicker Formulae Before going for quicker formula I suggest you to know the detail method of the solution as well. So, see the proofs of all the formulae carefully. Once you get familiar with the detail solution, you find it easier to understand the quicker method. Direct Formulae or Quicker Methods save your valuable time but they have very high potential of creating confusion in their usage. So you should know where the particular formula should be used. A little change in the questions may lead you to wrong solution. So be careful before using them. In case of any confusion, you are suggested to solve the questions without using direct formula or quicker method. Only frequent use of the quicker methods can make you perfect in Quicker Maths. After covering all the chapters and knowing all the methods, you should be prepared for practice.

3. Practice of Maths Paper (A) Pattern of paper You should know the pattern and style of the question paper of the exam for which you are going to appear. Suppose you are preparing for Bank PO exam. You should know that maths paper consists of 35 questions in prelims exam, 35 questions main exam and 50 questions in various PGDBF courses. Out of which 15-20 questions are from Data Analysis, 5 Questions from Data Sufficiency, 5-10 are from Numericals (Calculation based), and 20-25 are from Mathematical Chapters (like Profit & Loss, Percentage, Partnership, Mensuration, Time & Work, Train, Speed etc.). In other exams it may be different. The pattern can be known from previous papers.

(B) Collection of Previous Papers as well as sample papers If you can, you should arrange as many as possible numbers of previous and sample papers. There are many

4 considered completed unless you have achieved your goal. Part (iv): After completion of part (iii), you need to increase your speed. This part is penultimate stage of your final achievement. You should try to solve the complete paper of maths within a minimum

Quicker Maths possible time (say, 30-35 minutes for PO paper). This part may take 3 to 4 months. Keep patience and go on practicing. Part (v): This is the last part of your practice. After part (iv), you should take your test with complete fulllength paper for PO.

Chapter 2

Addition In the problem of addition we have two main factors (speed and accuracy) under consideration. We will discuss a method of addition which is faster than the method used by most people and also has a higher degree of accuracy. In the latter part of this chapter we will also discuss a method of checking and double-checking the results. In using conventional method of addition, the average man cannot always add a fairly long column of figures without making a mistake. We shall learn how to check the work by individual columns, without repeating the addition. This has several advantages: 1) We save the labour of repeating all the work; 2) We locate the error, if any, in the column where it occurs; and 3) We are certain to find error, which is not necessary in the conventional method. This last point is something that most people do not realise. Each one of us has his own weaknesses and own kind of proneness to commit error. One person may have the tendency to say that 9 times 6 is 56. If you ask him directly he will say “54”, but in the middle of a long calculation it will slip out as “56”. If it is his favourite error, he would be likely to repeat it when he checks by repetition.

Totalling in columns As in the conventional method of addition, we write the figures to be added in a column, and under the bottom figure we draw a line, so that the total will be under the column. When writing them we remember that the mathematical rule for placing the numbers is to align the right-hand-side digits (when there are whole numbers) and the decimal points (when there are decimals). For example: Right-hand-side-digits Decimal alignment alignment 4234 13.05 8238 2.51 646 539.652 5321 2431.0003

350 49.24 9989 The conventional method is to add the figures down the right-hand column, 4 plus 8 plus 6, and so on. You can do this if you wish in the new method, but it is not compulsory; you can begin working on any column. But for the sake of convenience, we will start on the righthand column. We add as we go down, but we “never count higher than 10”. That is, when the running total becomes greater than 10, we reduce it by 10 and go ahead with the reduced figure. As we do so, we make a small tick or checkmark beside the number that made our total higher than 10. For example: 4 8 4 plus 8,12: this is more than 10, so we subtract 10 from 12. Mark a tick and start adding again. 6 6 plus 2, 8 1 1 plus 8, 9 0 0 plus 9, 9 9 9 plus 9, 18: mark a tick and reduce 18 by 10, say 8. The final figure, 8, will be written under the column as the “running total”. Next we count the ticks that we have just made as we dropped 10’s. As we have 2 ticks, we write 2 under the column as the “tick figure”. The example now looks like this: 4234 8238 646 5321 350 9989 running total: 8 ticks: 2 If we repeat the same process for each of the columns we reach the result:

Quicker Maths

6 4234 8238 646 5321 350 9989 running total: 6558 ticks: 2222 Now we arrive at the final result by adding together the running total and the ticks in the way shown in the following diagram, running total: 0 6 5 5 8 0 ticks: 0 2 2 2 2 0 Total: 2 8 7 7 8 Save more time: We observe that the running total is added to the ticks below in the immediate right column. This addition of the ticks with immediate left column can be done in single step. That is, the number of ticks in the first column from right is added to the second column from right, the number of ticks in the 2nd column is added to the third column, and so on. The whole method can be understood in the following steps. 4 23 4 8 2 3 8 64 6 5321 350 9 9 8 9 Total: 8 [4 plus 8 is 12, mark a tick and add 2 to 6, which is 8; 8 plus 1 is 9; 9 plus 0 is 9; 9 plus 9 is 18, mark a tick and write down 8 in the first column of total-row.]

Step I.

Step II.

Total:

4234 8 2 3 8 6 4 6 5321 3 50 9 9 8 9 78

[3 plus 2 (number of ticks in first column) is 5; 5 plus 3 is 8; 8 plus 4 is 12, mark a tick and carry 2; 2 plus 2 is 4; 4 plus 5 is 9; 9 plus 8 is 17, mark a tick and write down 7 in 2nd column of total-row.] In a similar way we proceed for 3rd and 4th columns.

4 2 3 4 8 2 3 8 6 4 6 53 2 1 3 5 0 9 9 8 9 Total:

287 7 8

Note:We see that in the leftmost column we are left with 2 ticks. Write down the number of ticks in a column left to the leftmost column. Thus we get the answer a little earlier than the previous method. One more illustration : Q: 707.325 + 1923.82 + 58.009 + 564.943 + 65.6 = ? Solution: 707.325 1923.82 58.009 564.943 65.6 Total: 3319.697 You may raise a question:is it necessary to write the numbers in column-form? The answer is ‘no’. You may get the answer without doing so. Question written in a row-form causes a problem of alignment. If you get command over it, there is nothing better than this. For initial stage, we suggest you a method which would bring you out of the alignment problem. Step I. “Put zeros to the right of the last digit after decimal to make the no. of digits after decimal equal in each number.” For example, the above question may be written as 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 Step II. Start adding the last digit from right. Strike off the digit which as been dealt with. If you don’t cut, duplication may occur. During inning total, don’t exceed 10. That is, when we exceed 10, we mark a tick anywhere near about our calculation. Now, go ahead with the number exceeding 10. 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = __ _ _ . __7

5 plus 0 is 5; 5 plus 9 is 14, mark a tick in rough area and carry over 4; 4 plus 3 is 7; 7 plus 0 is 7, so write down 7. During this we strike off all the digits which are used. It saves us from confusion and duplica-tion. Step III. Add the number of ticks (in rough) with the digits in 2nd places, and erase that tick from rough. 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = _ _ _ . _97

1 (number of tick) plus 2 is 3; 3 plus 2 is 5; 5 plus 0

Addition is 5; 5 plus 4 is 9 and 9 plus 0 is 9; so write down 9 in its place. Step IV. 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = _ _ _ _ .697

3 plus 8 is 11; mark a tick in rough and carry over 1; 1 plus 0 is 1; 1 plus 9 is 10, mark another tick in rough and carry over zero; 0 plus 6 is 6, so put down 6 in its place. Step V. Last Step: Following the same way get the result: 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = 3319.697

Addition of numbers (without decimals) written in a row form Q. 53921 + 6308 + 86 + 7025 + 11132 = ? Soln: Step I: 53921 + 6308 + 86 + 7025 + 11132 = ___ _ 2 Step II: 53921 + 6308 + 86 + 7025 + 11132 = _ _ _ 72 Step III: 53921 + 6308 + 86 + 7025 + 11132 = _ _ 472 Step IV: 53921 + 6308 + 86 + 7025 + 11132 = _ 8472 Step V: 53921 + 6308 + 86 + 7025 + 11132 = 78472 Note: One should get good command over this method because it is very much useful and fast-calculating. If you don’t understand it, try again and again. Addition and subtraction in a single row Ex. 1: 412 – 83 + 70 = ? Step I: For units digit of our answer add and subtract the digits at units places according to the sign attached with the respective numbers. For example, in the above case the unit place of our temporary result is 2–3+0=–l So, write as: 412 – 83 + 70 = _ _ (–1) Similarly, the temporary value at tens place is 1 – 8 + 7 = 0. So, write as: 412 – 83 + 70 = _ (0) (–1) Similarly, the temporary value at hundreds place is 4. So, we write as: 412 – 83 + 70 = (4) (0) (–l) Step II: Now, the above temporary figures have to be changed into real value. To replace (–1) by a +ve digit we borrow from digits at tens or hundreds. As the digit at tens is zero, we will have to borrow from hundreds. We borrow 1 from 4 (at hundreds) which becomes 10 at tens leaving 3 at hundreds. Again we borrow 1 from tens which becomes 10 at units place, leaving 9 at tens. Thus, at units place 10 – 1 = 9. Thus our final result = 399. The above explanation can be represented as

7 (–1) (10) (–1) (10) (4) (0) (–1) (3) (9) (9) Note: The above explanation is easy to understand. And the method is more easy to perform. If you practise well, the two steps (I & II) can be performed simultaneously. The second step can be performed in another way like: (4) (0) (–l) = 400 – 1= 399 Ex. 2: 5124 – 829 + 731– 435 Soln: According to step I, the temporary figure is: (5) (– 4) (0) (– 9) Step II: Borrow 1 from 5. Thousands place becomes 5 – 1= 4.1 borrowed from thousands becomes 10 at hundreds. Now, 10 – 4 = 6 at hundreds place, but 1 is borrowed for tens. So digit at hundreds becomes 6 –1 = 5.1 borrowed from hundreds becomes 10 at tens place. Again we borrow 1 from tens for units place, after which the digit at tens place is 9. Now, 1 borrowed from tens becomes 10 at units place. Thus the result at units place is 10 – 9 = 1. Our required answer = 4591 Note: After step I we can perform like: 5 (– 4) (0) (– 9) = 5000 – 409 = 4591 But this method can’t be combined with step I to perform simultaneously. So, we should try to understand steps I & II well so that in future we can perform them simultaneously. Ex. 3: 73216 – 8396 + 3510 – 999 = ? Soln: Step I gives the result as: (7) (–2) (–5) (–16) (–9) Step II: Units digit = 10 – 9 = 1 [1 borrowed from (–16) results –16 –1 = –17] Tens digit = 20 –17 = 3 [2 borrowed from (–5) results –5 – 2 = –7] Hundreds digit = 10 – 7 = 3 [1 borrowed from –2 results –2 –1 = – 3] Thousands digit = 10 – 3 = 7 [1 borrowed from 7 results 7–1=6] So, the required value is 67331. The above calculations can also be started from the leftmost digit as done in last two examples. We have started from rightmost digit in this case. The result is the same in both cases. But for the combined operation of two steps you will have to start from rightmost digit (i.e. units digit). See Ex. 4. Note: Other method for step II: (–2) (–5) (–16) (–9) = (–2) (–6) (–6) (–9) = (–2669) Ans = 70000 – (2669) = 67331

Quicker Maths

8 Ex. 4: 89978 - 12345 - 36218 = ? Soln: Step I: (4) (1) (4) Step II: 4 1 4

(2) 1

(–5) 5

Single step solution: Now, you must learn to perform the two steps simultaneously. This is the simplest example to understand the combined method. At units place: 8 - 5 - 8 = (-5). To make it positive we have to borrow from tens. You should remember that we can’t borrow from -ve value i.e., from 12345. We will have to borrow from positive value i.e. from 89978. So, we borrowed 1 from 7 (tens digit of 89978): (-1) 8 9 9 7 8 - 12345 - 36218 = _ _ _ _ 5 Now digit at tens: (7 - 1 =) 6 - 4 - 1 = 1 Digit at hundreds: 9 - 3 - 2 = 4 Digit at thousands: 9 - 2 - 6 = 1 Digit at ten thousands: 8 - 1 - 3 = 4 the required value = 41415 Ex. 5: 28369 + 38962 - 9873 = ? Soln: Single step solution: Units digit = 9 + 2 – 3 = 8 Tens digit = 6 + 6 – 7 = 5 Hundreds digit = 3 + 9 – 8 = 4 Thousands digit = 8 + 8 - 9 = 7 Ten thousands digit = 2 + 3 = 5 required value = 57458 Ex. 6: Solve Ex. 2 by single-step method. Soln: 5124 – 829 + 731– 435 = Units digit: 4 – 9 + 1 – 5 = (–9). Borrow 1 from tens digit of the positive value. Suppose we borrowed from 3 of 731. Then –1 5124 – 829 + 731 – 435 = _ _ _ 1 Tens digit: 2 – 2 + 2 – 3 = (–1). Borrow 1 from hundreds digit of +ve value. Suppose we borrowed from 7 of 731. Then –1 –1 5124 – 829 + 731 – 435 = _ _ 9 1 Hundreds digit: 1 – 8 + 6 – 4 = (–5). Borrow 1 from thousands digit of +ve value. We have only one such digit, i.e. 5 of 5124. Then -1 -1 -1 5 1 2 4 – 829 + 731 - 435 = 4591 (Thousands digit remains as 5 - 1 = 4) Now you can perform the whole calculation in a single step without writing anything extra.

Ex. 7: Solve Ex. 3 in a single step without writing anything other than the answer. Try it yourself. Don’t move to next example until you can confidently solve such questions within seconds. Ex. 8: 10789 + 3946 – 2310 – 1223 = ? Soln: Whenever we get a value more than 10 after addition of all the units digits, we will put the units digit of the result and carry over the tens digit. We add the tens digit to +ve value, not to the –ve value. Similar method should be adopted for all digits. +1 +l +1 1 0 7 8 9 + 3946 – 2310 – 1223 = 11202 Note: 1. We put +1 over the digits of +ve value 10789. It can also be put over the digits of 3946. But it can’t be put over 2310 and 1223. 2. In the exam when you are free to use your pen on question paper you can alter the digit with your pen instead of writing +1, +2, –1, –2 .... over the digits. Hence, instead of writing 8, you should write 9 over 8 with your pen. 1

Similarly, write 8 in place of 7 . Ex. 9: 765.819 – 89.003 + 12.038 – 86.89 = ? Soln: First, equate the number of digits after decimals by putting zeros at the end. So, ? = 765.819 – 89.003 + 12.038 – 86.890 Now, apply the same method as done in Ex. 4, 5, 6, 7 & 8. -1 -1 -l -l +1 7 6 5. 8 1 9 - 89.003 + 12.038 – 86.890 = 601.964 Ex 10: 5430 - 4321 + 3216 - 6210 = ? Soln: The above case is different. The final answer comes negative. But as we don't know this in the beginning, we perform the same steps as done earlier. Step I: (-2) (1) (1) (5) Step II: The leftmost digit is negative. It can't be made positive as there is no digit at the left which can lend. So, our answer is

2000 115 1885 Note: The second step should be done mentally keeping in mind that except the leftmost digit all the other digits are positive. So, the final answer will be -ve but not (–)2115. It should be -2000 +115= –1885.

Addition

9

Ex 11: 2695 - 4327 + 3214 - 7350 = ? Soln: Step I: (-6) (2) (3) (2) So, required answer =

6000 232 5768

Method of checking the calculation: Digit sum Method This method is also called the nines-remainder method. The concept of digit-sum consists of this : I. We get the digit-sum of a number by “adding across” the number. For instance, the digit-sum of 13022 is 1 plus 3 plus 0 plus 2 plus 2 is 8. II. We always reduce the digit-sum to a single figure if it is not already a single figure. For instance, the digit-sum of 5264 is 5 plus 2 plus 6 plus 4 is 8 (17, or 1 plus 7 is 8). III. In “adding across” a number, we may drop out 9’s. Thus, if we happen to notice two digits that add up to 9, such as 2 and 7, we ignore both of them; so the digit-sum of 990919 is 1 at a glance. (If we add up 9’s we get the same result.) IV. Because “nines don’t count” in this process, as we saw in III, a digit-sum of 9 is the same as a digit-sum of zero. The digit-sum of 441, for example, is zero. Quick Addition of Digit-sum: When we are “adding across” a number, as soon as our running total reaches two digits we add these two together, and go ahead with a single digit as our new running total. For example: To get the digit-sum of 886542932851 we do like: 8 plus 8 is 16, a two – figure number. We reduce this 16 to a single figure: 1 plus 6 is 7. We go ahead with this 7; 7 plus 6 is 4 (13, or 1+3=4), 4 plus 5 is 9, forget it. 4 plus 2 is 6. Forget 9 .... Proceeding this way we get the digit-sum equal to 7. For decimals we work exactly the same way. But we don’t pay any attention to the decimal point. The digitsum of 6.256, for example, is 1. Note: It is not necessary in a practical sense to understand why the method works, but you will see how interesting this is. The basic fact is that the reduced digit-sum is the same as the remainder when the number is divided by 9. For example: Digit-sum of 523 is 1. And also when 523 is divided by 9, we get the remainder 1.

Checking of Calculation Basic rule: Whatever we do to the numbers, we also do to their digit-sum; then the result that we get

from the digit-sum of the numbers must be equal to the digit-sum of the answer. For example: The number: 23 + 49 + 15 + 30 = 117 The digit-sum: 5 + 4 + 6 + 3 = 0 Which reduces to : 0 =0 This rule is also applicable to subtraction, multiplication and upto some extent to division also. These will be discussed in the coming chapters. We should take another example of addition. 1.5 + 32.5 + 23.9 = 57.9 digit-sum: 6 + 1 + 5 =3 or, 3 =3 Thus, if we get LHS = RHS we may conclude that our calculation is correct. Sample Question: Check for all the calculations done in this chapter. Note: Suppose two students are given to solve the following question: 1.5 + 32.5 + 23.9 = ? One of them gets the solution as 57.9. Another student gets the answer 48.9. If they check their calculation by this method, both of them get it to be correct. Thus this method is not always fruitful. If our luck is against us, we may approve our wrong answer also. Addition of mixed numbers 1 4 1 Q. 3 4 9 ? 2 5 3 Solution:A conventional method for solving this question is by converting each of the numbers into pure fractional numbers first and then taking the LCM of denominators. To save time, we should add the whole numbers and the fractional values separately. Like here,

1 4 1 1 4 1 3 4 9 (3 4 9 ) 2 5 3 2 5 3 = 16

5 24 10 19 16 1 30 30

= (16 1)

19 19 19 17 17 30 30 30

2 1 3 1 Q. 5 4 2 1 3 6 4 4

Soln:

2 1 3 1 (5 4 2 1) 3 6 4 4

12 8 2 9 3 2 = 2 = 2 + 1 =3 12 12

10

Quicker Maths

Chapter 3

Multiplication Special Cases We suggest you to remember the tables up to 30 because it saves some valuable time during calculation. Multiplication should be well commanded, because it is needed in almost every question of our concern. Let us look at the case of multiplication by a number more than 10.

Multiplication by 11 Step I: The last digit of the multiplicand (number multiplied) is put down as the right-hand figure of the answer. Step II: Each successive digit of the multiplicand is added to its neighbour at the right.

The answer is 64812. As you see, each figure of the long number is used twice. It is first used as a “number”, and then, at the next step, it is used as a neighbour. Looking carefully, we can use just one rule instead of three rules. And this only rule can be called as “add the right neighbour” rule. We must first write a zero in front of the given number, or at least imagine a zero there. Then we apply the idea of adding the neighbour to every figure of the given number in turn: 05892 11 As there is no neighbour on the right, so 2 we add nothing.

05892 11 -------- As we did earlier 4812

Ex. 1. Solve 5892 × 11 = ? Soln: Step I: Put down the last figure of 5892 as the right-hand figure of the answer: 5892 11 2 Step II: Each successive figure of 5892 is added to its right-hand neighbour. 9 plus 2 is 11, put 1 below the line and carry over 1. 8 plus 9 plus 1 is 18, put 8 below the line and carry over 1. 5 plus 8 plus 1 is 14, put 4 below the line and carry over 1.

5892 11 (9 + 2 = 11, put 1 below the line and 12 carry over 1)

5892 11 (8 + 9 + 1 = 18, put 8 below the line and 812 carry over 1)

5892 11 (5 + 8 + 1 = 14, put 4 below the line and 4812 carry over 1) Step III: The first figure of 5892, 5 plus 1, becomes the left-hand figure of the answer: 5892 11 64812

05892 11 -------- Zero plus 5 plus carried-over 1 64812 is 6 This example shows why we need the zero in front of the multiplicand. It is to remind us not to stop too soon. Without the zero in front, we might have neglected the last 6, and we might then have thought that the answer was only 4812. The answer is longer than the given number by one digit, and the zero in front takes care of that. Sample Problems: Solve the following: 1) 111111 × 11 2) 23145 × 11 3) 89067 × 11 4) 5776800 × 11 5) 1122332608 × 11 Ans: 1) 1222221 2) 254595 3) 979737 4) 63544800 5) 12345658688

Multiplication by 12 To multiply any number by 12, we “Double each digit in turn and add its neighbour.” This is the same as multiplying by 11 except that now we double the “number” before we add its “neighbour.”

Quicker Maths

12 For example: Ex 1: Solve:5324 × 12 Soln: 05324 12 Step I. (double the right-hand figure 8 and add zero, as there is no neighbour)

05324 12 88 05324 12 Step III. 888 05324 12 Step IV. 3888 Step II.

Last Step.

05324 12 63888

(double the 2 and add 4) (double the 3 and add 2) (double the 5 and add 3 (=13), put 3 below the line and carry over 1) (zero doubled is zero, plus 5

plus carried-over 1) The answer is 63,888. If you go through it yourself you will find that the calculation goes very fast and is very easy. Practice Question: Solve the following: 1) 35609 × 12 2) 11123009 × 12 3) 456789 × 12 4) 22200007 × 12 5) 444890711 × 12 Ans: 1) 427308 2) 133476108 3) 5481468 4) 266400084 5) 5338688532

Multiplication by 13 To multiply any number by 13, we “Treble each digit in turn and add its right neighbour.” This is the same as multiplying by 12 except that now we “treble” the “number” before we add its “neighbour”. If we want to multiply 9483 by 13, we proceed like this: 09483 13 Step I. (treble the right-hand figure and 9 write it down as there is no neighbour on the right) 09483 13 Step II. (8 × 3 + 3 = 27, write down 7 79 and carry over 2) 09483 13 Step III. (4×3+8+2 = 22, write down 2 279 and carry over 2)

Step IV.

09483 13 (9×3 + 4 + 2 = 33, write down 3 3279

and carry over 3) 09483 13 Last Step. 123279 (0 × 3 + 9 + 3 = 12, write it down) The answer is 1,23,279. In a similar way, we can define rules for multiplication by 14, 15,....But, during these multiplications we will have to get four or five times of a digit, which is sometimes not so easy to carry over. We have an easier method of multiplication for those large values. Can you get similar methods for multiplication by 21 and 31? It is not very tough to define the rules. Try it.

Multiplication by 9 Step I: Subtract the right-hand figure of the long number from 10. This gives the right-hand figure of the answer. Step II: Taking the next digit from right, subtract it from 9 and add the neighbour on its right. Step III: At the last step, when you are under the zero in front of the long number, subtract one from the neighbour and use that as the left-hand figure of the answer. Ex. 1: 8576 × 9 = ? 08576 9 77184 Step I. Subtract the 6 of 8576 from 10, and we have 4 of the answer. Step II. Subtract the 7 from 9 (we have 2) and add the neighbour 6; the result is 8. Step III. (9 – 5) + 7 = 11; put 1 under the line and carry over 1. Step IV. (9 – 8) + 5 + l (carried over) = 7, put it down. Step V (Last step). We are under the left-hand zero, so we reduce the left-hand figure of 8576 by one, and 7 is the left-hand figure of the answer. Thus answer is 77184. Here are a few questions for you. 1) 34 × 9 = ? 2) 569 × 9 = ? 3) 1328 × 9 = ? 4) 56493 × 9 = ? 5) 89273258 × 9 = ? Answers: 1) 306 2) 5121 3)11952 4) 508437 5) 803459322 We don’t suggest you to give much emphasis on this rule. Because it is not very much easy to use. Sometimes it proves very lengthy also.

Soln:

Multiplication

13

Another method: Step I: Put a zero at the right end of the number; i.e., write 85760 for 8576. Step II: Subtract the original number from that number. Like 85760 – 8576 = 77184

Suppose you are given a large number like 125690258. And someone asks you to multiply that number by 25, What will you do? Probably you will do nothing but go for simple multiplication. Now, we suggest you to multiply that number by 100 and then divide by 4. To do so remember the two steps: Step I: Put two zeroes at the right of the number (as it has to be multiplied by 100). Step II: Divide it by 4. So, your answer is 12569025800 ÷ 4 = 3142256450. Is it easier than your method?

General Rule for Multiplication Having dealt in fairly sufficient detail with the application of special cases of multiplication, we now proceed to deal with the “General Formula” applicable to all cases of multiplication. It is sometimes not very convenient to keep all the above cases and their steps in mind, so all of us should be very much familiar with “General Formula” of multiplication.

Multiplication by a two-digit number Ex. 1: Solve (1) 12 × 13 = ? (2) 17 × 18 = ? (3) 87 × 92 = ? Soln: (1) 12 × 13 = ? Step I. Multiply the right-hand digits of multiplicand and multiplier (unit-digit of multiplicand with unit-digit of the multiplier). 1 2 1

3 6 (2 × 3) Step II. Now, do cross-multiplication, i.e., multiply 3 by 1 and 1 by 2. Add the two products and write down to the left of 6. 2

1

3 56

1

2

1

3

1 5 6

Multiplication by 25

1

Step III. In the last step we multiply the left-hand figures of both multiplicand and multiplier.

( 1 × 1)

(2) 17 × 18 = ? Step I. 1 1

7 8 6 (7 × 8 = 56, write down 6 carry over 5) Step II. 1 7 1 8 0 6 (1 × 8 + 7×1+5 = 20, write down 0 and carry over 2) Step III. 1 7 1 8 3 0 6 (1 × 1 + 2 = 3, write it down) (3) 87 × 92 = ? Step I. 8 7 9 2 4 (7 × 2 = 14, write down 4 and carry over 1) Step II. 8 7 9 2 04 (8 × 2 + 9 × 7 + 1 = 80, write down 0 and carry over 8) Step III. 8 7 9 2 8004 (8 × 9 + 8 = 80) Practice questions: 1) 57 × 43 2) 51 × 42 3) 38 × 43 4) 56 × 92 5) 81 × 19 6) 23 × 99 7) 29 × 69 8) 62 × 71 9) 17 × 37 10) 97 × 89 Answers: l) 2451 2) 2142 3) 1634 4) 5152 5) 1539 6) 2277 7) 2001 8) 4402 9) 629 10) 8633 Ex 2. Solve (1) 325 × 17 = ? (2) 4359 × 23 = ? Soln: (1): Step I. 3 2 5 1 7

(3 × 1 + 2 × 1)

5 (5 × 7 = 35, put down 5 and carry over 3)

Quicker Maths

14 We can write all the steps together:

3 2 5

Step II.

1 7 2 5 (2×7+5×1+3 = 22, put down 2 and carry over 2)

3 2 5

Step III.

1 7

5 2 5 (3 × 7 + 2 × 1 + 2 = 25, put down 5 and carry over 2) Note: Repeat the cross-multiplication until all the consecutive pairs of digits exhaust. In step II, we cross-multiplied 25 and 17 and in step III, we crossmultiplied 32 and 17. Step IV. 3

2

5

1 7 5 5 2 5 (3 × 1 + 2 = 5, Put it down) (2)

Step I. 4 3 5 9 2 3 7 (9 × 3 = 27, put down 7, carry over 2)

4 3 5 9 2 3 4 2 / 4 3 2 3/ 3 3 5 2 / 5 3 9 2 / 9 3 = 10 20 22 35 27 = 100257 Or, we can write the answer directly without writing the intermediate steps. The only thing we should keep in mind is the “carrying numbers”.

4359 23 10 2 0 2 2 35 2 7

4359 23 or, 100257

Note: You should try for this direct calculation. It saves a lot of time. It is a very systematic calculation and is very easy to remember. Watch the above steps again and again until you get that systematic pattern of crossmultiplication.

Multiplication by a three-digit number Ex: 1. Solve (1) 321 × 132 = ? (2) 4562 × 345 = ? (3) 69712 × 641 = ? Soln: (1) Step I.

3 2 1

Step II. 4 3 5 9 1 3 2

2 3

5 7

(5 × 3 + 9 × 2 + 2 = 35, put down 5 and carry over 3)

2 (1 × 2 = 2)

Step II.

3 2 1

Step III. 4 3 5 9 2 3

2 5 7

1 3 2

(3×3+5×2+3 = 22, put down 2, carry over 2)

7 2 (2 × 2 + 3 × 1= 7)

Step III.

Step IV. 4 3 5 9

3 2 1

2 3

0 2 5 7

1 3 2

(4×3+3×2+2 = 20, put down 0, carry over 2)

3 7 2

Step V. 4 3 5 9 Step IV.

2 3 100257

(4 × 2 + 2 = 10, put it down)

3 2 1 1 3 2 2 3 7 2

(2 × 3 + 3 × 2 + 1 × 1 = 13, write down 3 and carry over 1)

Multiplication Step V.

15 V. 6 9 7 1 2 VI. 6 9 7 1 2 VII. 6 9 7 1 2

3 2 1

641 641 641 For each of the groups of figures, you have to crossmultiply.

1 3 2 4 2 3 7 2 (1 × 3 + 1 = 4)

or,

3 21 13 2 1 3 / 3 3 1 2 / 2 3 3 2 11/ 2 2 3 1/1 2 = 4 12 = 42372 Soln: (2):

7

13

2

5

8

3

9

1

(5 × 6 = 30, write down 0 and carry over 3)

Step II.

4 6 3 2 / 5 6 4 2 / 2 5 3

2) 646329 × 8124

4 3 2 5

0

345 4 3/ 4 4 3 5 / 5 4 4 5 3 6 / 5 5

6

Example: Solve 1) 4325 × 3216 Soln: 1) 4325 × 3216 = ? Step I.

3 2 1 6

4562

= 15 37 = 1573890 Soln: (3):

Multiplication by a four-digit number

4 3 2 5

0 3 2 1 6 0 0 (2×6+1×5+3 = 20, write down 0 and carry over 2)

6 9 712 6 41 6 6 / 4 6 6 9 / 1 6 4 9 6 7 /1 9 4 7

Step III.

6 1/ 1 7 4 1 6 2 /11 4 2 /1 2 = 44 86 88 45 23 9 2 = 44685392 Note: Did you get the clear concept of crossmultiplication and carrying-cross-multiplication? Did you mark how the digits in cross-multiplication increase, remain constant, and then decrease? Take a sharp look at question (3). In the first row of the answer, if you move from right to left, you will see that there is only one multiplication (1×2) in the first part. In the second part there are two (1×1 and 4×2), in the 3rd part three (1×7, 4×1 and 6×2), in the 4th part three (1×9, 4×7 and 6 ×1), in the 5th part again three (1×6, 4×9 and 6×7), in the 6th part two (4 × 6 and 6 × 9) and in the last part only one (6 × 6) multiplication. The participation of digits in crossmultiplication can be shown by the following diagrams. I. 6 9 7 1 2 II. 6 9 7 1 2 III. 6 9 7 1 2 IV. 6 9 7 1 2 641

641

641

641

4 3 2 5 3 2 1 6 2 0 0

Step IV.

4 3 2 5

3 2 1 6 9 2 0 0 Step V.

4 3 2 5 3 2 1 6 0 9 2 0 0

Step VI. 4 3 2 5

3 2 1 6 9 0 9 2 0 0

(4× 2 + 3 × 3 + 2 = 19, write down 9 and carry over 1)

Quicker Maths

16 Step VII. 4 3 2 5

3 2 1 6 1390 9 2 0 0

(4 × 3 + 1 = 13, write it down)

Ans: 13909200 Soln: 1) 646329 × 8124 = ? Try this question yourself and match your steps with the given diagrammatic presentation of participating digits. Step I. 6 4 6 3 2 9 8124

Step II. 6 4 6 3 2 9 8124

Step III. 6 4 6 3 2 9 8124

Step IV. 6 4 6 3 2 9 8124

Step V. 6 4 6 3 2 9 8124

Step VI. 6 4 6 3 2 9 8124

Step VII. 6 4 6 3 2 9 8124

Step VIII. 6 4 6 3 2 9 8124

Step IX. 6 4 6 3 2 9 8124

Practice Problems: Solve the following: 1) 234 × 456 2) 336 × 678 3) 872 × 431 4) 2345 × 67 5) 345672 × 456 6) 569 × 952 7) 458 × 908 8) 541 × 342 9) 666 × 444 10) 8103 × 450 11) 56321 × 672 12) 1278 × 569 13) 5745 × 562 14) 4465 × 887 15) 8862 × 341 Answers: 1) 106704 2) 227808 3) 375832 4) 157115 5) 157626432 6) 541688 7) 415864 8) 185022 9) 295704 10) 3646350 11) 37847712 12) 727182 13) 3228690 14) 3960455 15) 3021942

Checking of Multiplication Ex 1:

15 × 13 = 195 digit-sum: 6 × 4 = 6 or, 24 = 6 or, 6 = 6. Thus, our calculation is correct. Ex. 2: 69712 × 641 = 44685392 digit-sum: 7 × 2 = 5 or, 14 = 5 or, 5 =5 Therefore, our calculation is correct. Ex. 3: 321 × 132 = 42372 digit-sum: 6 × 6 = 9 or, 36 = 9 or, 9 =9 Thus, our calculation is correct. But if someone gets the answer 43227, and tries to check his calculation with the help of digit-sum rule, see what happens: 321 × 132 = 43227 digit sum: 6×6 =9 or, 36 = 9 or, 9 =9 This shows that our answer is correct, but it is not true. Thus we see that if out luck is very bad, we can approve a wrong answer.

Chapter 4

Division We now go on to the Quicker Maths of at-sight division which is based on long-established Vedic process of mathematical calculations. It is capable of immediate application to all cases and it can be described as the “crowning gem of all” for the universality of its applications. To understand the at-sight mental one-line method of division, we should take an example and its explanation.

DIVISION BY A 2-DIGIT NUMBER Ex 1. Divide 38982 by 73. Soln: Step I. Out of the divisor 73, we put down only the first digit, ie, 7 in the divisor-column and put the other digit, ie, 3 “on top of the flag”, as shown in the chart below. 7 3 38 9 8 2 The entire division will be by 7. Step II. As one digit (3) has been put on top, we allot one place at the right end of the dividend to the remainder position of the answer and mark it off from the digits by a vertical line.

7

3

38 9 8 2

Step III. As the first digit from the left of dividend (3) is less than 7, we take 38 as our first dividend. When we divide 38 by 7, we get 5 as the quotient and 3 as the remainder. We put 5 down as the first quotientdigit and just prefix the remainder 3 before the 9 of the dividend.

7

3

38 3 9 8 2 5

Step IV. Now our dividend is 39. From this we, however, deduct the product of the indexed 3 and the first quotient-digit (5), ie, 3×5 = 15. The remainder 24 is our actual net-dividend. It is then divided by 7 and gives us 3 as the second quotient-digit and 3 as the remainder, to be placed in their respective places as was done in third step.

7

3

38 39 38 2 5 3

Step V. Now our dividend is 38. From this we subtract the product of the index (3) and the 2nd quotientdigit (3), ie, 3 × 3 = 9. The remainder 29 is our next actual dividend and divide that by 7. We get 4 as the quotient and 1 as the remainder. We put them at their respective places.

7

3

38 3 9 38 12 5 3 4

Step VI. Our next dividend is 12 from which, as before, we deduct 3 × 4 ie, 12 and obtain 0 an the remainder

7

3

38 39 38 12 5 3 4 0

Thus we say : Quotient (Q) is 534 and Remainder (R) is 0. And thus finishes the whole procedure; and all of it is one-line mental arithmetic in which all the actual division is done by the single-digit divisor 7. The procedure is very simple and needs no further exposition and explanation. A few more illustrations with running comments will be found useful and helpful and are therefore given below : Ex 2: Divide 16384 by 128 (As 12 is a small number to handle with, we can treat 128 as a two-digit number). Soln:

12

8

16 43 118 64 1 2 8 0

Step I. We divide 16 by 12. Q = 1 & R = 4. Step II. 43 – 8 × 1 = 35 is our next devidend. Dividing it by 12, Q = 2, R = 11. Step III. 118 – 8 × 2 = 102 is our next dividend. Dividing it by 12, Q = 8, R = 6 Step IV. 64 – 8 × 8 = 0

Quicker Maths

18 Then our final quotient = 128 & remainder = 0 Ex 3: Divide 601325 by 76.

7

Soln :

6

60 111 63 22 7 9 1 2

25 13 (=25 - 6×2)

Step I. Here, in the first division by 7, if we put 8 down as the first quotient-digit, the remainder then left will be too small for the subtraction expected at the next step. We get -ve dividend in next step which is absurd. So, we take 7 as the quotient-digit and prefix the remainder 11 to the next dividend-digit. All the other steps are similar to the previously mentioned steps in Ex 1 & 2. Our final quotient is 7912 and remainder is 13. If we want the values in decimal, we go on dividing as per rule instead of writing down the remainder. Such as;

76

60 1 3 2 7

9 1 2

5 60 50 1 7 1

Ans = 7912.171 Note: The vertical line separating the remainder from the quotient part may be the demarcating point for decimal. Ex 4: Divide 7777777 by 38 Soln:

38 7 2

17 17 57 0 4 6

77 87 77 7 8 13 (=77 - 8×8)

Q = 204678, R = 13 You must go through all the steps of the above solution. Try to solve it. Did you find some difference? Ex 5: Divide 8997654 by 99. Try it step by step. Ex 6:(i) Divide 710.014 by 39 (to 4 places of decimals) (ii) 718.589 ÷ 23 = ? (iii) 718.589 ÷ 96 = ? Soln. (i) Since there is one flag-digit the vertical line is drawn such that one digit before the decimal comes under remainder portion.

3

9

7 41 1 8

80. 20 2 0

21 64 70 5 4

For the last section, we had 64 - 45 = 19 as our dividend, divided by 3 we choose 4 as our suitable quotient. If we take 5 as a quotient it leaves 4 as remainder (19 – 15). Now the next dividend will be 40 – 9 × 5 = –5, which is not acceptable.

The vertical line separating the remainder from quotient part may be demarcation point for decimal. Therefore, ans = 18.2054 (ii)

2

3

71 1 3 1

08. 15 2 4

18 09 3 0

8

3

Ans = 31.2430 (iii)

7

4

5

2

Ans = 7.4853 Ex 7 : Divide 7.3 by 53 Soln.

5 3 7. 23 50 6 0 40 40 1 3 7 7 3 5

Ans = 0.137735

DIVISION BY A 3-DIGIT NUMBER Ex 8 : Divide 7031985 by 823. Soln: Step I. Here the divisor is of 3 digits. All the difference which we make is to put the last two digits(23) of divisor on top. As there are two flag-digits (23), we will separate two digits (85) for remainder.

8

23

70 3 1 9 85

Step II. We divide 70 by 8 and put down 8 and 6 in their proper places.

8

23

70 63 1 9 85 8 Step III. Now, our gross dividend is 63. From that we subtract 16, the product of the tens of the flagdigits, ie 2, and the first quotient-digit, ie 8, and get the remainder 63 – 16 = 47 as the actual dividend. And, dividing it by 8, we have 5 and 7 as Q & R respectively and put them at their proper places.

8 23 70 63 71 9 85 8 5 Step IV. Now our gross dividend is 71, and we deduct the cross-products of two flag-digits 23 and the two quotient digits (8 & 5) ie 2 × 5 + 3 × 8 = 10 + 24 = 34; and our remainder is 71 – 34 = 37. We then continue to divide 37 by 8 . We get Q = 4 & R =5

Division

19

8 23 70 63 71 59 85 8 5 4 Step V. Now our gross dividend is 59. And actual dividend is equal to 59 minus cross-product of 23 and 54, ie, 59 – (2 × 4 + 3 × 5) = 59 – 23 = 36. Dividing 36 by 8, our Q = 4 and R = 4.

8 23 70 63 71 59 485 8 5 4 4 Step VI. Actual dividend = 48 – (3 × 4 + 2 × 4) = 48 – 20 = 28. Dividing it by 8, our Q = 3 & R = 4

8 23 70 63 71 59 48 45 8 5 4 4 3 Step VII. Actual dividend = 45 – (3 × 4 + 2 × 3) = 45 – 18 = 27. Dividing 27 by 8, we have Q = 3 & R = 3.

8 23 70 63 71 59 48 45 3 8 5 4 4 3 3 The vertical line separating the remainder from the quotient part may be a demarcation point for decimal. Ans = 8544.33 Our answer can be 8544.33, but if we want the quotient and remainder, the procedure is somewhat different. In that case, we do not need the last two steps, ie, the calculation upto the stage

8 23 70 63 71 59 48 5 8 5 4 4 is sufficient to answer the question. Quotient = 8544; Remainder = 485 – 10 × (Cross multiplication of 23 and 44)* – unit digit of flagged number × unit digit of quotient. = 485 –10 (4 × 2 + 4 × 3) – 3×4 = 485 – 200 –12 = 273 * Cross-multiplication of the two flag-digits and last two digits of quotient. Ex 9: Divide 1064321 by 743 ( to 4 places of decimals). Also find the remainder. Soln:

7 43 10 36 44 43 52 1 1 4 3 2

Q = 1432, Remainder = 521 – 10 (Cross-multiplication of 43 & 32) – 3×2 = 521 – 10 × 17 – 6 = 345 For decimals :

7 43 10 3 6 44 43 1 4 3 2

52 71 70 60 50 4 6 4 3

Ans = 1432.4643 Ex 10: Divide 888 by 672 ( to 4 places of decimals).

6

72

8 2 8 38 30 40 5 0 1 3 2 1 4 2

Ans = 1.3214 Note : Vertical line separating the remainder from the quotient part is the demarcation point for decimal. Can you find the quotient and remainder? Try it. Ex 11: Divide 4213 by 1234 to 4 places of decimals. Also find quotient and remainder. Soln: Although 1234 is a four-digit number, we can treat it as a 3-digit number because 12 is small enough to handle with.

12

34

42 61 3 3 Q=3, R = 613-10 (cross multiplication of 03 and 34) – 4 × 3 = 613 – 90 – 12 = 511

12

34

42 3

6 1 43 70 30 2 0 4 1 4 1 0

Ans = 3.4141 Note:Division by 4-digit or 5-digit number is not of much use. So these are not being discussed here. Now you must have seen all the possible cases which you may come across in mathematical division. Don’t escape any of the examples discussed above. Having a broad idea of at-sight mathematical division, you should solve as many questions yourself as you can.

Checking of Calculation Rule of digit-sum fails in some cases here. Ex 1: We will check the calculations of the examples one by one. 38982 ÷ 73 = 534 (Since remainder is zero) digit sum: 3 ÷ 1 = 3 or, 3 = 3 Therefore, our calculation is correct.

Quicker Maths

20 Ex 2: Check yourself. Ex 3: 601325 ÷ 76 gives quotient = 7912 and remainder = 13 We may write the above calculation as 7912 × 76 + 13 = 601325 Now, check the correctness. digit sum : 1 × 4 + 4 = 8 or 4 + 4 = 8 or, 8 = 8 Therefore, our calculation is correct. Ex 4: Check yourself. Ex 5: Check yourself. Ex 6: (i) 710.014 ÷ 39 = 18.2054... We can’t apply the digit-sum method to check our calculation because our calculation is not complete. (ii) 718.589 ÷ 23 = 31.2430 This calculation is complete because in the end we get 0 as quotient and remainder. We can apply the digit-sum rule in this case. digit-sum = 2 ÷ 5 = 4 or, 0.4 = 4 or, 4 = 4 Our calculation is correct. Note : During digit-sum (forget-nine method) we don’t take decimals into account. Ex 7: Tell whether digit-sum rule is applicable to this calculation or not. Note : Thus, we see the limitation of digit-sum rule.

Practice Problems Q.1. Divide ‘x’ by ‘y’ where a) x=135921 & y = 89

b) x=64932 & y = 99 c) x=8899 & y = 101 d) x=9995 & y =122 e) x=13596289 & y = 76 f) x= 89325 & y = 132 g) x=89 & y = 71 h) x=96 & y = 95 i) x=53 & y = 83 j) x=93 & y = 109 Q.2. Divide ‘x’ by ‘y’ where a) x= 359281 & y = 567 b) x= 8932 & y = 981 c) x= 99899 & y = 789 d) x= 1053 & y = 989 e) x= 738 & y = 895 f) x= 13569 & y = 1051 g) x= 69325 & y = 1163 h) x= 935 & y = 1259 i) x= 100002 & y = 777 j) x= 12345 & y = 567 Answers: 1. a) 1527.2022 c) 88.10891 e) 178898.539 g) 1.2535 i) 0.6385 2. a) 633.6525 c) 126.6147 e) 0.8245 g) 59.6087 i) 128.7027

b) 655.87878 d) 81.926229 f) 676.7045 h) 1.0105 j) 0.8532 b) 9.1049 d) 1.0647 f) 12.9105 h) 0.7426 j) 21.7724

Chapter 5

Divisibility We now take up the interesting question as to how one can determine whether a certain given number, however large it may be, is divisible by a certain given divisor. There is no defined general rule for checking the divisibility. For different divisors, the rules differ at large. We will discuss the rule for divisors from 2 to 19.

Divisibility by 2 Rule: Any number, the last digit of which is either even or zero, is divisible by 2. For example: 12, 86, 130, 568926 and 5983450 are divisible by 2 but 13, 133 and 596351 are not divisible by 2.

Divisibility by 3 Rule: If the sum of the digits of a number is divisible by 3, the number is also divisible by 3. For example: 1) 123: 1 + 2 + 3 = 6 is divisible by 3; hence 123 is also divisible by 3. 2) 5673: 5 + 6 + 7 + 3 = 21; therefore divisible by 3. 3) 89612: 8 + 9 + 6 + 1 + 2 = 26 = 2 + 6 = 8 is not divisible by 3. Therefore, the number is not divisible by 3.

Divisibility by 4 Rule: If the last two digits of a number is divisible by 4, the number is divisible by 4. The number having two or more zeros at the end is also divisible by 4. For example: 1) 526428: 28 is divisible by 4. Therefore, the number is divisible by 4. 2) 5300: There are two zeros at the end, so it is divisible by 4. 3) 134000: As there are more than two zeros, the number is divisible by 4. 4) 134522: As the last two-digit number (22) is not divisible by 4, the number is not divisible by 4. Note: The same rule is applicable to check the divisibility by 25. That is, a number is divisible by 25 if its last two digits are either zeros or divisible by 25.

Divisibility by 5 Rule: If a number ends in 5 or 0, the number is divisible by 5. For example: 1) 1345: As its last digit is 5, it is divisible by 5. 2) 1340: As its last digit is 0, it is divisible by 5. 3) 1343: As its last digit is neither 5 nor 0, it is not divisible by 5.

Divisibility by 6 Rule: If a number is divisible by both 3 and 2, the number is also divisible by 6. So, for a number to be divisible by 6, 1) the number should end with an even digit or 0 and 2) the sum of its digits should be divisible by 3. For example: 1) 63924 : The first condition is fulfilled as the last digit (4) is an even number and also (6+3+9+2+4=)24 is divisible by 3; therefore, the number is divisible by 6. 2) 154 : The first condition is fulfilled but not the second; therefore, the number is not divisible by 6. 3) 261 : The first condition is not fulfilled; therefore, we need not to check for the second condition.

Special Cases The rules for divisibility by 7, 13, 17, 19 ... are very much unique and are found very rarely. Before going on for the rule, we should know some terms like “one-more” osculator and negative osculator. “One-more” osculator means the number needs one more to be a multiple of 10. For example: osculator for 19 needs 1 to become 20 (=2 × 10), thus osculator for 19 is 2 (taken from 2 × 10 = 20). Similarly osculator for 49 is 5 (taken from 5 × 10 = 50). Negative osculator means the number should be reduced by one to be a multiple of 10. For example:

Quicker Maths

22 Negative osculator for 21 is 2 (taken from 2 × 10 = 20). Similarly, negative osculator for 51 is 5 (taken from

5 × 10 = 50). Note: (1) What is the osculator for 7? Now, we look for that multiple of 7 which is either less or more by 1 than a multiple of 10. For example 7 × 3 = 21, as 21 is one more than 2 × 10; our negative osculator is 2 for 7. And 7 × 7 = 49 or 49 is one less than 5 × 10; our ‘one-more’ osculator is 5 for 7. Similarly, osculators for 13, 17 and 19 are: For 13: 13 × 3 = 39, “one more” osculator is 4 (from

4 × 10) For 17: 17 × 3 = 51, negative osculator is 5 (from

5 × 10) For 19: 19 × 1. “one-more” osculator is 2 (from 2 × 10) (2) Can you define osculators for 29, 39, 21, 31, 27 and 23. (3) Can you get any osculator for an even number or a number ending with ‘5’? (No. But why?)

Divisibility by 7 First of all we recall the osculator for 7. Once again, for your convenience, as 7 × 3 = 21 (one more than 2 × 10), our negative osculator is 2. This oscuator ‘2’ is our key-digit. This and only this digit is used to check the divisibility of any number by 7. See how it works: Ex 1: Is 112 divisible by 7? Soln: Step 1. 11 2: 11 – 2 × 2 = 7 As 7 is divisible by 7, the number 112 is also divisible by 7. Ex 2: Is 2961 divisible by 7? Soln: Step I: 296 1: 296 – 1 × 2 = 294 Step II: 29 4: 29 – 4 × 2 = 21 As 21 is divisible by 7, the number is also divisible by 7. Ex 3: Is 55277838 is divisible by 7? Soln: 5527783 8: 5527783 – 8 × 2 = 5527767 552776 7: 552776 – 7 × 2 = 552762 55276 2: 55276 – 2 × 2 = 55272 5527 2: 5527 – 2 × 2 = 5523 552 3: 552 – 3 × 2 = 546 54 6: 54 – 6 × 2 = 42

As 42 is divisible by 7, the number is also divisible by 7. Note: (1) In all the examples, each of the numbers obtained after the equal sign (=) is also divisible by 7. Whenever you find a number which looks divisible by 7, you may stop there and conclude the result without any hesitation. (2) The above calculations can be done in one line or even mentally. Try to do so.

Divisibility by 8 Rule: If the last three digits of a number is divisible by 8, the number is also divisible by 8. Also, if the last three digits of a number are zeros, the number is divisible by 8. Ex. 1. 1256: As 256 is divisible by 8, the number is also divisible by 8. Ex. 2. 135923120: As 120 is divisible by 8, the number is also divisible by 8. Ex. 3. 139287000: As the number has three zeros at the end, the number is divisible by 8. Note: The same rule is applicable to check the divisibility by 125.

Divisibility by 9 Rule: If the sum of all the digits of a number is divisible by 9, the number is also divisible by 9. Ex. 1. 39681: 3 + 9 + 6 + 8 + 1=27 is divisible by 9, hence the number is also divisible by 9. Ex. 2. 456138: 4 + 5 + 6 + 1 + 3 + 8= 27 is divisible by 9, hence the number is also divisible by 9.

Divisibility by 10 Rule: Any number which ends with zero is divisible by 10. There is no need to discuss this rule.

Divisibility by 11 Rule: If the sums of digits at odd and even places are equal or differ by a number divisible by 11, then the number is also divisible by 11. Ex. 1. 3245682: S1 = 3 + 4 + 6 + 2= 15 and

S2 = 2 + 5 + 8 = 15 As S1 = S2 , the number is divisible by 11. 1. Ex. 2. 283712: S1 = 2 + 3 + 1= 6 and S2 = 8 + 7 + 2 = 17. As S1 and S2 differ by 11 (divisible by 11), the number is also divisible by 11.

Divisibility Ex. 3. 84927291658: S1 = 8 + 9 + 7 + 9 + 6 + 8 = 47 and S2 = 4 + 2 + 2 + 1 + 5 = 14 As (S1 – S 2 ) 33 is divisible by 11, the number is also divisible by 11.

Divisibility by 12 Rule: Any number which is divisible by both 4 and 3, is also divisible by 12. To check the divisibility by 12, we 1) first divide the last two-digit number by 4. If it is not divisible by 4, the number is not divisible by 12. If it is divisible by 4 then 2) check whether the number is divisible by 3 or not. Ex. 1. 135792: 92 is divisible by 4 and also (l + 3 + 5 + 7 + 9 + 2=)27 is divisible by 3; hence the number is divisible by 12. Remark: Recall the method for calculation of digit-sum. What did you do earlier (in the 1st chapter)? “Forget nine”. Do the same here. For example: digit-sum of 135792 __ 1 plus 3 plus 5 is 9, forget it. 7 plus 2 is 9, forget it. And finally we get nothing. That means all the “forget nine” adds to a number which is multiple of 9. Thus, the number is divisible by 9.

Divisibility by 13 Oscuator for 13 is 4 (See note). But this time, our osculator is not negative (as in case of 7). It is ‘onemore’ osculator. So, the working principle will be different now. This can be seen in the following examples. Ex 1: Is 143 divisible by 13? Soln: 14 3: 14 + 3 × 4 = 26 Since 26 is divisible by 13, the number 143 is also divisible by 13. or, This working principle may further be simplified as: Step I : 1 4 3 16 [4 × 3 (from 143) + 4 (from 143)] Step II : 1 4 3 26/16 [4 × 6 (from 16) +1 (from 16) +1 (from 143) = 26] As 26 is divisible by 13, the number is also divisible by 13. Note: The working of second method is also very systematic. At the same time, it is more acceptable because it has less writing work.

23 Ex 2: Check the divisibility of 24167 by 13. 2 4 1 6 7 26/6/20/34 [4 × 7 (from 24167) + 6 (from 24167) = 34] [4 × 4 (from 3 4) +3 (from 34) +1 (from 24167) = 20] [4 × 0 (from 20) +2 (from 20) +4 (from 24167) = 6] [4 × 6 (from 6) +2 (from 24167) = 26] Since 26 is divisible by 13 the number is also divisible by 13. Remark: Have you understood the working principle? If your answer is no, we suggest you to go through each step carefully. This is very simple and systematic calculation. Ex 3: Check the divisibility of 6944808 by 13. Soln: 6 9 4 4 8 0 8 39/18/12/41/19/32 4 × 8 + 0 = 32 4 × 2 + 3 + 8 = 19 4 × 9 + 1 + 4 = 41 4 × 1 + 4 + 4 = 12 4 × 2 + 1 + 9 = 18 4 × 8 + 1 + 6 = 39 Since 39 is divisible by 13, the given number is divisible by 13. Note: (1) This method is applicable for “one-more” osculator only. So we can’t use this method in the case of 7. (2) This is a one-line method and you don’t need to write the calculations during exams. These are given merely to make you understand well.

Divisibility by 14 Any number which is divisible by both 2 and 7, in also divisible by 14. That is, the number’s last digit should be even and at the same time the number should be divisible by 7.

Divisibility by 15 Any number which is divisible by both 3 and 5 is also divisible by 15.

Divisibility by 16 Any number whose last 4 digit number is divisible by 16 is also divisible by 16.

Quicker Maths

24

Divisibility by 17

Divisibility by 18

Negative osculator for 17 is 5 (see note). The working for this is the same as in the case of 7. Ex. 1: Check the divisibility of 1904 by 17. Soln: 190 4: 190 – 5 × 4 = 170 Since 170 is divisible by 17, the given number is also divisible by 17. Note: Students are suggested not to go upto the last calculation. Whenever you find the number divisible by the given number on right side of your calculation stop further calculation and conclude the result. Ex. 2: 957508: 95750 8: 95750 – 5 × 8 = 95710 9571 0: 9571 – 5 × 0 = 9571 957 l: 957 – 5 × l = 952 95 2: 95 – 5 × 2 = 85 Since 85 is divisible by 17, the given number is divisible by 17. Ex. 3: 8971563: 897156 3: 897156 – 5 × 3 = 897141 89714 1: 89714 – 5 × 1 = 89709 8970 9: 8970 – 5 × 9 = 8925 892 5: 892 – 5 × 5 = 867 86 7: 86 – 5 × 7 = 51 Since 51 is divisible by 17, the given number is also divisible by 17.

Rule: Any number which is divisible by 9 and has its last digit (unit-digit) even or zero, is divisible by 18. Ex. 1. 926568: Digit-sum is a multiple of nine (ie, divisible by 9) and unit-digit (8) is even, hence the number is divisible by 18. Ex. 2. 273690: Digit-sum is a multiple of nine and the number ends in zero, so the number is divisible by 18. Note: During the calculation of digit-sum, follow the method of “forget nine”. If you get zero at the end of your calculation, it means the digit-sum is divisible by 9.

Divisibility by 19 If you recall, the ‘one-more’ osculator for 19 is 2. The method is similar to that of 13, which is well known to you. Let us take an example. Ex. 1: 149264 Soln: 1 4 9 2 6 4 19/9/12/11/14 Thus, our number is divisible by 19. Note: You must have understood the working principle (see the case of 13).

Chapter 6

Squaring Squaring of a number is largely used in mathematical calculations. There are so many rules for special cases. But we will discuss a general rule for squaring which is capable of universal application. The method of squaring is intimately connected with a procedure known as the “Duplex Combination” process. We now go on to a brief study of this procedure.

Duplex Combination Process The first one is by squaring; and the second one is by cross-multiplication. In the present context, it is used in both senses (a2 and 2ab). In the case of a single central digit, the square is meant; and in the case of an even number of digits equidistant from the two ends, double the cross-product is meant. A few examples will elucidate the procedure. Ex. 1: For 2, Duplex (D) = 22 = 4 Ex. 2: For 8, D = 82 = 64 Ex. 3: For 34, D = 2 × (3 × 4) = 24 Ex. 4: For 79, D = 2 × (7 × 9) = 126 Ex. 5: For 103, D = 2(1 × 3) + 02 = 6 Ex. 6: For 346, D = 2(3 × 6) + 42 = 52 Ex. 7: For 096, D = 2(0 × 6) + 92 = 81 Ex. 8: For 1342, D = 2(1 × 2) +2(3 × 4) = 28 Ex. 9: For 5156, D = 2(5 × 6) + 2(1 × 5) = 70 Ex.10: For 23564, D = 2(2 × 4) +2(3 × 6) + 52 = 77 Ex.11: For 123456, D = 2(1 × 6) +2 (2 × 5) +2 (3 × 4) = 56 Now, we see the method of squaring in the following examples. Ex. 1. 2072 = ? Soln: 2072 = D for 2 / D for 20 / D for 207 / D for 07 / D for 7 = 22/ 2(2 × 0) / 2(2 × 7) + 02/ 2(0 × 7) / 72 =4 / 0 / 28 / 0 / 49 =4 / 0 / 8 / 0 / 49 2 = 4 / 0 + 2 / 8 / 0 + 4 / 9 = 42849 If you have understood the duplex method and its use in squaring, you may get the answer in a line. For example: 2072 = 4228449.

Explanations. 1. Duplex of 7 is 72 = 49. Put the unit digit (9) of duplex in answer line and carry over the other (4). 2. 2 × 0 × 7 + 4(carried) = 4; write it down at 2nd position. 3. 2 × 2 ×7 + 02 = 28; write down 8 and carry over 2. 4. 2 × 2 × 0 + 2(carried) = 2; write it down. 5. 22 = 4; write it down. Note: (1) If there are n digits in a number, the square will have either 2n or 2n-l digits. (2) Participation of digits follows the same systematic pattern as in multiplication. Ex. 2: (897)2 = 8016420613049 = 804609 Explanations: 1. 72 = 49; write down 9 and carry over 4. 2. 2 × 9 × 7 + 4 (carried) = 130; write down 0 and carry over 13. 3. 2 × 8 × 7 + 92 + 13 = 206; write down 6 and carry over 20. 4. 2 × 8 × 9 + 20 = 164; write down 4 and carry over 16. 5. 82 + 16 = 80; write it down. Ex. 3: (1 4 3 2)2 = 210253026124 = 2050624 Explanations: 1. 22 = 4; write it down. 2. 2 × (3 × 2) = 12; write down 2 and carry over 1. 3. 2 × (4 × 2) + 32 + l = 26; write down 6 and carry over 2. 4. 2(1 × 2) + 2(4 × 3) + 2 = 30; write down 0 and carry over 3. 5. 2(1 × 3) + 42 +3 = 25; write down 5 and carry over 2. 6. 2(1 × 4) + 2 = 10; write down 0 and carry over 1. 7. 12 +1 = 2; write it down. Ex 4: (73214)2 = 53464032682917916 = 5360289796 Ex 5: (5432819)2 = 29455155115102162122128661472681 = 29515522286761

26 Practice problem: Q: Find the squares of the following numbers: 1) 835 2) 8432 3) 45321 4) 530026 5) 73010932 Answers: 1) 697225 2) 71098624 3) 2053993041 4) 280927560676 5) 5330596191508624 Note: To find the square of a fractional (decimal) number, we square the number without looking at decimal. After that we count the number of digits after the decimal in the original value. In the squared value, we place the decimal after double the number of digits after decimal in the original value. For example: (12.46)2 = 151532455l36 = 155.2516 Some special cases derived with help of Duplex Combination Process 1. Square of numbers from 51 to 59. We take a general representative of the numbers (from 51 to 59), say 5A. Now, (5A)2 = 52 / 2 × 5 × A / A2 = 25 / 10 × A / A2 We have; 10 × A = A0 [like 10 × 4 = 40, 10 × 6 = 60, etc] (5A)2 = 25 / A0 / A2 = (25 + A) / A2 ; where A2 should be written as a two-digit number Now, we see that our duplex combination process reduces to a simplier form. Using the above equation: Ex. 1: (51)2 = 25 + 1 / (1)2 = 26 / 01 = 2601 Ex. 2: (52)2 = 25 + 2 / (2)2 = 27 / 04 = 2704 Ex. 3: (54)2 = 25 + 4 / (4)2 = 29 / 16 = 2916 Ex. 4: (59)2 = 25 + 9 / (9)2 = 34 / 81 = 3481 2. Square of a number with unit digit as 5. We take a general representative of such number, say A5. Now, (A5)2 = A2/2 × A × 5/52 = A2/10 × A / 25

Quicker Maths = A2 / A0 / 25 [ 10 × A = A0] = A2 + A / 25 = A (A + 1) / 25 Using the above equation: Ex. 1: (15)2 = 1 × (1 + 1) / 52 = 2 / 25 = 225 Ex. 2: (25)2 = 2 × (2 + 1) / 52 = 6 / 25 = 625 Ex. 3: (85)2 = 8 × (8 + 1) / 52 = 72 / 25 = 7225 Ex. 4: (115)2 = 11 × (12) / 52 = 132 / 25 = 13225 Ex. 5: (225)2 = 22 × (23) / 52 = 506 / 25 = 50625 3. Square of a number wich is nearer to 10x We use the algebraic formula x2 = (x2 – y2) + y2 = (x + y) (x – y) + y2 Ex. 1: (98)2 = (98 + 2) (98 – 2) + 22 = 9600 + 4 = 9604 Ex. 2: (103)2 = (103 – 3) (103 + 3) + 32 = 10600 + 9 = 10609 Ex. 3: (993)2 = (993 + 7) (993 – 7) + 72 = 986000 + 49 = 986049 Ex. 4: (1008)2 = (1008 – 8) (1008 + 8) + 82 = 1016000 + 64 = 1016064 To check the calculation We use the digit-sum method for checking calculations in squaring. For example: In Ex 1: (207)2 = 42849 digit-sum: (0)2 = (0)2 Hence, our calculation is correct. In Ex 2: (897)2 = 804609 digit-sum: (6)2 = 18 or, 36 =18 or, 0 = 0 Thus, our calculation is correct. In Ex 3: (1432)2 = 2050624 digit-sum: 12 = 1 or, 1 = 1 Thus, our calculation is correct. Note: 1. Follow the “forget-nine” rule during the calculation of digit-sum. 2. Check all the calculations mentally. 3. Check the correctness of calculations in other examples without using pen.

Chapter 7

Cube Cubes of large numbers are rarely used. During our mathematical calculations, we sometimes need the cube value of two-digit numbers. So, an easy rule for calculating the cubes of 2-digit numbers is being given. In its process the cube values of the “first ten natural numbers”, ie, 1 to 10, are used. Readers are suggested to remember the cubes of only these “first ten natural numbers.” 13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216, 73 = 343, 83 = 512, 93 = 729 and 103 = 1000. To calculate the cube value of two-digit numbers we proceed like this: Step I: The first thing we have to do is to put down the cube of the tens-digit in a row of 4 figures. The other three numbers in the row of answer should be written in a geometrical ratio in the exact proportion which is there between the digits of the given number. Step II: The second step is to put down, under the second and third numbers, just two times of second and third number. Then add up the two rows. For example: Ex 1. 123 = ? Soln. Step I: We see that the tens-digit in the number is 1, so we write the cube of 1. And also as the ratio between 1 and 2 is 1 : 2, the next digits will be double the previous one. So, the first row is 1 2 4 8 Step II: In the above row our 2nd and 3rd digits (from right) are 4 and 2 respectively. So, we write down 8 and 4 below 4 and 2 respectively. Then add up the two rows. 1 2 4 8 4 8 1 7 12 8 = 1728 Ex. 2: 113 = ? (Solve it yourself.) Ex. 3: 163 = ? Soln: 1 6 36 216 12 72 4 30 9 6 = 4096 12 21 Explanations: 13 (from 16) = 1. So, 1 is our first digit in the first row. Digits of 16 are in the ratio 1 : 6, hence our other digits should be 1 × 6 = 6, 6 × 6 = 36, 36 × 6 = 216. In the second row, double the 2nd and 3rd number

is written. In the third row, we have to write down only one digit below each column (except under the last column which may have more than one digit). So, after putting down the units-digit, we carry over the rest to add up with the left-hand column. Here, (i) Write down 6 of 216 and carry over 21. (ii) 36 + 72 + 21 (carried) = 129, write down 9 and carry over 12. (iii) 6 + 12 + 12 (carried) = 30, write down 0 and carry over 3. (iV) 1 + 3 (carried) = 4, write down 4. Ex 4: 183 = ? Soln:

1

8

64

512

5

16 48

128 24 3

512

= 5832

(i) Write down 2 and carry over 51 of 512. (ii) 64 +128 + 51 = 243, write down 3 and carry over 24. (iii) 8 + 16 + 24 = 48, write down 8 and carry over 4. (iv) 1 + 4 = 5 write it down. Ex 5: 173 = ? (Solve it yourself) Ex 6: 193 = ? (Solve it yourself) Ex 7: 213 = ? Soln: 8 4 2 1 [8 = 23, 8 ÷ 2 = 4, 4 ÷ 2 = 2, 2 ÷ 2 = 1, since ratio is 2:1] 8 4 [4 × 2 = 8, 2 × 2 = 4, double is written below] 9 12 6 1 = 9261 Do you mark the difference? If no, go through the following explanations. Step I: (i) 23 = 8 is the first figure (from left) in the first row. (ii) Ratio between the two digits is 2 : 1, ie, the number should be halved subsequently. Therefore, the next three numbers in the first row should be 4, 2 & 1.

Quicker Maths

28 Step II:It should be clear to all of you because it has nothing new. Ex 8: 233 = ? Soln: 8 12 18 27 24 36 12 41 12167 56 27 Explanations: Step I: (i) 23 = 8 ---------- the first figure (from left) in the first row. 3 (ii) 2 : 3 the next numbers should be of 2 the previous ones. So,

we have 8 ×

3 3 3 = 12, 12 × =18, 18 × = 27. 2 2 2

Ex 9: 333 = ? Soln:

27

27 54

27 54

27

35

89

83

27

= 35937

Explanations: Step i) 33 = 27 ------ the first figure in first row. ii) 3 : 3 = 1 : 1 the subsequent numbers should be the same. Ex 10: 343 = ? Soln:

27

36 72

48 96

64

39

12 3

15 0

64

= 39304

(ii) Ratio is 3 : 4, ie, the next numbers should be of their previous ones. Here, 27 ×

24 5

27

4 3

4 = 36, 3

4 4 36 × = 48, 48 × = 64. 3 3 3 Ex 11: 93 = ? Soln: 729 243 81 27 486 162 753

Soln:

729

567 1134

441 882

343

912

1836

1357

34 3

= 912673

Explanations: (i) 93 = 729 ---------- first figure (from left) in the first row. 7 (ii) Ratio = 9 : 7 Next numbers should be of 9 7 the previous ones. Therefore, 729 × = 567, 9 7 7 567 × = 441, 441 × = 343. 9 9 Explanation of the above method: (It leads us to more Quicker Method.) Any two-digit number can be represented as 10x + y. Now,

10x y 3 103 x 3 3102 x 2 y 310xy 2 y 3

Explanations: (i) 33 = 27 --------- the first figure (from left) in the first row.

804

Explanations: (i) 93 = 729 --------- the first figure (from left) in the first row, (ii) 9 : 3 3 : 1 the subsequent figures should be 1 of their previous ones. 3 Ex. l2: 973 = ?

= 1000x 3 100 3x 2 y 10 3xy 2 y 3 The above expansion gives us four parts. The leftmost part (1000x3) has three zeroes at the end. Similarly, the second part has two zeroes and the third part has one zero. This implies that the firstpart leaves three places for the other parts. Similarly, second part leaves two places for the other parts and the third part leaves one place for the last part. You can better understand by an example: Suppose we have to find (23)3. Now ,

233 20 33 203 3202 3 32032 33 = 2 3 1000 322 3 100 32 32 10 33 = 8000 + 36 × 100 + 54 × 10 + 27 The answer is in the form

= 804357

8 – – – 36 – – 54 – 27

Cube

29

523 140

8 (36) (54) (27) 12

1

6

7

Note: In place of 27 we have to write only a single digit, ie 7, and carry forward 2 to the ten’s position. Now at the ten’s position we have 54 + 2 = 56. So, we write 6 at the ten’s position and carry forward 5 to the hundred’s position, and so on. The summary of the above explanation is as follows:

a b 3 233

a3

3a 2 b

3ab 2

b3

23 3(2)2 (3) 3(2)(3)2 (3)3

= 8

36

54

27

= 12

1

6

7

Now, we reach at the stage where we can write the answer directly if a b a 3 3a 2 b 3ab 2 b 3 is at our fingertip. We start from the rightmost position and move towards the leftmost positions. Take another example: 3

213

1

6

Step I:

3

1

3 2 1 6 2

Step II:

2

1 1

6

1

Step III: 3 2 2 1 12 Put 2, carry over 1

9

2

6

1

Step IV: 23 = 8; 8 + 1(carried)= 9 Combining all the steps into a single step:

343

39

3

0

4

253 15

6

2

5

6

0

8

Practice Problems. Q. Find the cubes of the following 1) 17 2) 26 4) 32 5) 41 7) 49 8) 51 10) 55 11) 57 13) 67 14) 69 16) 77 17) 88 19) 95 20) 99 Answers: 1) 4913 4) 32768 7) 117649 10) 166375 13) 300763 16) 456533 18) 857375

2) 17576 5) 68921 8) 132651 11) 185193 14) 328509 17) 681472 19) 970299

numbers. 3) 27 6) 43 9) 53 12) 64 15) 73 18) 92

3) 19683 6) 79507 9) 148877 12) 262144 15) 389017 18) 778688

Note: Don't use this method for getting the cubes of 20, 30, 40, --- Do you know the other quick method?

Checking the correctness (with the help of digitsum) Ex. 1: 123 1728 digit sum:

(3) 3 0 (7 2 9, forget it. 1 8 9, forget it.) or,

0 = 0, thus the cube value is correct.

Ex. 2: 16 4096 3

digit-sum:

73 1 or, (49) 7 1 or, 4×7 =1 or, 28 = 1 or, 1 = 1, Thus cube value is correct. Practice-Problem: Check all the calculations (from Ex 2 to Ex 12) done in this chapter.

30

Quicker Maths

Chapter 8

HCF and LCM Factor: One number is said to be a factor (Gunankhand) of another when it divides the other exactly. Thus, 6 and 7 are factors of 42. Common Factor: A common factor of two or more numbers is a number that divides each of them exactly. Thus, 3 is a common factor of 9, 18,21 and 33. Highest Common Factor (HCF): HCF of two or more numbers is the greatest number that divides each of them exactly. Thus, 6 is the HCF of 18 and 24. Because there is no number greater than 6 that divides both 18 and 24. Note: The terms Highest Common Divisor and Greatest Common Measure are often used in the sense of Highest Common Factor (HCF).

To find the HCF of two or more numbers Method I: Method of Prime Factors Rule: Break the given numbers into prime factors and then find the product of all the prime factors common to all the numbers. The product will be the required HCF. Ex. 1. Find the HCF of 42 and 70. Soln: 42 = 2 × 3 × 7 70 = 2 × 5 × 7 HCF = 2 × 7 = 14 Ex. 2. Find the HCF of 1365, 1560 and 1755. Soln: 1365 = 3 × 5 × 7 × 13 1560 = 2 × 2 × 2 × 3 × 5 × 13 1755 = 3 × 3 × 3 × 5 × 13 HCF = 3 × 5 × 13 = 195 Note: (1) In finding the HCF, we need not break all the numbers into their prime factois. We may find the prime factors of one of the numbers. Then the product of those prime factors which divide each of the remaining numbers exactly will be the required HCF. In Ex. (1), the prime factors of 42 are 2 × 3 ×7. Of these three factors, only 2 and 7 divide 70 exactly. Hence, the required HCF = 2 × 7 = 14 In Ex. (2) the prime factors of 1365 are 3 × 5 × 7 × 13. Of these four factors, only 3,5 and

13 divide the other two numbers 1560 and 1755 exactly. Hence, the required HCF = 3 × 5 × 13 = 195 (2) We must remember that the quotient obtained by dividing the numbers by their HCF are prime to each other. In Ex. (1) 42 ÷ 14 = 3 70 ÷ 14 = 5. We see that 3 is prime to 5, i.e., 3 can’t divide 5 exactly. In Ex. (2), 1365 ÷ 195 = 7, 1560 ÷ 195 = 8 and 1755 ÷ 195 = 9, We see that 7, 8 and 9 are prime to one another, i.e. none divides the other.

Method II: Method of Division Rule: Divide the greater number by the smaller number, divide the divisor by the remainder, divide the remainder by the next remainder, and so on until no remainder is. left. The last divisor is the required HCF. Ex. 1. 42) 70 (1 42 28) 42 (1 28 14) 28 (2 28 0 HCF =14 Note: The above rule for finding the HCF of numbers is based on the following two principles: (i) Any number which divides a certain number also divides any multiple of that number; for example, 6 divides 18 therefore, 6 divides any multiple of 18. (ii) Any number which divides each of the two numbers also divides their sum, their difference and the sum and difference of any multiples of that numbers. Thus 5, being a common factor of 25 and 15, is also a factor of (25 + 15), and (25 – 15). Again, 5 is also a factor of (25 × a + 15 × b) and of (25 × a – 15 × b), where a and b are integers.

Quicker Maths

32 In accordance with these principles, HCF of 42 and 70 = HCF of 28 and 42 [ 28 = 70 – 42 ] = HCF of 14 and 28 [14 = 42 – 28] = HCF of 14 and 14 [ 14 = 28 – 14 ] HCF of 42 and 70 = 14 HCF of 13281 and 15844 = HCF of 2563 and 13281 [2563 = 15844 –13281] = HCF of 466 and 2563 [466 = 13281 – 5 × 2563] = HCF of 233 and 466 [233 = 2563 – 5 × 466 ] = HCF of 233 and 233 [233 = 466 – 233] HCF of 13281 and 15844 = 233 The above discussed method is very much interesting. It gives results very quickly. But one should have a good understanding of this method.

4199) 4693 (1 4199 494 494 is divisible by 2 but 4199 is not. We, therefore, divide 494 by 2 and proceed with 247 and 4199 (by rule iii). 247) 4199 (17 247 1729 1729 0 The HCF of 4199 and 4693 is 247. Hence, the HCF of the original numbers is 247 × 9 = 2223. Note: If the HCF of two numbers be unity, the numbers must be prime to each other.

To find the HCF of more than two numbers

HCF of smaller numbers

Rule: Find the HCF of any two of the numbers and then find the HCF of this HCF and the third number and so on. The last HCF will be the required HCF. Ex. 1. Find the HCF of 1365, 1560 and 1755. Soln: 1365) 1560 (1 1365 195) 1365 (7 1365 0 Therefore, 195 is the HCF of 1365 and 1560. Again, 195 ) 1755 ( 9 1755 0 the required HCF = 195 Method III: The work of finding the HCF may sometimes be simplified by the following devices: (i) Any obvious factor which is common to both numbers may be removed before the rule is applied. Care should however be taken to multiply this factor into the HCF of the quotients. (ii) If one of the numbers has a prime factor not contained in the other, it may be rejected. (iii) At any stage of the work, any factor of the divisor not contained in the dividend may be rejected. This is because any factor which divides only one of the two cannot be a portion of the required HCF. Ex. Find the HCF of 42237 and 75582. Soln: 42237 = 9 × 4693 75582 = 2 × 9 × 4199 We may reject 2 which is not a common factor (by rule (i). But 9 is a common factor. We, therefore, set it aside (by rule ii) and find the HCF of 4199 and 4693.

If the numbers are not very large, we can follow the following steps to get the HCF very quickly. For example, (i) Find the HCF of 8, 20, 28 and 44. Soln: Step I: As we know that HCF is the highest common factor of all the numbers, it cannot be larger than the smallest number. So, take the smallest number, ie 8. Step II: Divide the other numbers by 8. As it does not divide 20, we reject 8 as our HCF. Step III: Take the second highest factor of 8, ie 8 ÷ 2 = 4. Check the divisibility of the other numbers by 4. As it divides all the other numbers, our HCF is 4. (ii) Find the HCF of 21, 15 and 36. Soln: Step I: Take the smallest number 15. As it does not divide the other numbers it is rejected as our HCF. Step II: Take the second highest factor of 15, ie 15 ÷ 3 = 5. As it does not divide the other numbers, it is rejected as our HCF. Step III: Take the next highest factor, ie 15 ÷ 5 = 3. As it divides the other numbers also, our HCF is 3. (iii) Find the HCF of 72, 126 and 198. Soln: Step I: Take the smallest number, ie 72. As it does not divide any other number, it is not our HCF. Step II: Take the second highest factor of 72, ie 72 ÷ 2 = 36. It is rejected as it does not divide other numbers.

HCF and LCM Step III: Take the next fact, ie 72 ÷ 3 = 24. It is also rejected, as it does not divide 126. Step IV: Take the next factor ie 72 ÷ 4 = 18. As it divides all the other numbers, our HCF is 18. (iv) Find the HCF of 24, 60, 84 and 108. Soln: The smallest number 24 is rejected as HCF. The next largest factor of 24, ie 12 is the required HCF as it divides all the other numbers.

33

Thus, the fractions are

HCF of 6, 60, 12 6 HCF LCM of 1, 17, 17 17 Note: We see that each of the numbers is perfectly divisible by

HCF of decimals Rule: First make (if necessary) the same number of decimal places in all the given numbers; then find their HCF as if they are integers and mark off in the result as many decimal places as there are in each of the numbers. Ex. 1: Find the HCF of 16.5, 0.45 and 15. Soln: The given numbers are equivalent to 16.50, 0.45 and 15.00 Step I: First we find the HCF of 1650, 45 and 1500. Which comes to 15. Step II: The required HCF = 0.15 Ex. 2: Find the HCF of 1.7, 0.51 and 0.153. Soln: Step I: First we find the HCF of 1700, 510 and 153. Which comes to 17. Step II: The required HCF = 0.017

HCF of vulgar fractions Def: The HCF of two or more fractions is the highest fraction which is exactly divisible by each of the fractions. Rule: First express the given fractions in their lowest terms: HCF of numerators Then, HCF = LCM of denominators Note: The HCF of a number of fractions is always a fraction (but this is not true with LCM).

54 9 36 , 3 and 9 17 51 54 6 9 60 36 12 ,3 Soln: Here, and 9 1 17 17 51 17 Ex. 1: Find the HCF of

6 . 17

Ex. 2: Find the greatest length that is contained an exact

1 3 number of times in 3 m and 8 m. 2 4

To find the HCF of two or more concrete quantities First, the quantities should be reduced to the same unit. Ex. Find the greatest weight which can be contained exactly in 1 kg 235 gm and 3 kg 430 gm. Soln: 1 kg 235 gm = 1235 gm 3 kg 430 gm = 3430 gm The greatest weight required is the HCF of 1235 and 3430, which will be found to be 5 gm.

6 60 12 , and 1 17 17

Soln:

3

1 7 3 35 and 8 2 2 4 4

The greatest length will be the HCF of

7 35 and . 2 4

the required length

HCF of 7 and 35 7 3 1 m LCM of 2 and 4 4 4

Miscellaneous Examples on HCF Ex. 1: What is the greatest number that will divide 2400 and 1810 and leave remainders 6 and 4 respectively? Soln: Since on dividing 2400 a remainder 6 is left, the required number must divide (2400 – 6) or 2394 exactly. Similarly, it must divide (1810 – 4) or 1806 exactly. Hence, the greatest required number should be the HCF of 2394 and 1806, ie., 42. Ex. 2. What is the greatest number that will divide 38, 45 and 52 and leave remainders as 2, 3 and 4 respectively? Soln: The required greatest number will be the HCF of (38 – 2), (45 – 3) and (52 – 4) or 36, 42 and 48. Ans = 6 Ex. 3: Find the greatest number which will divide 410, 751 and 1030 so as to leave remainder 7 in each case? Soln: The required greatest number = HCF of (410 – 7), (751 – 7) and (1030 – 7). Ans = 31 Ex. 4: Find the greatest number which is such that when 76,151 and 226 are divided by it, the remainders are all alike. Also find the common remainder.

34 Soln:

Let k be the remainder, then the numbers (76 – k), (151 – k) and (226 – k) are exactly divisible by the required number. Now, we know that if two numbers are divisible by a certain number, then their difference is also divisible by that number. Hence, the numbers (151 – k) – (76 – k), (226 – k) – (151 – k) and (226 – k) – (76 – k) or 75, 75 and 150 are divisible by the required number. Therefore, the required number = HCF of 75, 75 and 150 = 75 And the remainder will be found after dividing 76 by 75, as 1. Ex. 5: The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find that number of three digits. Soln: The required number must be a factor of (11284 – 7655) or 3629. Now, 3629 =19 × 191 191 is the required number.. Ex 6: The product of two numbers is 7168 and their HCF is 16; find the numbers. Soln: The numbers must be multiples of their HCF. So, let the numbers be 16a and 16b where a and b are two numbers prime to each other. 16a × 16b = 7168 or, ab = 28 Now, the pairs of numbers whose product is 28 are (28), (1); (14), (2); (7, 4). 14 and 2 which are not prime to each other should be rejected. Hence, the required numbers are 28 × 16, 1 × 16; 7 × 16, 4 × 16 or, 448, 16; 112, 64 Common multiple: A common multiple of two or more numbers is a number which is exactly divisible by each of them. Thus, 30 is a common multiple of 2, 3, 5, 6, 10 and 15. Least common multiple (LCM): The LCM of two or more given numbers is the least number which is exactly divisible by each of them. Thus, 15 is a common multiple of 3 and 5. 30 is a common multiple of 3 and 5. 45 is a common multiple of 3 and 5. But 15 is the least common multiple (LCM) of 3 and 5.

Quicker Maths

To find the LCM of two or more given numbers Method I: Method of Prime Factors Rule: Resolve the given numbers into their prime factors and then find the product of the highest power of all the factors that occur in the given numbers. This product will be the LCM. Ex. 1: Find the LCM of 8,12, 15, and 21. Soln: 8 = 2 × 2 × 2 = 23 12 = 2 × 2 × 3 = 22 × 3 15 = 3 × 5 21 = 3 × 7 Here, the prime factors that occur in the given numbers are 2, 3, 5 and 7 and their highest powers are respectively 23 , 3, 5 and 7. Hence, the required LCM = 23 × 3 × 5 × 7 = 840 Ex. 2: Find the LCM of 18, 24, 60 and 150. Soln: 18 = 2 × 3 × 3 = 2 × 32 24 = 2 × 2 × 2 × 3 = 23 × 3 60 = 2 × 2 × 3 × 5 = 22 × 3 × 5 150 = 2 × 3 × 5 × 5 = 2 × 3 × 52 Here, the prime factors that occur in the given numbers are 2, 3 and 5, and their highest powers are 23 , 32 and 52 respectively . Hence, the required LCM = 23 × 32 × 52 = 1800 Note: The LCM of two numbers which are prime to each other is their product. Thus, the LCM of 15 and 17 is 15 × 17 = 255 Method II: The LCM of several small numbers can be easily found by the following rule: Write down the given numbers in a line separating them by commas. Divide by any one of the prime numbers 2,3,5,7, etc., which will exactly divide at least any two of the given numbers. Set down the quotients and the undivided numbers in a line below the first. Repeat the process until you get a line of numbers which are prime to one another. The product of all divisors and the numbers in the last line will be the required LCM. Note: To simplify the work, we may cancel, at any stage of the process, any one of the numbers which is a factor of any other number in the same line.

HCF and LCM

35

Ex. 1: Find the LCM of 12, 15, 90, 108, 135, 150. Soln:

2 12, 15, 90, 108, 135, 150 ....(1) 3 45, 54, 135, 75 ....(2) 3 18, 45, 25 ....(3) 5 6, 15, 25 ....(4) 6, 3, 5 ....(5) the required LCM = 2 × 3 × 3 × 5 × 6 × 5 = 2700 In line (1), 12 and 15 are the factors of 108 and 90 respectively, therefore, 12 and 15 are struck off. In line (2), 45 is a factor of 135, therefore 45 is struck off. In line (5), 3 is a factor of 6, therefore 3 is struck off. Note: The product of two numbers is equal to the product of their HCF and LCM. For example: The LCM and HCF of 12 and 15 are 60 and 3 respectively. Multiplication of two numbers =12 × 15 = 180; HCF × LCM = 3 × 60 =180 Thus, we see that the product of two numbers is equal to the product of their LCM and HCF.

Rule: First make (if necessary) the same number of decimal places in all the given numbers; then find their LCM as if they were integers, and mark in the result as many decimal places as there are in each of the numbers. Ex. Find the LCM of 0.6, 9.6 and 0.36. Soln: The given numbers are equivalent to 0.60, 9.60 and 0.36. Now, find the LCM of 60, 960 and 36. Which is equal to 2880. the required LCM = 28.80

LCM of fractions The LCM of two or more fractions is the least fraction or integer which is exactly divisible by each of them. Rule: First express the fractions in their lowest terms, then

LCM of numerator HCF of denominator

Ex. 1: Find the LCM of a)

1 5 , 2 8 1 2

c) 4 , 3, 10

a) The required LCM

b)

b)

1 2

108 17 54 , 1 , 375 25 55

LCM of 1 and 5 5 1 2 HCF of 2 and 8 2 2

108 36 17 42 ,1 375 125 25 25

Thus, the fractions are

36 42 54 , and 125 25 55

the required LCM =

LCM of 36, 42 and 54 756 1 151 HCF of 125, 25, 55 5 5

1 9 1 21 c) 4 , 10 2 2 2 2 Thus, the fractions are

9 3 21 , and 2 1 2

the required LCM

LCM of decimals

LCM

Soln:

LCM of 9, 3 and 21 63 63 HCF of 2, 1, 2 1

Note: In Ex. 1 (c), we see that the LCM of fractions is an integer. Thus, we may conclude that LCM of fractions may be a fraction or an integer.

Miscellaneous Examples on LCM Ex. 1: The LCM of two numbers is 2079 and their HCF is 27. If one of the numbers is 189, find the other. Soln: The required number

LCM HCF 2079 27 297 = First number 189 Ex. 2: Find the least number which, when divided by 18, 24, 30 and 42, will leave in each case the same remainder 1. Soln: Clearly, the required number must be greater than the LCM of 18, 24, 30 and 42 by 1. Now, 18 = 2 × 32 24 = 23 × 3 30 = 2 × 3 × 5 42 = 2 × 3 × 7 LCM = 32 × 23 × 5 × 7 = 2520 the required number = 2520 + 1 = 2521 Ex 3. What is the least number which, when divided by 52, leaves 33 as the remainder, and when divided by 78 leaves 59, and when divided by 117 leaves 98 as the respective remainders?

36 Soln:

Since 52 – 33 = 19, 78 – 59 = 19, 117 – 98 = 19 We see that the remainder in each case is less than the divisor by 19. Hence, if 19 is added to the required number, it becomes exactly divisible by 52,78 and 117. Therefore, the required number is 19 less than the LCM of 52, 78 and 117. The LCM of 52,78 and 117 = 468 the required number = 468 – 19 = 449 Ex 4: Find the greatest number of six digits which, on being divided by 6, 7, 8, 9 and 10, leaves 4, 5, 6, 7 and 8 as remainders respectively. Soln: The LCM of 6, 7, 8, 9 and 10 = 2520 The greatest number of 6 digits = 999999 Dividing 999999 by 2520, we get 2079 as remainder. Hence, the 6-digit number divisible by 2520 is (999999 – 2079), or 997920. Since 6 – 4 = 2, 7 – 5 = 2, 8 – 6 = 2, 9 – 7 = 2, 10 – 8 = 2, the remainder in each case is less than the divisor by 2. the required number = 997920 – 2 = 997918 Ex 5: Find the greatest number less than 900, which is divisible by 8,12 and 28. Soln : The least number divisible by 8,12 and 28 is 168. Clearly, any multiple of 168 will be exactly divisible by each of the numbers 8, 12 and 28. But since the required number is not to exceed 900, it is 168 × 5 = 840. Ex 6: Find the least number which, upon being divided by 2, 3, 4, 5,6 leaves in each case a remainder of 1, but when divided by 7 leaves no remainder. Soln: The LCM of 2, 3, 4, 5, 6 = 60 the required number must be = 60k +1, where k is a positive integer. = (7 × 8 + 4)k + 1 = (7 × 8k) + (4k + 1) Now, this number is to be divisible by 7. Whatever may be the value of k, the portion (7 × 8k) is always divisible by 7. Hence, we must choose that least value of k which will make 4k + 1 divisible by 7. Putting k = 1, 2, 3, 4, 5 etc. in succession, we find that k should be 5.

Quicker Maths the required number = 60k + 1 = 60 × 5 + 1 = 301 Note: The above example could also be worded as follows. A person had a number of toys to distribute among children. At first, he gave 2 toys to each child, then 3, then 4, then 5, then 6, but was always left with one. On trying 7 he had none left. What is the smallest number of toys that he could have had? Ex. 7: What least number must be subtracted from 1936 so that the remainder when divided by 9, 10, 15 will leave in each case the same remainder 7? Soln: The LCM of 9, 10 and 15 = 90 On dividing 1936 by 90, the remainder = 46 But 7 is also a part of this remainder . the required number = 46 – 7 = 39 Ex. 8: What greatest number can be subtracted from 10,000 so that the remainder may be divisible by 32, 36, 48 and 54? Soln: LCM of 32, 36, 48, 54 = 864 the required greatest number = 10,000 – 864 = 9,136 Ex. 9: What is the least multiple of 7, which when divided by 2, 3, 4, 5 and 6, leaves the remainders 1, 2, 3, 4, 5 respectively? Soln: LCM of 2, 3,4, 5, 6 = 60 Other numbers divisible by 2, 3, 4, 5, 6 are 60k, where k is a positive integer. Since 2 – 1 = 1, 3 – 2 = 1, 4 – 3 = 1, 5 – 4 = 1 and 6 – 5 = 1, the remainder in each case is less than the divisor by 1, the required number = 60k – 1 = (7 × 8k) + (4k × 1) Now, this number is to be divisible by 7. Whatever may be the value of k the portion 7 x 8k is always divisible by 7. Hence, we must choose the least value of k which will make (4k –1) divisible by 7. Putting k equal to 1, 2, 3, etc. in succession, we find that k should be 2. the required number = 60k –1=60 × 2 – 1 = 119

HCF and LCM

37

EXERCISES 1. What is the greatest number that will divide 2930 and 3250 and will leave as remainders 7 and 11 respectively? 2. What is the least number by which 825 must be multiplied in order to produce a multiple of 715? 3. The LCM of two numbers is 2310 and their HCF is 30. If one of the numbers is 7 × 30, find the other number. 4. Three bells commence tolling together and they toll after 0.25, 0.1 and 0.125 seconds. After what interval will they again toll together? 5. What is the smallest sum of money which contains 2.50, 20, 1.20 and 7.50? 6. What is the greatest number which will divide 410, 751 and 1030 so as to leave the remainder 7 in each case? 7. What is the HCF and LCM of

4 5 7 , and ? 5 6 15

8. Three men start together to travel the same way around a circular track of 11 km. Their speeds are 4, 5.5 and 8 km per hour respectively. When will they meet at the starting point? 9. Find the smallest whole number which is exactly

1 2

1 3

1 4

divisible by 1 , 1 , 2 , 3

1 1 and 4 2 5

10. How many times is the HCF of 48, 36, 72 and 24 contained in their LCM? 11. Find the least square number which is exactly divisible by 4, 5, 6, 15 and 18. 12. Find the least number which, when divided by 8, 12 and 16, leaves 3 as the remainder in each case; but when divided by 7 leaves no remainder. 13. Find the greatest number that will divide 55, 127 and 175 so as to leave the same remainder m each case. 14. Find the least multiple of 11 which, when divided by 8,12 and 16, leaves 3 as remainder. 15. What least number should be added to 3500 to make it exactly divisible by 42, 49, 56 and 63? 16. Find the least number which, when divided by 72, 80 and 88, leaves the remainders 52, 60 and 68 respectively. 17. Find the greatest number of 4 digits which, when divided by 2, 3, 4, 5, 6 and 7, should leave remainder 1 in each case.

18. Find the greatest possible length which can be used to measure exactly the lengths 7m, 3m 85cm, 12m 95cm. 19. Find the least number of square tiles required to pave the ceiling of a hall 15m 17cm long and 9m 2cm broad. 20. What is the largest number which divides 77, 147 and 252 to leave the same remainder in each case? 21. The traffic lights at three different road crossings change after every 48 sec., 72 sec., and 108 sec. respectively. If they all change simultane-ously at 8:20:00 hrs, then at what time will they again change simulta-neously? 22. The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is devided by 2, the quotient is 44. What is the other number? 23. The product of two numbers is 2160 and their HCF is 12. Find the possible pairs of numbers. 24. Find the greatest number of 4 digits and the least number of 5 digits that have 144 as their HCF. 25. Find the least number from which 12, 18, 32 or 40 may be subtracted, each an exact number of times. 26. Find the least number that, being increased by 8, is divisible by 32, 36 and 40. 27. The sum of two numbers is 528, and their HCF is 33. How many pairs of such numbers can be formed? 28. In a school, 391 boys and 323 girls have been divided into the largest possible equal classes, so that each class of boys numbers the same as each class of girls. What is the number of classes ? 29. Is it possible to divide 1000 into two parts such that their HCF may be 15? 30. Show that 2205 and 4862 are prime to each other. 31. WTiat least number must be subtracted from 1294 so that the remainder, when divided by 9, 11, 13 will leave in each case the same remainder 6? 32. Find the sum of the numbers between 300 and 400 such that when they are divided by 6, 9 and 12 (a) it leaves no remainder; and (b) it leaves remainder as 4 in each case. 33. Three friends J, K and L jog around a circular stadium and complete one round in 12, 18 and 20 seconds respectively. In how many minutes will all the three meet again at the starting point?

Quicker Maths

38

Solutions (Hints) 1. The greatest such number will be the HCF of (2930 – 7) and (3250 –11), i.e. 79. 2. 825 = 3 × 5 × 5 × 11; 715 = 5 × 11 × 13 Any multiple of 715 must have factors of 5, 11 and 13. So, 825 should be multiplied by the factor(s) of 715, which is (are) not present in 825. the required least number = 13 3. 4.

5.

6. 7.

2310 30 330 7 30 They will toll together after an interval of time equal to the LCM of 0.25 sec, 0.1 sec and 0.125 sec. LCM of 0.25, 0.1 and 0.125 = (LCM of 250, 100 and 125) × 0.001 = 500 × 0.001 = 0.5 sec. LCM of 2.5, 20, 1.2 and 7.5 = (LCM of 25, 200,12 and 75) × 0.1 = 600 × 0.1 = 60 The required number willl be the HCF of (410 – 7), (751 – 7), (1030 – 7), i.e. 31. HCF of Numerators 1 HCF = = LCM of Denominators 30 The required number

LCM =

LCM of Numerators 140 = = 140 HCF of Denominators 1

8. Time taken by them to complete one revolution

11 11 11 11 2 11 , and hrs respectively , and 4 5.5 8 4 1 8

LCM of

11 2 11 , and 4 1 8

LCM of 11, 2, 11 22 22 hrs. HCF of 4, 1, 8 1

they will meet after 22 hrs. 9. The required smallest number = LCM of the given numbers 10. HCF of 48, 36, 72, 24 = 12; LCM of 48, 36, 72, 24 = 144 LCM = 12 x HCF 11. LCM of 4, 5, 6, 15, 18 = 180, which is exactly divisible by the given numbers. 180 = 2 × 2 × 3 × 3 × 5 = 22 × 32 × 5 Therefore, if 180 is multiplied by 5 (180 × 5 = 900) then the number will be a perfect square as well as divisible by 4, 5, 6, 15 and 18. 12. The reast number which, when divided by 8, 12 and 16, leaves 3 as remainder = (LCM of 8, 12, 16)+ 3 = 48 + 3 = 51

Other such numbers are 48 × 2 + 3 = 99, 48 × 3 + 3 = 147,.... the required number which is divisible by 7 is 147. Note: This is a hit-and-trial method. Can you get the answer by the defined method? (see Ex. 6). 13. Let x be the remainder, then the numbers (55 – x), (127 – x) and (175 – x) are exactly divisible by the required number. Now, we know that if two numbers are divisible by a certain number, then their difference is also divisible by the number. Hence the numbers (127 – x) – (55 – x), (175 – x) – (127– x) and (175 – x) – (55 – x) or, 72, 48 and 120 are divisible by the required number. HCF of 48, 72 and 120 = 24, therefore the required number = 24. Note: Find the HCF of the positive differences of numbers. It will serve your purpose quickly. 14. LCM of 8, 12 and 16 = 48. Such numbers are (48 × 1 + 3) = 51, (48 × 2 + 3) = 99, which is divisible by 11. the required number = 9. Note: This is a hit-and-trial method. Try to solve by the detailed method. 15. LCM of 42, 49, 56, 63 = 3528; therefore, the required least number = 3528 – 3500 = 28 16. 72 – 52 = 20, 80 – 60 = 20, 88 – 68 = 20. We see that in each case, the remainder is less than the divisor by 20. The LCM of 72, 80 and 88 = 7920, therefore, the required number = 7920 – 20 = 7900 17. The greatest number of 4 digits = 9999. LCM of 2, 3, 4, 5, 6, 7 = 420 On dividing 9999 by 420, we get 339 as remainder. the greatest number of 4 digits which is divisible by 2, 3, 4, 5, 6 and 7 = 9999 – 339 = 9660 the required number = 9660 + 1 = 9661 18. The required length = HCF of 7m, 3.85m and 12.95m = (HCF of 700, 385, 1295) × .01m = 35 × .01m = 0.35m = 35 cm 19. Side of each tile = HCF of 1517 and 902 = 41 cm. Area of each tile = 41 × 41 cm2 1517 902 814 the number of tiles 41 41 20. Solve as in Ex. 13. The required number = HCF of (147 – 77), (252 – 147) and (252 – 77) = HCF of 70, 105, 175 = 35 21. LCM of 48, 72, 108 = 432 The traffic lights will change simultaneously after 432 seconds or 7 min = in 12 secs.

HCF and LCM they will change simultaneously at 8 : 27 : 12 hrs. 22. The first number = 2 × 44 = 88 The second number HCF LCM 44 264 132 88 88 FICF = 12. Then let the numbers be 12x and 12y. Now l2x × 12y = 2160 xy = 15 Possible values of x and y are (1, 15); (3, 5); (5, 3); (15, 1) the possible pairs of numbers (12, 180) and (36, 60) The required numbers shcld be multiples of 144. We have the greatest number of 4 digits = 9999. On dividing 9999 by 144, we get 63 as the remainder. the required greatest number of 4 digits = 9999 – 63 = 9936 Again, we have the least number of 5 digits = 10000 On dividing 10,000 by 144, we get 64 as the remainder. the required least number of 5 digits = 10,000 + (144 – 64) = 10,080 The required number = LCM of 12, 18, 32, 40 =1440 LCM of 32, 36 and 40 = 1440, therefore, the required number = 1440-8 = 1432 Let the numbers be 33a and 33b. Now, 33a + 33b = 528 or, 33 (a + b) = 528 a + b = 16 The possible values of a and b are (1, 15); (2, 14); (3, 13); (4, 12); (5, 11); (6, 10); (7, 9); and (8, 8). Of these the pairs of numbers that are prime to each other are (1, 15); (3, 13); (5, 11); and (7, 9). the possible pairs of numbers are (33, 495); (99, 429); (165, 363); (231,297)

23.

24.

25. 26. 27.

39 28. Number of classes = HCF of 391 and 323 = 17 29. If two numbers are divisible by a certain number, then their sum is also divisible by that number. According to this rule: if 15 is the HCF of two parts of 1000, then 1000 must be divisible by 15. But it is not so. Therefore, it is not possible to divide 1000 into two parts such that their HCF may be 15. 30. If the numbers are prime to each other, then their HCF should be unity. Conversely, if their HCF is unity, the numbers are prime to each other. In this case, the HCF is 1, so they are prime to each other. 31. LCM of 9, 11 and 13 = 1287 Therefore, the number which, after being divided by 9, 11 and 13, leaves in each case the same remainder 6 = 1287 + 6 = 1293 the required least number = 1294 – 1293 = 1. 32. The LCM of 6, 9 and 12 = 36 (a) Multiples of 36 which lie between 300 and 400 are 324, 360 and 396. the required sum = 324 + 360 + 396= 1080 (b) Here, the remainder is 4 in each case. So, the numbers are (324 + 4 =) 328 and (360 + 4 =) 364. (The no. 396 + 4 = 400 does not lie between 300 and 400 so it is not acceptable.) the required sum = 328 + 364 = 692. 33. J, K and L will meet again at the starting point after LCM of 12, 18 and 20. LCM of 12, 18 and 20 = 180 seconds = 3 min Note: Why LCM? Because to meet again at the starting point all of them should take a certain number of complete rounds. We can see LCM (180 sec) is exactly divisible by 12, 18 or 20 seconds. So, J completes (180 ÷ 12 =) 15 rounds, K completes (180 ÷ 18 =) 10 rounds and L completes (180 ÷ 20 =) 9 rounds when they meet at starting point.

40

Quicker Maths

Chapter 9

Fractions If any unit be divided into any number of equal parts, one or more of these parts is called a fraction of the unit. The fraction one-fifth, two-fifths, three-fourths are written

1 2 3 , and respectively.. 5 5 4 The lower number, which indicates the number of equal parts into which the units is divided, is called denominator. The upper number, which indicates the number of parts taken to form the fraction, is called the numerator. The numerator and the denominator of a fraction are called its terms. Note: 1. A fraction is unity when its numerator and denominator are equal. 2. A fraction is equal to zero when its numerator alone is zero. The denominator of a fraction is always assured to be non-zero. 3. A fraction is also called a rational number. 4. The value of a fraction is not altered by multiplying or dividing the numerator and the denominator by the same number. as

2 25 24 5 55 5 4 5. When the numerator and the denominator of a fraction have no common factor, the fraction is said to be in its lowest terms. Ex.:

15 3 5 20 4 5 The numerator and the denominator have a Ex.:

other i.e. they have no common factor. 6. If the numerator and the denominator are large numbers, or if their common factors cannot be guessed easily, we may find their HCF. After dividing the terms by their HCF, the fraction is reduced to its lowest term.

385056 ; HCF of 385056 and 715104 715104 = 55008 Ex.:

Now,

385056 385056 55008 7 = 715104 715104 55008 13

7 is in its lowest term because the terms 13 have no common factor. 7. An integer can be expressed as a fraction with any denominator we want. For example: if we want to express 23 as a fraction whose denominator is 17, the process will be as follows: Here,

23

23 17 391 17 17

Proper fraction: A proper fraction is one whose numerator is less than the denominator.

3 17 21 , , are proper fractions. The 4 19 42 value of a proper fraction is always less than 1. For example:

common factor 5, so

Improper fraction: A fraction whose numerator is equal to or greater than the denominator is called an improper fraction.

3 3 , which has no common factor. Hence is 4 4

17 12 18 , , are improper fractions. 12 7 5 The value of an improper fraction is always more than or equal to 1. In the above examples,

15 in its lowest terms. 20 When a fraction is reduced in its lowest terms, its numerator and denominator are prime to each

17 5 12 5 18 3 1 , 1 , 3 12 12 7 7 5 5 Thus, we see that an improper fraction is made up of a whole number and a proper fraction. When an improper

15 is not in its lowest 20 terms. If we cancel out 5 by dividing both the numerator and the denominator by 5, we get

the fraction

For example:

Quicker Maths

42 fraction is changed to consist of a whole number and a proper fraction, it is called a mixed number.

5 5 3 In the above examples 1 , 1 and 3 are mixed 12 7 5 numbers. Fractions in which the denominators are powers of 10 are called decimal fractions. e.g.

3 7 , , 10 10

3 9 , , etc. 100 100 Fractions in which the denominators are the same are called like fractions.

13 19 8 20 1 5 17 , , , ; , , etc. For example: 17 17 17 17 12 12 12 In this case, the fraction having the greatest numerator is the greatest. 20 19 13 8 1 5 17 and 17 17 17 17 12 12 12 Fractions in which the denominator are different are called unlike fractions. So,

For example:

13 15 379 , , are unlike fractions. 17 8 1000

Solved Examples: Ex. 1: Find the sum of a) b)

1 2 3 , and 2 3 4

125 50 48 1 , , and 1 100 36 45 2

Soln: a) All the fractions are in their lowest terms. Now, LCM of 2, 3, 4 = 12 Then, 1 2 3 (12 2) 1 (12 3) 2 (12 4) 3 2 3 4 12

6 8 9 23 11 1 12 12 12 Note: Students should not write the step =

(12 2) 1 (12 3) 2 (12 4) 3 12 It should be done mentally. b) First reduce the given fractions into their lowest terms. 125 5 50 25 48 16 ; ; 100 4 36 18 45 15

Now, change the fractions into mixed numbers.

5 1 25 7 16 1 1 ; 1 ; 1 4 4 18 18 15 15 Thus,

the

given

expression

becomes:

1 7 1 1 1 1 1 1 4 18 15 2 Now, add all the whole numbers together and all the fraction together. Thus, 1 7 1 1 1 7 1 1 1 1 1 1 (1 1 1 1) 4 18 15 2 4 18 15 2

4

45 70 12 90 180

217 37 37 4 1 5 180 180 180 Ex. 2: Solve : 4

a)

7 11 13 1 – – 9 12 16 8

b) 3

10 7 9 9 5 – 2 – 4 11 15 22 10

5 17 c) 3 81 9 16 5 d) 10 91 6 4 e) 50 14 7 f)

15 3 4 20 4 5

g)

6 3 7

h)

6 4 7

Soln: a) LCM of 9, 12, 16, 8 = 144

7 11 13 1 16 7 – 11 12 13 9 – 18 – – 9 12 16 8 144

112 – 132 117 – 18 229 – 150 79 144 144 144

Fractions b) 3

10 7 9 9 5 – 2 – 4 11 15 22 10

9 9 10 7 (3 5 – 2 – 4) – – 11 15 22 10

300 154 – 135 – 297 2 330 22 1 1 2 2 2 330 15 15 5 17 c) 3 81 9 16 Change the mixed fraction into an improper fraction. Then the expression becomes: 32 17 81 306 9 16 5 65 65 1 5 91 d) 10 91 6 6 6 91 42 4 410 410 1 205 9 14 4 e) 58 14 7 7 7 14 49 49 Or, if the integral part of the mixed number be greater than the divisor, we proceed as follows:

43 Soln: "To divide a fraction by a fraction, multiply the fraction by the reciprocal of the divisor." a) First change the mixed number into an improper fraction. Then multiply the fraction by the reciprocal of the divisor.

90 108 90 31 155 17 1 23 31 23 108 138 138 4 1 85 3 5 b) 9 11 9 3 9 34 6 Compound Fractions: A fraction of a fraction is called a compound fraction. Thus,

1 3 1 3 3 of 2 7 2 7 14 Combined Operations In simplifying fractions involving various signs and brackets the following points should be remembered. (i) The operations of multiplication and division should be performed before those of addition and subtraction. (ii) Each of the signs × or ÷ should be applied only to the number which immediately follows it.

4 4 18 1 9 9 58 14 56 2 14 4 4 4 7 7 7 14 49 49

f)

15 3 4 9 20 4 5 20

6 3 ; when the numerator is perfectly divisible 7 by the divisor, divide it without changing the ÷ sign into the × sign. 63 2 Ans. = 7 7 6 h) 4 7 In this case, the numerator 6 is not perfectly divisible by 4. Hence, 6 6 1 3 4 7 7 4 14 Ex. 3: Simplify: g)

a) 3

21 15 3 23 31

4 1 b) 9 11 9 3

1 3 of is a compound fraction. And 2 7

Ex. 1.

4 7 5 4 7 24 56 5 12 24 5 12 5 25

Ex. 2.

4 7 5 4 12 5 2 5 12 24 5 7 24 7

4 7 5 4 12 24 1152 5 12 24 5 7 5 175 (iii) The operations within brackets are to be carried out first. (iv) The rule of ‘BODMAS’ is applied for combined operations. Complex Fractions: A complex fraction is one in which the numerator or denominator or both are fractions. For example: Ex. 3.

5 7, 8

8 , 5 7

8 9, 7 9

4 15 Ex. 1: Simplify (i) 2 5

1 2 2 3 are complex fractions. 3 2 – 4 5 1 2 2 3 (ii) 3 2 – 4 9

Quicker Maths

44 4 15 4 2 4 5 2 Soln: (i) 2 15 5 15 2 3 5 1 2 3 4 7 2 3 6 6 7 36 42 2 4 (ii) 3 2 27 – 8 19 6 19 19 19 – 4 9 36 36

1 1 5 24 10 1 5 25 – 1 – 24 25 – 24 1 Soln: 1 1 5 2 1 25 5 5 Continued Fractions: Fractions that contain an additional fraction in the (numerator or the) denominator are called continued fraction. Fractions of the form (i) 4

or, multiply the numerator and denominator by the LCM of the denominators 2, 3, 4, 9 namely 36. Thus,

1 2 1 2 36 36 18 24 42 4 2 32 3 2 3 2 3 2 19 – 36 – 36 27 – 8 19 4 9 4 9 Note: The second method works quickly, so we suggest that you adopt this method. Ex. 2: Simplify:

3– 7 8 5– 3

2

3

4 3 – 5 7 6 5 21 5 7 10 5 6 5 2

1

or (ii)

1

5

4–

2 5

Rule: To simplify a continued fraction, begin at the bottom and work upwards. Ex. 1: Simplify:

1 2

1

1 1 1 of 555 5 5 5 1 1 – 5 – Ex. 3: Simplify: 5 5 5 1 of 1 1 5 2 5 5 5 10

1

3

1

2

1 5 2 19

1 1 3 5 4

1 8

1 6 81 10

5

2

1 6

1

1

1 19 43 43 19

Ex. 2: Simplify: 5

Soln: 5

1 4

1

Soln: The fraction

3–

2 5 49 9 7 2 – – 1 3 7 10 2 2 2

2 3

are called continued fractions.

2

1–

5 2 5 7 21 3 – 3 5 7 10 5

3

1

1

5

1 1 3 3 –2 3– 1 6 2 2 –5 of 3 of 3 5 47 3 7 1 – 2 4– 21 7 2

2 3 10 5 3 – 7 5 7 6 3 2 2 – Soln: 7 5 7 10 5 47 – 2 3 2 21

1

5

1 10

1 10 6 81

1 81 81 5 5 496 496 496 81

1 4 3 5

1 2

1 19 5

Fractions

45

Note: No shortcut method has been derived for solving these questions. But you can solve these questions more quickly by (i) more mental calculations; and (ii) skipping the steps.

Miscellaneous Solved Examples on Fractions Ex. 1: One-quarter of one-seventh of a land is sold for 30,000. What is the value of an eight thirty-fifths of land?

1 1 1 Soln: One-quarter of one-seventh 4 7 28 Now,

1 of a land costs = 30,000 28

8 30, 000 28 8 35 of the land will cost 35 = 1,92,000 1 1 of its contents, 5 12 of the remainder was found to be 7.40. What sum did the purse contain at first?

Ex. 2: After taking out of a purse

Soln: After taking out remains with

1 of its contents, the purse 5

4 of contents. 5

1 4 1 of = 7.40 or, = 7.40 12 5 15 1 = 1111 Ex. 3: A sum of money when increased by its seventh part amounts to 40. Find the sum. Now,

S 8S = 40 = 40 S = 135 7 7 Ex. 4: A train starts full of passengers. At the first station, it drops one-third of these and takes in 96 more. At the next, it drops half of the new total and takes in 12 more. On reaching the next station, there are found to be 248 left. With how many passengers did the train start? Soln: Let the train start with x passengers. After dropping one-third and taking in 96 x 2x 96 passengers, the train has x – 96 3 3 Soln: S

passengers

2x 288 passengers 3

At the second station, the number of passengers 2x 288 12 6

2x 288 12 248 6 or, 2 x 288 1416 x = 564 Ex. 5: Determine the missing figures (denoted by stars) in the following equations, the fractions being given in their lowest terms. 3 2 (i) 6 * 30 * 3 1 1 ** –* 1 (ii) 4 6 * 17 6* Soln: (i) Since (6 + a fraction) is contained (4 + a fraction) times in 30 (because 6 × 4 is just less than 30), the integral portion of the second mixed number must be 4. Then, 3 2 6 4 30 * 3 Now,

3 30 3 90 45 3 6 or 6 * 14 14 7 7 Hence, the missing digits are 7 and 4. 1 1 ** (ii) 4 6 * – * 17 1 6 * The denominator in the first mixed number and denominator in the third mixed number are in sixties, which must be multiple of 17. Thus, the equation becomes: 1 1 ** 4 – * 1 68 17 68 1 ** 1 or, (4 – *) – 1 68 17 68

Since

1 1 1 , will borrow 1 from 4. 68 17 68

** 69 1 – 1 Then 3 – * 68 68 17 or, (3 – 2)

65 ** 65 65 1 1 1 68 68 68 68

Hence, the equation becomes: 4

1 1 65 – 2 1 68 17 68

Quicker Maths

46

EXERCISES 1. Which one of the following is the smallest fraction?

2.

3.

4.

5.

6 13 15 19 5 , , , , 11 18 22 36 6 What smallest fraction should be added to 2 7 9 1 3 6 4 5 7 to make the sum a whole 3 12 36 12 number? What must be subtracted from the sum of 7 5 13 and 4 to have a remainder equal to their 66 66 difference? Find the smallest fraction which, when added to 2 15 7 3 gives a whole number.. 5 21 10 8 Find out the missing figures (denoted by stars) in the following equations, the fractions being given in their lowest terms: 3 3 * i) * 2 14 7 * 14

1 1 1 1 144 3 1 8 * 9 143 v) 2 5 4 5 15

5 hour 6 service time every 90 days, while after overhauling, 5 it requires hour service time every 120 days. What 6 fraction of the pre-overhauling service time is saved in the latter case? 7. If the numerator of a fraction is increased by 150% and the denominator of the fraction is increased by 5 300%, the resultant fraction becomes . What is the 18 original fraction? 6. A motorcycle, before overhauling, requires

8.

* 5 * ii) 7 – * 3 3 11 *3

3 4 th of the girls and th of the boys of a primary 8 9 school participated in the annual sports. If the number of participating students is 155 out of which 92 are boys, what is the total number of students in the primary school?

2 3 of what he received to his mother. Vijay’s mother gave 5 of the money she received from Vijay to the grocer.. 8 Vijay's mother is now left with 600. How much money did Abhay have initially?

9. Abhay gave 30% of his money to Vijay. Vijay gave

9 2 16 iii) 8 * 7 ** 27 17

1 2 5 2 iv) 2 – 3 1 – 0 2 3 6 *

Solutions (Hints) 1. Out of the first two fractions, we see that 6 6 × 18 < 11 × 13, so, is smaller.. 11 6 15 and Now, from , we see that 6 22 11 15 , 11 22

6 is smaller.. 11 6 19 , we see that 19 11 6 36 , Now, from and 11 36 19 so is smaller.. 36 so

Now, from so

19 5 , and , we see that 19 6 5 36, 36 6

19 is smaller.. 36

19 we conclude that 36 is the smallest. Note: The above method does not need accurate calculations. You can decide which of the multiplications is greater by your keen observation only.

2.

2 7 9 1 3 6 4 5 7 3 12 36 12

Fractions

47

Add only the fractional parts:

2 7 9 1 3 12 36 12

24 21 9 3 57 19 7 1 36 36 12 12 Thus, to make the expression a whole number we should

add 1 –

7 5 12 12

3. Difference 13

7 5 5 7 – 4 (13 – 4) – 66 66 66 66

9

2 1 9 66 33

7 5 12 2 4 17 17 66 66 66 11 the required answer Sum 13

2 1 6 1 5 –9 17 – 9 8 11 33 33 33 33 Direct formula: The required answer = 2 × smaller value 17

24 4.

5 5 8 66 33

2 15 7 3 3 5 21 10 8 40 3 37 the required fraction = 1 – 40 40

3 3 13 5. i) 5 2 14 7 4 14 2 5 7 ii) 7 – 4 3 3 11 33 iii) 8

9 2 16 1 7 17 27 17

1 2 5 2 iv) 2 – 3 1 – 0 2 3 6 3 1 3 6. LCM of 90 and 120 = 360 So, in 360 days, the pre-overhauling service time V) * 7

5 360 10 hrs and after overhauling, the service 6 90 3

5 360 5 hrs. 6 120 2 10 5 5 Time saved – hrs 3 2 6 time

5 6 5 3 1 The required answer 10 6 10 4 3 7. Let the original fraction be So,

x . y

x 2.5 5 = y4 18 x 5 4 4 = = y 18 2.5 9

Note: Increase by 150% means the original numerator becomes (100 + 150=) 250% or 2.5 times. And similarly the original denominator becomes 4 times. Now, we can find a direct formula as given below: 5 2.5 5 4 4 Original fraction = 18 4 18 2.5 9 8. Total students who participated in annual sports =155 Boys = 92 Girls = 155 – 92 = 63

Total number of boys in school =

92 9 207 4

63 8 168 3 Total number of students = 207+ 168 = 375 9. Suppose Abhay has initially x.

Total number of girls =

30 3x Then Vijay got = x = 100 10 3x 2 x = 10 3 5 Vijay’s mother gave money to the grocer

Vijay’s mother got

x5 x = 5 8 8 Money left with Vijay’s mother

=

=

x x 8x 5x 3x = 5 8 40 40

Now,

3x = 600 40

x=

600 40 = 8000 3

Quicker Maths

48 Logical Approach: In such questions, we should think in reverse. Like 5 3 the mother is left with 1 8 th of the money,, 8 which is equal to 600. So, she must have

1 1 8 600 5 = 600 3 / 8 = 600 3 = 1600. 1 8 1 3 Similarly, Vijay must have 1600 2 / 3 1600 2

= 2400

1 100 and Abhay must have 2400 30% = 2400 30

= 8000. Combining all the above steps, we get the following quicker method: Quicker Method: 1 Abhay's initial money = 600 1 5 / 8 8 3 10 = 600 3 2 3 = 8000

1 1 2 / 3 30%

Chapter 10

Decimal Fractions Decimal Fraction: Fractions in which the denominators are powers of 10 are called decimal fractions. 1 7 9 , , , etc. 10 10 1000 Reading a decimal: In reading a decimal, the digits are named in order. Thus 0.457 is read as zero point (or decimal) four, five, seven. Decimal places: The number of figures which follow the decimal point is called the number of decimal places. Thus, 1.432 has three decimal places and 7.82 has two decimal places.

For example:

Converting a decimal into a vulgar fraction Rule: Write down the given number without the decimal point, for the numerator, and for the denominator write 1 followed by as many zeros as there are figures after the decimal point. Ex. 1: Reduce the following decimal values into the vulgar fractions. a) 0.63 b) 0.0032 c) 3.013 Soln: a) 0.63

63 100 0032 32 10000 10000

b) 0.0032

013 13 3 1000 1000 Note: Adding zeros to the extreme right of a decimal fraction does not change its value. For example: 0.9 = 0.90 = 0.9000 = 0.90000000 (Why is this so?) If numerator and denominator of a fraction contain the same number of decimal places, then we may remove the decimal sign. 0.53 053 53 Ex. 2: i) 3.21 321 321 9.83051 983051 ii) 18.53342 1853342

c) 3.013 3

iii) Change

1.53 into vulgar fraction. 2.4321

1.53 1.5300 15300 2.4321 2.4321 24321 Thus, to remove the decimal, we equate the number of figures in the decimal part of the numerator and the denominator (by putting zeros) and then remove the decimals. Note: An integer may be expressed as a decimal by putting zeros in the decimal part. Ex: 17 = 17.0 = 17.00 Addition and Subtraction of Decimals Rule: Write down the numbers under one another, placing the decimal points in one column. The numbers can now be added or subtracted in the usual way. Ex. 3: Add together 5.032, 0.8, 150.03 and 40. Soln: 5.032 Soln:

0.8 150.03 40.00 195.862

Ex. 4: Subtract 19.052 from 24.5. Soln: Here, we write two zeros to the right of 24.5 and then subtract as in the case of integers. 24.500 19.052 5.448

Multiplication of decimals I: To multiply by 10, 100, 1000, etc. Rule: Move the decimal point by as many places to the right as many as there are zeros in the multiplier. Ex.: (i) 39.052 × 100 = 3905.2 (ii) 42.63 × 1000 = 42630 II. To multiply by a whole number Rule: Multiply as in the case of integers, and in the product mark as many decimal places as there are in the multiplicand, prefixing zeros if necessary. Ex.: (i) 0.9 × 12 = ? Soln: Step I: 9 × 12 = 108 Step II: As there is one decimal place in multiplicand, the product should also have only one decimal place. So, the answer = 10.8

Quicker Maths

50 Ex.: (ii) 0.009 × 12 = ? Soln: Step I: 9 × 12 = 108 Step II: As there are three decimal places in the multiplicand, the product shuld also have three decimal places. So, the answer = 0.108 Ex.: (iii) 0.00009 × 12 = ? Soln: Step I: 9 × 12 = 108 Step II: As there are five decimal places in the multiplicand, the product should also have five decimal places. So, we prefix two zeros to get the answer. Ans = 0.00108 III: To multiply a decimal by a decimal Rule: Multiply as in integers, and in the product mark as many decimal places as there are in the case of the multiplier and the multiplicand together, prefixing zeros, if necessary. Ex.: (i) 0.61 × 0.07 = ? Soln: Step I: 61 × 7 = 427 Step II: As there are (2 + 2 =)4 decimal places in the multiplier and the multiplicand together, the product should also have 4 decimal places. But there are only three digits in the product; so we prefix one zero to the product before placing the decimal. So, the answer = 0.0427 Ex.: (ii) 0.2345 × 0.24 = ? Soln: Step I: 2345 × 24 = 56280 Step II: There should be (4 + 2 =)6 decimal places in the product. Thus, answer = 0.056280 = 0.05628

Division of decimals I: When the divisor is 10, 100, 1000, etc. Rule: To divide a decimal by 10, 100, 1000 etc., move the decimal point 1, 2, 3 etc. places to the left respectively. Thus, (i) 463.8 ÷ 10 = 46.38 (ii) 4.309 ÷ 100 = 0.04309 (iii) 0.003 ÷ 10000 = 0.0000003 (iv) 234.789 ÷ 1000000 = 0.000234789 (v) 5.08 ÷ 100000000 = 0.0000000508 II: When the divisor is a decimal fraction Rule: Move the decimal points as many places to the right in both the divisor and dividend as will make the divisor a whole number, annexing zeros to the dividend, if necessary. Divide as in simple divison and when you take a figure from the decimal part (if any) of the dividend so altered, set down the decimal point in the quotient. Ex.

32.5 325000 5078.125 (i) 0.0064 64

0.0323 3230 190 0.00017 17

(ii)

Recurring Decimals A decimal in which a figure or set of figures is repeated continually is called a recurring or periodic or circulating decimal. The repeated figures or set of figures is called the period of the decimal. For example: (1) (2)

1 0.333..... 3 1 0.142857142857142857...... 7

13 0.29545454...... 44 Recurring is expressed by putting a bar (or dots) on the set of repeating numbers. So, in the above examples:

(3)

(1) 0.333 ... = 0.3 (or , 0.3 ) (2) 0.142857 142857.......

0.142857 (or, 0.142857) (3) 0.295454... 0.2954 (or , 0.295 4 ) Pure Recurring Decimal: A decimal fraction in which all the figures after the decimal point are repeated, is called a pure recurring decimal. For example, 0.142857 is a pure recurring decimal. Mixed Recurring Decimal: A decimal fraction in which some figures do not recur is called a mixed recurring decimal. For example: 0.2954 is a mixed recurring decimal. In example (1), the period is 3, in (2) it is 142857 and in (3) it is 54. Note: 1. If the denominator of a vulgar fraction in its lowest terms be wholly made up of powers of 2 and 5, either alone or multiplied together, the fraction is convertible into a terminating decimal. Ex.:

(i)

12 12 0.48 25 52

(ii)

19 19 0.38 50 52 2

(iii)

13 13 2 0.65 20 2 5

(iv)

3 3 0.375 8 23

Decimal Fractions

51

2. A pure recurring decimal is produced by a vulgar fraction in its lowest terms, whose denominator is neither divisible by 2 nor by 5. Ex.:

2 (i) 0.6 3

3 0.27 (ii) 11

4 5 0.571428 (iv) 0.5 7 9 3. A mixed recurring decimal is produced by a vulgar fraction in its lowest terms, whose denominator contains powers of 2 or 5 in addition to other factors.

(iii)

Ex.:

By note 2, a pure recurring decimal will be produced. 1 1 18 9 2 By note 3, a mixed recurring decimal will be produced.

e)

7 7 45 9 5 By note 3, a mixed recurring decimal will be produced.

f)

(i)

1 1 0.16 6 23

1 1 g) 80 (2)4 5

(ii)

1 1 0.06 15 5 3

By note 1, no recurring decimal will be produced.

(iii)

To convert a recurring decimal fraction into a vulgar fraction

7 7 0.093 75 52 3

8 8 0.53 15 5 3 Question: Which of the following vulgar fractions will produce recurring decimal fractions?

(iv)

a)

12 50

b)

12 75

c)

3 18

d)

8 14

e)

1 18

f)

7 45

1 80 Soln: a) Reduce the fraction to its lowest terms, so

g)

Case I: Pure Recurring Decimals Rule: A pure recurring decimal is equal to a vulgar fraction which has for its numerator the period of the decimal, and for its denominator the number which has for its digits as many nines as there are digits in the period. Ex.: Express the following recurring decimals into vulgar decimals. a) 0.5

b) 0.45

c) 0.532

Soln: a) We have 0.5 0.555..... -------(1) Multiplying both sides by 10, we get

10 0.5 5.55... ----- (2) Subtracting (1) from (2), we get

9 0.5 5

5 0.5 9

12 6 6 2 50 25 (5)

Directly from the rule, we also get the same result.

By note 1, the above vulgar fraction will not produce recurring decimal.

0.5 has its period 5, so the numerator is 5. And as there is only one digit in the period, the denominator

12 4 4 b) 75 25 (5)2

will have one nine. Thus, the vulgar fraction

By note 1, the above vulgar fraction will not produce recurring decimal. 3 1 1 18 6 3 2 By note 3, a mixed recurring decimal will be produced.

c)

d)

8 4 14 7

5 9

b) 0.45 By the rule, numerator is the period (45) and the denominator is 99 because there are two digits in the period. Ans

45 99

c) 0 .532 Numerator = period = 532

Quicker Maths

52 Denominator = as many nines as the number of digits in the denominator 532 999 Try to solve Ex. (b) & (c) by detailed method. Ans

Case II: Mixed Recurring Decimals Rule: A mixed recurring decimal is equal to a vulgar fraction which has for its numerator the difference between the number formed by all the digits to the end of first period, and that formed by the digits which do not recur; and for its denominator the number formed by as many nines as there are recurring digits, followed by as many zeros as there are non-recurring digits. Ex.: Express the following as vulgar fractions. a) 0.18

b) 0.43213

Soln: a) 0.18 0.18888

c) 5.00983 --------- (1)

10 0.18 1.8888

--------- (2)

and 100 × 0.18 18.8888 -------- (3) Subtracting (2) from (3), we have,

90 0.1 8 18 – 1 17 0.18 90

Also, by the rule, numerator = 18 – 1 (all-digitnumber – non-recurring-number) and denominator = one nine (as there is one recurring digit) followed by one zero (as there is one non-recurring digit) = 90

Vulgar fraction =

18 – 1 17 90 90

b) 0.43213 ; Numerator = 43213 – 43 = 43170 Denominator = 3 nines (as there are three recurring digits) followed by 2 zeros (as there are two nonrecurring digits) = 99900 43170 4317 Vulgar fraction = 99900 9990 00983 – 00 983 5 c) 5.00983 5 99900 99900 Solve Ex. (b) and (c) by the detailed method.

Addition and Subtraction of Recurring Decimals Addition and subtraction of recurring decimals can be understood well with the help of given examples. Example: Add and subtract 3.76 and 1.4576 .

3.7 676767 67 1.4 576576 57 (+)5.2 253344 24

5.2253344

(-)2.3 100191

2.3100191

10

Explanation: Step I: We separate the expansion of recurring decimals into three parts. In the left-side part, there is integral value with non-recurring decimal digits. In the above case, 1.4576 has 1 as integral part and 4 is as non-recurring decimal digit. So, 1.4 has been separated from 1.4576576576...... Although the first value does not have any nonrecurring decimal digit (i.e. in 3.76 , both the digits 7 and 6 are recurring), we put as many digits in the left-side part as there are nonrecurring digits in the second value. Thus, we have put 3.7 in the left-side part. Step II:In the middle part, the number of digits is equal to the LCM of the number of recurring digits in two given values. In this case, the first value has 2 recurring digits and the second value has 3 recurring digits, so the middle part has 6 digits (as LCM of 2 and 3 = 6). Step III: In the right-side part take two digits. Now, add or subtract as you do with simple addition and subtraction. Step IV: Now, leave the right-side part and put a bar on the middle part of the resultant. The left-side part will remain as it is in the resultant. You can see the same in the above example.

Verification: 76 76 3 and 99 99 4576 – 4 4572 1.4576 1 1 9990 9990 76 4572 3.76 1.4576 3 1 99 9990 3.76 3

4 4

76 9990 4572 99 99 9990

76 1110 4572 11 11 9990 134652 24762 4 4 1 109890 109890

5

24762 109890

-------- (1)

Decimal Fractions Now, 5.2253344 5

53 2253344 – 2 2253342 5 9999990 9999990

24762 91 24762 5 5 ------ (2) 109890 91 109890 From (1) and (2); it is verified that our addition is correct. Now, for subtraction: 76 4572 3.76 – 1.4576 3 – 1 99 9990 2

76 9990 – 4572 99 99 9990

2

76 1110 – 4572 11 11 9990

2

34068 11 9990

Now, 2.3100191 2

Ex. 2: Find 324 .786 – 10.193 Soln: 324.78

Soln: 324.78

10.19

678 333

67 33

334.98

012

00

10.193 10

786 999

193 – 19 174 10 900 900

786 174 324.786 10.193 324 10 999 900 786 900 174 999 334 999 900 78600 174 111 97914 334 334 99900 99900

334

98012 – 98 334.98012 99900

345

34

786 900 – 174 999 999 900 786 100 – 174 111 314 99900 59286 59345 – 59 314 314 314.59345 99900 99900

Ex. 3: Find 17.83 0.007 310.0202 Soln:

17.838 0.007 310.020 327.866

38 77 22 38

38 77 22 37

Thus, the answer = 327.86638 Ex. 4: Find 17.1086 – 7.9849 Soln: 17.108

Thus, the answer = 334.98012 Verification: 324.786 324

314.59

314

91 34068 34068 2 = 2 ------(2) 91 11 9990 11 9990 From (1) and (2); it is verified that our subtraction is correct.

Ex. 1: Find 324.786 10.193

67 33

Thus, the answer = 314 .59345 Verification: 786 174 324.786 – 10.193 324 – 10 999 900

------ (1)

3100191 3 3100188 2 9999990 9999990

10.19

678 333

7.984

68 99

9.123

68

68 99 69

Thus, the answer = 9.12368

Multiplication of Recurring Decimals Case A: While multiplying a recurring decimal by a multiple of 10, the set of repeating digits is not altered. For example: 1.

4.03 10 4.030303... 10 40.30303... 40.303

2.

124.427 1000 124.427427... 1000 124427.427

3.

0.006379 10000 0.006379379... 10000

63.79379... 63.79379 Case B: While multiplying a recurring decimal by a number which is not a multiple of 10, first of all, the recurring decimal is changed into the vulgar fraction and then the calculation is done. For example:

Quicker Maths

54 63 7 84 11 7 11 11 84 99 11 11

1.

7.63 11 7

2.

13.345 15 13

345 – 34 311 15 13 15 900 900

11700 311 12011 12000 11 11 15 200 900 60 60 60

200 0.1833... 200 0.183 200.183

3.

27 1.22 5.5262 1.6

27 1

22 – 2 5262 – 52 6 5 90 9900 9

2 521 11 5471 5471 18 1 5 18 9 990 9 990 45 121.5777... 121.57

Division of Recurring Decimals Case A: When the divisor is a multiple of 10. 1.

0.06 1000 0.060606... 1000

0.000060606... 0.00006

2. 16.45379 10 1.645379 Case B: When the divisor is not a multiple of 10. Ex. 1: 17.26 2 17

26 – 2 24 1 4 1 2 17 17 90 90 2 15 2

259 240 19 19 8 30 30 30

57 63 – 6 8 8 0.63 8.63 90 90 Another way of calculation is 8

17.26 2 8

19 6.33... 8 8 0.633... 30 10

8 0.63 8.63 Second Method: 2) 17.266... (8.633... 16 12 12 06 6 06 Thus, the answer = 8.63

Ex. 2: 36.343 7 36.34343 ... 7 7) 36.34343... (5.19 35 13 7 64 63 13 The process of division repeats; so, our quotient becomes 5.1919.... i.e. 5.19 .

Approximation and Contraction Rule: Increase the last figure by 1 if the succeeding figure be 5 or greater than 5. For Example: (i) The approximate value of 0.3689 up to three decimal places is 0.369. (ii) The approximate value of 0.3684 up to three decimal places is 0.368. (iii) The approximate value of 0.3685 up to three decimal places is 0.369. (iv) The approximate value of 0.3689 up to two decimal places is 0.37. (v) The approximate value of 0.3468 up to one decimal place is 0.3. Note: Decimals which are correct at one, two, three ... places are said to be correct to the nearest tenth, hundredth, thousandth, ... respectively. Significant figures: The following examples are given to explain the meaning of the term significant figures. (a) The population of a certain place is 189000 correct to the nearest thousand. Here, the assumed unit is one thousand, and the population is stated to be 189 such units correct to the nearest unit. The figure 189, which gives the number of units, is said to be significant while the three zeros, which indicate magnitude of the unit, are said to be non-significant. (b) The distance between two places is 1400 km correct to the nearest hundred km. The figure 14 is significant and the two zeros are nonsignificant. (c) The distance between two places is 1400 km correct to the nearest km. The figure 1400 is significant but there is no non-significant figure. (d) The length of a string is 0.07 cm correct to the 2nd decimal place. This means 7 is the significant figure but the zero at the beginning is non-significant.

Decimal Fractions

55

Note:

Zeros at the beginning of a decimal are always non-significant. Thus, significant figures are those which in any approximate result express the number of units correct to the nearest such unit.

Contracted Addition Rule: Set down the decimals under one another. Then add in the usual way, taking care that the last figure retained be increased by 1 if the succeeding figure be 5 or greater than 5. Ex: Find the sum of 320.4321, 29.042934, 0.0085279 and 0.3412 correct to 3 decimal places. Soln:

Remark: If you are asked to get the answer correct to three decimal places, then use not more than 5 (two more) decimal places during calculation.

Contracted Subtraction Rule: Write down the subtrahend under the minued in the usual way, retaining two places of decimals more than what is required. Ex: Find 160.342195 – 32.0048326 correct to four decimal places. Soln: As we are asked to get the solution correct to four decimal places, we will use not more than 6 decimal places during our calculation.

160.342195 32.004832

320.4321 29.04293

128.337363

0.00852 0.3412

Answer = 128.3374

349.82475

Answer = 349.825

EXERCISES 1. Express the following decimals as fractions in their lowest terms:a) 0.0375 b) 0.00625 c) 1.008125 2. Simplify: a) 0.25 + 0.036 + 0.0075 b) 34.07 + 0.007 + 0.07 c) 30.9 + 3.09 + 0.309 + 0.039 d) 35 – 7.892 + 0.005 – 10.345 e) 0.6 + 0.66 + 0.066 – 6.606 + 66.06 3. Remove decimals a) 0.35 × 106 b) 0.275 × 10–7 c) 0.0034 × 10,000 d) 0.132 500 e) 5.302 × 513 4. Divide: a) 28.9 17 b) 0.457263 ÷ 18 c) 64 ÷ 800 d) 64 ÷ 0.008 e) 2.375 ÷ 0.0005 f) 0.1 ÷ 0.0005 g) 0.1 ÷ 5000

5. Find the quotient to three places of decimals:a) 0.5 ÷ 0.71 b) 4.321 ÷ 0.77 c) 5.002 ÷ 0.00078 6. Simplify the following: a) 12 ÷ 0.09 of 0.3 × 2 b)

0.0025 1.4 0.0175

c)

9.5 0.085 0.0017 0.19

d)

3 of 0.176 11

7. What should come in place of question mark (?) ? a) 3 × 0.3 × 0.03 × 0.003 × 300 = ? b) 0.25 0.0025 0.025 of 2.5 ? c) 0.00033 ÷ 0.11 of 30 × 100 = ? d) 0.8 × ? = 0.00004 e)

3420 ? 7 19 0.01

f)

17.28 ? 0.2 400 3.6

Quicker Maths

56 d) 2.432

0.324 0.081 4.624 1.5625 0.0289 72.9 64

g)

e) 10 .036

20 8 0.05 16 40 – ? i) ?% of 10.8 = 32.4 j) 3.79 × 31 + 3.79 × 37 + 3.79 × 32 = ? k) 321 × 11.54 – 203 × 11.54 – 105 × 11.54 = ?

1 0.0276 , then what is the value of 36.18

h)

10. If

l) (1.27)3 3(1.23)2 1.27 3(1.27)2 1.23 (1.23)3 ?

1 ? 0.0003618 11. If 13324 ÷ 145 = 91.9, then what is the value of 133.24 ÷ 9.19?

m) (2.3) 3 – 3 (2.3) 2 (0.3) 3( 2.3)(0.09) – (0.3) 3 ? 8. State in each case whether the equivalent decimal is terminating or non-terminating: a)

1 6

b)

8 625

c)

17 90

d)

104 111

33 e) 165 9. Express the following recurring decimals in their vulgar fractions:-

12. If

3 5 5 2.24 , then what is the value of 2 5 – 0.48 ?

13. If

15 3.88 , then what is the value of

14. If

2916 54 , then what is the value of

29.16 0.2916 0.002916 0.00002916 ? 15. What decimal of an hour is a second? y–x 16. If 1.5x = 0.05y, then what is the value of y x ?

a) 0.3

17. (0.6 0.7 0.8 0.3) 9000 ?

b) 0.037

18. 0.3467 (0.45 0.5) 11 ?

c) 0.09

Answers 375 3 1. a) 0.0375 10, 000 80 625 1 b) 0.00625 1, 00,000 160 8125 13 c) 1.008125 110, 00,000 11600 2. a) 0.2935 b) 34.147 c) 34.338 d) 16.768 e) 60.78 3. a) 35 10 4 b) 275 10 –10 c) 34

5 ? 3

d) 66 e) 2719926 10 3 4. a) 1.7 b) 0.0254035 c) 0.08 d)8000 e) 4750 f) 200 g) 0.00002 5. a) 0.704 b) 5.612 c) 6412.821 6. a) 888.88 b) 0.2 c) 2500 d) 0.048

Decimal Fractions

57

7. a) 0.0243 b) 6.25 c) 0.01 d) 0.00005 e) 0.257 f) 0.0024

d) 2.432 2

e) 10.036 10

0.324 0.081 4.624 1.5625 0.0289 72.9 64

g)

(0.0276) 105 2760 This question should not be solved by detailed method. If you observe the following points you will be able to answer within seconds.

324 81 4624 10 15625 289 729 64 10 –9 –9

18 9 68 3 0.024 125 17 27 8 125 h) 38.725 i) 300 j) Given expression = 3.79 (31+37+32) = 3.79(100) = 379 k) Given expression = 11.54 (321 – 203 – 105) = 11.54 × 13 = 150.02 l) Given expression = (1.27 + 1.23)3 = (2.5)3 = 15.625 m) Given expression = (2.3 – 0.3)3 = 23 = 8 8. If the denominator of a vulgar fraction in its lowest terms be wholly made up of powers of 2 and 5,either alone or multiplied together, the fraction is convertible into a terminating decimal. Following the same rule:

b)

1 is non-terminating since its denominator has a 6 factor (3) other than 2 and 5.

8 8 4 is terminating since its denominator 625 (5) is made up of powers of 5 only.

c)

d)

17 17 is non-terminating since its 90 2 3 3 5 denominator has factors other than 2 and 5. 104 104 is non-terminating. 111 3 37

33 1 is terminating. e) 165 5

9. a) 0.3

3 1 9 3

b) 0.037 c) 0.09

37 999

9 1 99 11

36 – 3 33 11 10 10 900 900 300

1 1 (10)5 10. 0.0003618 36.18 (10) –5 36.18

a)

432 – 4 428 214 2 2 990 990 495

1 1 from the decimal in 0.0003618 36.18 denominator is moved 5 positions left. ii) So, to get the answer from 0.0276 the decimal should be moved 5 places right. Note: Without going into details, we can get the

i) To get

answer. You should know that

1 is larger 0.0003618

1 . Therefore the required answer should 36.18 also be larger than 0.0276. So, the required larger value should be obtained by moving the decimal to the right. 11. We have, 13324 ÷ 145 = 91.9 or, 13324 ÷ 91.9 = 145. Then we have to find 133.24 ÷ 9.19 = ? To get the answer divide the whole number of dividend by the whole number of divisor and observe the number of digits in the whole number of quotient. In this case, when 133 is divided by 9, the quotient (whole number) will have two digits. So, our answer should have a 2-digit whole number. required answer = 14.5

than

12.

13.

3 5 2 5 – 0.48

6.72 1.68 4

5 5 3 15 3.88 1.2933 3 3 3 3 3

14. If 2916 54 , then the required expression = 5.4 + 0.54 + 0.054 + 0.0054 = 5.9994 15.

1 0.00028 (approx) 60 60

Quicker Maths

58 16. 1.5x 0.05y

x 0.05 or, y 1.5 By the rule of componendo-dividendo, we have,

y – x 1.5 – 0.05 0.935 y x 1.5 0.05

6 7 8 3 17. The given expression 9000 9 9 9 9

24 9000 24, 000 9

3467 – 3 45 5 18. The given expression 11 9990 99 9

3464 225 93642 31214 15607 . 9990 81 9990 3 9990 4995

Chapter 11

Elementary Algebra Algebraic Expressions: A number, including literal numbers, along with the signs of fundamental operations is called an algebraic expression. They may be monomials, having only one term as +6x, –3y etc. They could also be binomials, i.e. having two terms as p2 +2r, –x2 –2x etc. They may even be polynomials, i.e. having more than two terms. Addition and Subtraction of Polynomials: The sum or difference of coefficients of like terms is performed. For example, let the two polynomials be 3x2 + 2x – 3, 5x3 – 2x2 – x. The sum of these two expressions is 5x3 + (3x2 – 2x2) + (2x + x) – 3 = 5x3 + x2 + 3x – 3. The difference between these two is the subtraction of smaller expression from the greater expression i.e. 5x3 – 2x2 + x – (3x2 + 2x – 3) = 5x3 – 2x2 + x – 3x2 – 2x +3 = 5x3+ (–2x2 – 3x2) + (x – 2x) + 3 = 5x3 – 5x2 – x + 3. Ex. 1: Find the sum of

15a 2 3ab 6b2 , a 2 5ab 11b2 , 7a 2 18ab 13b2 and 26a 2 16ab 7b 2 Soln: (15a 2 3ab 6b 2 ) (a 2 5ab 11b 2 )

( 7a 2 18ab 13b 2 ) (26a 2 16ab 7b2 ) = (15a 2 a 2 7a 2 26a 2 )

(3ab 5ab 18ab 16ab)

(6b 2 11b 2 13b2 7b2 ) = 5a 2 36ab 15b 2

Ex. 2: If P a 4 a 3 a 2 6 , Q a 2 2a 3 2 3a and

R 8 3a 2a 2 a 3 Find the value of P + Q + R Soln: P + Q + R = (a 4 a 3 a 2 6)

(a 2 2a 3 2 3a) (8 3a 2a 2 a 3 ) = a 4 (a 3 2a 3 a 3 ) (a 2 a 2 2a 2 )

(3a 3a) (–6 2 8) = a4 + 0 + 0 + 0 + 0 = a 4

Remainder Theorem: This theorem represents the relationship between the divisor of the first degree in the form (x–a) and the remainder r (x). If an integral function of x is divided by x – a, until the remainder does not contain x, then the remainder is the same as the original expression with ‘a’ put in place of ‘x’. In other words, if f(x) is divided by x – a, the remainder is f(a); e.g. when f (x) x 3 2x 2 3x 4 is divided by x – 2, the remainder is f(2).

f 2 2 3 22 2 32 4 2 Note: (1) If f(x) is divided by x + a, the remainder is f(–a) (2) If f(x) is divided by ax + b, the remainder is b f a (3) If f(x) is divided by ax – b, the remainder is

b f a (4) The method of finding the remainder is as follows: I: Put the divisor equal to zero. Like, if the divisor b a II: The remainder will be a function of the value of

is ax + b then ax b 0 x

b x, i.e., f . a Similarly, you can find the other remainders. Ex. 1: Without using the division process, find the remainder when x3 + 4x2 + 6x – 2 is divided by (x + 5). Soln: Step I: Put divisor equal to zero and find the value of x. x+5=0 x=–5 Step II: The remainder will be f(–5). f 5 5 4 5 6 5 2 3

2

= –125 + 100 – 30 – 2 = – 57.

Quicker Maths

60 Ex. 2: Find the remainder when 27x 3 9x 2 3x 8 is divided by 3x + 2. Soln: Step I: 3x + 2 = 0 x =

2 3

2 Step II: Remainder is f 3

2 3

2 3

3

2 3

2

2 3

f 27 9 3 8 = – 8 – 4 – 2 – 8 = – 22

Ex. 3: If the expression Px 3 3x 2 3 and 2x 3 5x P when divided by x – 4 leave the same remainder, find the value of P. Soln: The remainder are:

R1 f (4) P(4)3 3(4)2 3 64P 45 R 2 f 4 2 4 5 4 P P 108 3

Since, R 1 R 2 , we have 64P + 45 = P + 108 or, 63P = 63 P=1 Ex. 4: Find the values of p and q when px 3 x 2 2x q is exactly divisible by (x – 1) and (x + 1). Soln: When the expression is exactly divisible by any divisor, the remainder will be zero. Now, the remainder, when the divisor is x – 1, is f(1) = p + 1 – 2 – q = 0 p – q = 1 .............. (1) and the remainder, when the divisor is x + 1, is

f 1 p 13 12 2 1 q 0 –p + 1 + 2 – q = 0 p + q = 3 .............. (2) Solving (1) & (2), we have, p = 2, q = 1 Factorisation of Polynomials: The factor theorem based on the remainder theorem is useful in the factorisation of polynomials. Factor Theorem: Let f(x) be a polynomial and a be a real number. Then the following two results hold: (i) If f(a) = 0 then (x – a) is a factor of f(x). (ii) If (x – a) is a factor of f(x) then f(a) = 0. Ex. 1: Let f x x 3 12x 2 44x 48 Find out whether (x – 2) and (x – 3) are factors of f (x).

Soln: (a) To check whether (x – 2) is a factor of f (x) we find f (a) i.e., f(2). If this becomes zero then (x – 2) is a factor of f(x) according to the factor theorem (i) of division algorithm. f (2) = 23 – 12 × 22 + 44 × 2 – 48 = 0 Hence, by the factor theorem, (x – 2) is a factor of f (x). (b) To check whether (x – 3) is a factor of f (x), we repeat the above process with f (3). f (3) = 33 – 12 × 32 + 44 × 3 – 48 = 3 f(a) = f(3) 0. Hence (x – 3) is not a factor of f (x). Ex. 2: Find out whether (3x – 1) is a factor of 27x 3 – 9x 2 – 6x 2 by the above rule. Soln: We have, 3x – 1 = 0 x

1 3

1 If (3x – 1) is a factor of f(x) then f should be 3 equal to zero. 3

2

1 1 1 1 Here, f 27 9 6 2 3 3 3 3 = 1–1–2+2=0 (3x – 1) is a factor of the above expression. Note: (1) We can see how the factor theorem has been derived from the remainder theorem: “When remainder is zero after dividing an expression.” (2) If f(a) = 0, then x – a is a factor of f(x) (3) If f(–a) = 0, then x + a is a factor of f(x) (4) If f(x) = 0, when x = a and x = b then f(x) is exactly divisible by (x – a) (x – b) i.e., (x – a) and (x – b) both are the factors of f(x). (5) If an integral function of two or more variables is equal to zero when two of these variables are supposed to be equal, then the function is exactly divisible by the difference of these variables; e.g. (b – c), (c – a) and (a – b) are the factors of

a ( b – c ) 3 b (c – a ) 3 c (a – b ) 3 Proof: Put b = c in the given expression. a(c – c)3 c(c a)3 c(a c)3 0 c(c – a)3 c(c a)3 0 (b – c) is a factor of the given expression. Similarly, it can be shown that c – a and a – b are the factors of the given expression.

Elementary Algebra

61

Conditions of Divisibility 1. x n a n is exactly divisible by (x + a) only when n is odd. e.g. a 5 b5 is exactly divisible by a + b. 2. x n a n is not exactly divisible by (x + a) when n is even. e.g. a 8 b8 is not exactly divisible by a + b. 3. x n a n is never divisible by (x – a). e.g. a 7 b 7 or a10 b10 is not divisible by a – b. 4. x n a n is exactly divisible by x + a when n is even. e.g. x 6 a 6 is exactly divisible by x + a 5. x n a n is exactly divisible by x – a (whether n is odd or even). e.g. x 9 a 9 and x10 a10 are exactly divisible by x – a. Proof: All the above statements can be proved. We are going to prove only statements (1) and (5). You should try to prove the remaining three yourself. (1) x n a n is exactly divisible by x + a. Put x + 4 = 0 x = –a Then f(–a) = (–a)n + an = 0 This is possible only when n is odd.

Soln: (26 )2n – (62 )2n (64)2n – (36) 2n Since 2n is an even integer, the above expression must be divisible by (64 + 36) {By condition (4)}. Hence, 64 + 36 = 100 is the factor of the above expression. Hence, the last two digits of its expansion must be 00. Ex. 6: For all integral values of n, the expression 7 2n – 33n is a multiple of 1) 22 2) 12 3) 10 4) 11 5) 24 2n

3n

2 n

3 n

n

n

Soln: 1; 7 – 3 (7 ) – (3 ) ( 49) – ( 27 ) By the condition (5), 49 – 27 = 22 is a factor of the expression 7 2n – 33n . Ex. 7: What should be subtracted from 3 2 27x 9x 6x 5 to make it exactly divisible by (3x – 1)? Soln: We will find the remainder by the rule of remainder. We have, 3x – 1 = 0 x

(5) x n a n is exactly divisible by x – a. Put x – a = 0 x = a

1 3

Thus, the remainder is 3

2

1 1 1 1 f 27 – 9 – 6 5 3 3 3 3 =1–1–2–5=–7 If we reduce the given expression by the remainder (–7), the expression will be exactly divisible by the given divisor. Hence, our required value is –7.

n n Then f(a) = a a 0 This is true in all cases.

Ex. 3: The expression 5 2 n 2 3n has a factor 1) 3 2) 7 3) 10 4) 17 5) None of these Soln: 4; 52n 23n (52 )n (23 ) n (25)n (8)n As we don’t know about n, by the condition (5), (25 – 8) is a factor, i.e. 17 is a factor of 5

is a factor, the last digit in the expansion should be 0. Ex. 5: Find the last two digits of the expansion of 212n – 64n when n is any positive integer..

2n

3n

2 .

n

Ex. 4: The last digit in the expansion of (41) – 1 when n is any +ve integer is 1) 2 2) 1 3) 0 4) –1 5) None of these Soln: 3; The last digit in the (41)n for any value of n is 1. Thus, the last digit in (41)n – 1 is 0. or, (41 – 1) is a factor or exact divisor of (41)n – (1) n for any +ve integer of n. Now, when 41 – 1 = 40

Theorem for Zero of a Polynomial Let f ( x ) a 0 x n a1x n –1 a 2 x n – 2 ..... a n be a polynomial with integral co-efficients. If an integer k is a zero polynomial then k is a factor of a n . Solved Example: Find all integral zeros of the polynomial,

f ( y) y 3 – 2 y 2 y 4 . Solution: Suppose (k) is an integral zero of the polynomial f(y). Then by the above theorem, k is a factor of a n i.e., 4. Hence, possible values of k are 1, –1, 2, –2, 4 and –4. Now, test each of them to see whether it is zero of the polynomial or not.

Quicker Maths

62 (i) f (1) 13 – 2 12 1 4 4. Since f (1) 0 , so 1 is not a zero of f(y). (ii) f(–1) = ( 1)3 (2)( 1) 2 ( 1) 4 0 . Since f(–1) = 0, therefore –1 is the zero of f(y). (iii) Similarly, 2, –2, 4 and –4 are not the zeros of f(y). Thus, the only integral zero of f(y) is –1. If there are other zeros of the f(y), they are not integers. Quadratic Equations: An equation in which the highest power of the variable is 2 is called a quadratic equation. For example, the equation of the type ax 2 bx c 0 denotes a quadratic equation. The product of multiplication of two linear polynomials also gives a quadratic polynomial. Let the two linear polynomials be (lx + m) and (px + q), where 1 0, p 0. Then the product of these polynomials is the quadratic polynomial, lpx2 + (lq + mp)x + mq. This is written in the standard form as ax2 + bx + c. Then a = lp, b = (lq + mp) and c = mq. Factorisation of Quadratic Equations: The quadratic polynomial ax2 + bx + c can be factorized only if there exist two numbers r and s such that (i) r = lq, s = mp (ii) r + s = b = co-efficient of x = lq + mp (iii) r × s = ac = l.p.m.q = co-efficient of x2 × constant Solved Example: Factorize 2 x 2 11x 5. Solution: (i) Here a = 2, b = 11 and c = 5 (ii) Find two numbers r and s such that r + s = b = 11 and r × s = a × c = 2 × 5 = 10 So, the numbers are 10 and 1. (iii) Now, break up the middle term 11x of the given polynomial as 10x + 1x

2x 2 11x 5 2x 2 10x lx 5 = 2x(x + 5) + l(x + 5) = (2x + 1) (x + 5) Conditions for Factorisation of a Quadratic Equation: All quadratic expressions cannot be factorized. To test whether it can be factorized or not follow the points given below:

1.

a 2 b2 (a b)(a b)

2.

(a b)2 a 2 2ab b 2

3.

(a – b)2 a 2 2ab b 2

4.

a 3 b3 (a b)(a 2 ab b 2 )

5.

a 3 b3 (a b)(a 2 ab b2 )

6.

a 3 b3 (a b)3 3ab(a b)

a 3 b3 (a b)3 3ab(a b) 8. (a b)2 (a b)2 4ab 9. (a b) 2 (a b)2 4ab 10. a 3 b3 c3 3abc (a b c)(a 2 b2 c2 – ab bc ca) 7.

1 (a b c)[(a – b) 2 (b – c) 2 (c – a) 2 ] 2 11. (a b c)3 a 3 b3 c3 3(b c)(c a)(a b) 12. a 2 b2 (a b)2 2ab

13. a 2 b 2 (a – b)2 2ab 14. If a b c 0 then the value of a 3 b3 c3 is 3abc. 15. x 2 x(a b) ab (x a)(x b) 16. a 2 (b c) b2 (c a) c2 (a b) 3abc (a b c)(ab bc ca) 17. ab(a b) bc(c b) ca(c a) 2abc (a b)(b c)(c a) 18. a 2 (b c) b 2 (c a) c2 (a b) (a b)(b c)(c a) 19. (a b c)2 a 2 b2 c2 2(ab bc ca) Theory of Indices: The expression of a m means the product of m factors, each equal to ‘a’ . When m is a positive integer, m is called the exponent or index or the power of ‘a’. Any quantity raised to the power zero is always equal to 1. For example,

a 0 1(where a 0) 0

(1) If b 2 4ac 0 , then the quadratic equation can be factorized.

1 10 1, ( 1) 0 1, 1, (3)0 1, (1000) 0 1 etc. 2 Whenever any index of the power is taken from the numerator to denominator or from the denominator to numerator, the sign of the power changes. For example,

(2) If b 2 4ac 0 , then the quadratic equation cannot be factorized. Some important formulae that are used in basic operations and in finding the factors of an expression are summarized below:

1 am a m , n a m a n a m n m a a a Law of Indices: While solving the problems of exponents, the following laws are useful: am

1

m

,

Elementary Algebra

63

(1) x m x n x m n n (2) x m x mn (3) x mn x (mn ) (4) (xy)m x m ym

xm xm 1 x m n , if m > n and n n m ; if n > m n x x x Note: The difference between (2) & (3) can be seen in the following examples: (2) (33 )2 332 729 2 (3) 33 39 36 33 = 729 × 27 = 19683 In all the above laws, m and n are positive integers. In algebra, the root problems are solved by laws of indices, (5)

for example 4

2

a means a

1

2

,

4

a means a

1

4

,a

3

4

Step IV: Divide the second divisor by the second remainder. Step V: Continue the process till the remainder become zero. Step VI: The divisor which does not leave remainder is the HCF of the two numbers and thus, the last divisor is required (HCF) of given number. L.C.M.: The minimum multiplied quantities out of two or more algebraic expressions is called LCM. If only two expressions are given, then at first we find their HCF by the factor method or by the division method and apply the following formula to find their LCM. LCM of two expressions =

means

a . 3

H.C.F.: The highest common factor of two or more algebraic expressions can be determined by the following methods: 1. By the factor method 2. By the division method Factor Method: In case of the factor method, at first, the factors of the given expressions are found separately. Then the maximum number of common factors are taken out and multiplied, the product of which becomes the H.C.F. For example, if we are asked to find the H.C.F. of 8(x 2 5x 6) and 12(x 2 9) we shall first find the factors of the given expressions separately. Thus, 8(x 2 – 5x 6) 4 2(x 3)(x 2) and 12(x 2 9) 4 3(x 3)(x 3) Then the maximum number of common elements of the two will give the HCF of the expressions. Therefore, HCF = 4 (x – 3) = 4x –12 Division Method: To find highest common factor by using division method in discussed here: Finding highest common factor (HCF) by prime factorization for larger number is not very convenient. The method of log division is more useful for large numbers. We use the repeated division method for finding HCF of two or more number. By following steps: Step I: Divide the larger number by smaller one. Step II: Then remainder is treated as divisor and the divisor or dividend. Step III: Divide the first divisor by the first remainder.

Product of the two expressions Their HCF

Thus, LCM of 8(x 2 5x 6) and 12(x 2 9) will be equal to

8(x 2 5x 6) 12(x 2 9) (by the formula) 4(x 3) 24(x 2 5x 6)(x 3) = 24(x 3 5x 2 6x 3x 2 15x 18)

24(x 3 2x 2 9x 18) Solving Quadratic Equations: A quadratic equation when solved will always give two values of the variable. These values of the variable are called the roots of the equation. Any quadratic equation can either be solved by the factor method or by a formula. (1) By the Factor Method: When we solve a quadratic equation by the factor method, we first find the factors of the given equation making the right-hand side equal to zero and then by equating the factors to zero, we get the values of the variable. For example, in solving the quadratic equation x2 – 5x + 6 = 0 we first find the factors of x2 – 5x + 6 which are (x – 3) and (x – 2). Then we say that (x – 3) (x – 2) = 0 After that, we put each bracket equal to zero and find the values of x. i.e. when x – 3 = 0, x = 3 and when x –2 = 0, x = 2 Therefore, solution is x = 3 or x = 2. (2) By Formula: When we want to solve a quadratic equation ax 2 bx c 0, we apply the following formula

b b2 4ac 2a Taking + and – signs separately, we get two values of x. The quantity b2 – 4ac is called the discriminant. x

Quicker Maths

64 The two values of x obtained from a quadratic equation are called the roots of the equation. These roots are denoted by and . It is seen that the sum of the roots b of a quadratic equation ax 2 bx c 0 is equal to , a b c i.e., and product of the root is equal to a a c i.e., . a For a quadratic equation ax2 – bx + c = 0. (i) The roots will be equal if b2 = 4ac. (ii) The roots will be unequal and real if b2 > 4ac. (iii) The roots will be unequal and unreal if b2 < 4ac. Formulating a quadratic equation from given Roots: Whenever we are given the roots of a quadratic equation x2 – x (sum of the roots) + product of the roots = 0. For example, if the roots are given as 2 and 3, then the quadratic equation will be as follows: x 2 (3 2)x 3 2 0 x 2 5x 6 0 Note: (i) A quadratic equation ax 2 bx c 0 will have reciprocal roots, if a = c. (ii) When a quadratic equation ax 2 bx c 0 has one root equal to zero, then c = 0. (iii) When the roots of the quadratic equation ax 2 bx c are negative reciprocals of each other, then c = – a. (iv) When both the roots are equal to zero, b = 0 and c = 0. (v) When one root is infinite, then a = 0 and when both the roots are infinite, then a = 0 and b = 0. (vi) When the roots are equal in magnitude but opposite in sign, then b = 0. (vii)If two quadratic equations ax2 + bx + c = 0 and a 1x 2 b1x c1 0 and have a common root (i.e., one root common), then (bc 1 – b 1c) (ab1 – a1b) = (ca 1 c1a ) 2 . (viii) If they have both the roots common, then a b c . a1 b1 c1 (ix) The square root of any negative number is an

imaginary number (e.g.

–2, –4, –25,

13 , etc), we do not 17 have to deal with the problems regarding –0.04, 20.2, 34 ,

imaginary numbers. So a simple introduction is sufficient i 2 1 , so, i 1 So, 4 ( 1) 4 1 4 2i , which is an imaginary number. Solved examples Ex. 1: Find the discriminant of the following quadratic equations. Tell the nature of the roots of the equations. Verify them. (i) x2 – 4x + 3 = 0 (ii) 3x2 + 4x – 3 = 0 (iii) –5x2 + 7x = 0 (iv) 2x2 – 6x + 7 = 0 Soln: (i) In x2 – 4x + 3 = 0, a = 1, b = –4, c = 3 2 2 D b 4ac (4) 4 1 3 16 12

4 0 and also a perfect square. So, the roots will be distinct rational numbers.

b D (4) 4 4 2 2a 2 1 2 2 1 3 and 1. (These are distinct rational numbers.) (ii) In 3x2 + 4x – 3 = 0, a = 3, b = 4, c = –3 2 2 D b 4ac (4) 4 3 (3) 16 36 52 0 and is not a perfect square. So, the roots will be distinct irrational numbers. Now, the roots are Now, the roots are

b D 4 52 4 2 13 2a 23 23 2 13 2 13 2 13 and 3 3 3 (iii) Here, a = –5, b = 7, c = 0. =

2 D b 4ac becomes b 2 7 2 49 0 and also a perfect square. So, the roots will be distinct rational numbers.

Now, the roots are

b D 7 7 2 7 7 2a 2 ( 5) 10

0 14 and i.e. 0 and 1.4. 10 10 As c = 0, we can simply solve this equation as –5x2 + 7x = 0 or, x (–5x + 7) = 0 either x = 0 or –5x + 7 = 0, i.e., 5x = 7 7 i.e. x 1.4 5 Finally, we get x = 0 and 1.4.

Elementary Algebra

65

(iv) Here a = 2, b = –6, c = 7

D b 4ac (6) 4 2 7 36 56 2

2

20 0; so the roots will be imaginary.. Now, the roots are

b D (6) 20 2a 22

6 2 5i 3 5i 3 5 3 5 i and i 2 2 2 2 2 2 2 Ex. 2: Form a quadratic equation whose roots are (i) 3 and –5 (ii) 2 and 7, and verify them. Soln: (i) The quadratic equation will be x2 – sum of the roots × x + products of the roots = 0 =

or, x 2 {3 ( 5)} x 3 (5) 0 or, x 2 (2)x (15) 0 or, x2 + 2x – 15 = 0 Verification: D = 4 + 60 = 64

The roots are

2 64 2 8 1 4 i.e. 2 1 2

3 and (–)5.

27 3 6 4 4 Ex. 4: Find the roots of the equations (i) 2x2 + 3x – 5 = 0 (ii) x2 – 8x + 7 = 0 Note: Whenever we get a + b + c = 0, one of the roots will always be 1. Soln: (i) Here a + b + c = 2 + 3 + (–5) = 0, so, 1 is one of the roots of the equation 2x2 + 3x – 5 = 0 The other root = sum of the roots – one of the 3 5 roots = 1 2 2 5 So, the required roots are 1 and 2 (ii) Try yourself.

i.e.

Ex. 5: A motorcycle travels 20 km an hour faster than a cycle over a journey of 600 km. The cycle takes 15 hours more than the motorcycle. Find their speeds. Soln: Let the speed of the cycle be x kmph then that of the motorcycle = x + 20.

distance 600 speed x

(ii) x 2 ( 2 7) x 2 7 0

Time taken by the cycle =

or, x 2 9x 14 0 Verification: D = 81 – 56 = 25

and the time taken by the motorcycle

The roots are

– (–9) 25 9 5 14 and 2 1 2 2

4 7 and 2. 2 Ex. 3: (i) If one of the roots of the equation x2 – 19x + 88 = 0 is 8, find the other root. (ii) If one of the roots of the equation 4x2 – 27x + 18 = 0 is 6,find the other root.

Soln: (i) We have the product of the roots =

600 15 hours less than the time taken by x 20 the cycle.

c 88 88 a 1

88 The other root = one root i.e. 8 11

19 b 19 a 1 the other root = 19 – 8 = 111 (ii) The required other root = sum of the roots Or, the sum of the roots =

b 27 one of the roots (= 6) a 4

600 600 15 x 20 x

i.e.

40 40 40 x 1 x 20 x x or, 40x = 40(x + 20) – x(x + 20) or, 40x = 40x + 800 – x2 – 20x or, x2 + 20x – 800 = 0

or,

x

20 (20)2 4 1(800) 2 1

20 400 3200 2

20 60 10 30 20 or 40 2 As x, the speed of the cycle, cannot be negative, so x = –40 is not acceptable. x, the speed of the cycle = 20 kmph and the speed of the motorcycle = x + 20 = 40 kmph

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66

It is the real number line. As we move right, the value becomes greater. So, 2 < 3, –3 < –2, (on the real number line as –2 is in the right side of –3, –2 is greater than –3) Also –2 < 0, –2 < 1, –1.5 < –0.5, –1.999 > –2, and so on.

Ex. 6: Solve the equation 32x 1 3x 3x 3 32 Soln: 32 x 3 3x 3x 33 32 or, 3(3x)2 – 3x = 27 × 3x – 9 or, 3m2 – m = 27m – 9 Where m = 3x or, 3m2 – 28m + 9 = 0 28 (28) 2 4 3 9 28 676 23 23 28 26 14 13 1 9, 23 3 3 When m = 9, then 3x = 9 = 32 x = 2

m

and when m =

1 1 x 1 then 3 3 3 3

x = –1 Ex. 7: For what value of m can the equation –9x2 + 12x – m = 0 be a perfect square of a linear expression? Soln: A linear expression is of the form ax + b = 0; (a 0), a and b are constants. A quadratic equation whose roots are and is given by (x – ) (x – ) = 0 If = then the equation becomes (x – )2 = 0. For both the roots to be equal, we have D = 0. So, the given equation –9x2 + 12x – m = 0 can be a perfect square of a linear expression if D = 0 or, b2 – 4ac = (12)2 – 4 × (–9) (–m) = 0 or, 144 – 36m = 0 144 m = 36 4 Verification: If we put m = 4 in the equation –9x2 + 12x – m = 0, the equation becomes –9x2 + 12x – 4 = 0 or, 9x2 – 12x + 4 = 0 or, (3x – 2)2 = 0 Clearly, 3x – 2 is a linear expression of the form ax + b. Here a = 3 and b = –2.

Quadratic Expression An expression of the form ax2 + bx + c, ( a 0 ) where a, b, c are real numbers is called a quadratic expression in x. The corresponding equation of the expression ax2 + bx + c is ax2 + bx + c = 0 Before discussing the sign scheme for the quadratic expression, let us study the concept of real number line. -ve side

+ve side

– -3 -2 -1 0 (left)

smaller no.

1 2 3 greater no.

(right)

Sign scheme for the quadratic expression Note: 1. The sign sheme for the quadratic expression is always meant for the real values of x. We cannot compare any two imaginary numbers. So to say that ri > 2i or 4i < wi is absolutely incorrect. Let and be the roots of the corresponding quadratic equation (i.e. ax2 + bx + c = 0) of the quadratic expression ax2 + bx + c (= y, suppose). Then, we have, for all the real values of x: Case I: If and are real and equal or both imaginary i.e., D 0 then y i.e., ax2 + bx + c will have the same sign as that of a, the coefficient of x2. That is, if D 0 and a is +ve, y will always be +ve, and if a is -ve y will always be -ve. Case II: If and are real and unequal i.e., if D > 0, the sign scheme for y i.e. ax2 + bx + c is as follow:

Ex. 1: Find the sign scheme for the quadratic expression x2 – 4x + 7. Soln: The corresponding equation is x2 – 4x + 7 = 0

D (4)2 4 1 7 16 – 28 –12 0 i.e. roots are imaginary. Here, a = 1 > 0. Therefore, for all the real values of x, the given expression x2 – 4x + 7 is always positive. Note: 1.For all the real values of x (x may be 0.2, 32.07, 3 , –4, – 32.07, 2 5 , etc) x2 – 4x + 7 > 0. 8 You can put x = any real number and verify that x2 – 4x + 7 > 0 Let us suppose x = –4.2, then x2 – 4x + 7 = (4.2)2 – 4 × (–4.2) + 7 = 17.64 + 16.8 + 7 > 0 When x = 0.07 then x2 – 4x + 7 = (0.07)2 – 4 × 0.07 + 7 = 0.07(0.07 – 4) + 7 = 0.07 × (3.93) + 7 = 7 – 0.2751 > 0 When x = 2 – 5 then x2 – 4x + 7 = (2 – 5 )2 – 4(2 – 5 ) + 7

Elementary Algebra

67 So, the sign scheme for the given expression –x2 + 3x + 28 is as follows:

=4+5– 4 5 –8+ 4 5 +7=1+7>0 2.Here, D = –12 < 0 i.e., the roots are imaginary.

b D i.e. The roots are 2a –(–4) 12 4 2 3i 2 3i 2 1 2 So, if we put x = 2 3 i or 2 3 i , the given expression will become zero. The expression cannot be +ve or -ve. When x = 2 3i , x2 – 4x + 7 = ( 2 3 i ) 2 4( 2 3 i ) 7

4 3i 2 4 3 i 8 4 3 i 7 = 4 + 3(–1) – 8 + 7 (we have i 2 1 ) = 4 – 3 – 1= 0 Similarly, when x 2 – 3 i, x 2 4 x 7 0 3. The sign scheme is not valid for imaginary values of x. If we put x = 4i in the given expression, we get x2 – 4x + 7 = (4i)2 – 4(4i) + 7 2

16i 16i 7 = 16 ( 1) 16i 7 = –16 –16i + 7 = –9 –16i which is imaginary and it cannot be compared with any real or imaginary number. So the purpose of the sign scheme becomes meaningless except for the roots ( 2 3 i which are imaginary) Ex. 2: For what values of x, 9x2 + 42x + 49 > 0? Soln: The corresponding equation is 9x2 + 42x + 49 = 0

D ( 42) 2 4 9 49 1764 – 1764 0 Here, the coefficient of x2 = a = 9 > 0 So, for all real values of x, 9x2 + 4x + 49 > 0 i.e. the given expression is always positive. Ex. 3: Find the sign scheme for –x2 + 3x + 28. Soln: The corresponding equation is –x2 + 3x + 28 = 0 D (3) 2 4 ( 1) 28 9 112 121 0 The roots of the equation are 3 121 3 11 8 14 , = – 4, 7 2 ( 1) 2 2 2 Here, the coefficient of x2 is –1 < 0.

(+ve) – 7(–ve)

(–ve)–4

That is, Case (i): if –4 < x < 7, –x2 + 3x + 28 > 0 Case (ii): if x < –4 or x > 7, –x2 + 3x + 28 < 0 Case (iii): if x = –4 or 7, –x2 + 3x + 28 = 0 Note: You can verify the above cases (i) and (ii) by putting such values of x as may be easier for calculation. If we put x = 0, –4 < 0 < 7 we see, –x2 + 3x + 28 = 0 +0 + 28 = 28. If we put x = –5 < –4, –x2 + 3x + 28 = –25 – 15 + 28 = –12 < 0 If we put x = 10 > 7, –x2 + 3x + 28 = –100 + 30 + 28 = –42 < 0 Ex. 4: For what range of the values of x, is 2x2 – 4x – 7 non-positive? Soln: The corresponding equation is 2x2 – 4x – 7 = 0 D ( 4)2 4 2 ( 7) 16 56 72 0 Here, the coefficient of x2 is 2 > 0 The roots

are

–(–4) 72 4 6 2 2 3 2 22 22 2 So, the sign scheme for the given expression 2x2 – 4x + 7 is as follows: (–ve) –

(– ve) for x =

23 2 2

23 2 (– ve) 2

23 2 , the given expression will be zero 2

2–3 2 23 2 x the given 2 2 expression will be negative. For other values, the expression will be +ve. and for

2–3 2 23 2 x the given 2 2 expression is non-positive i.e., the expression is either zero or negative. Note: If you want to check the sign scheme, you simply Thus, for

take an approximate value of For example, we take

2 and then proceed.

2 = 1.4 (approximately)

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68

Now, the roots are

2 3 1.4 1 3 0.7 2

(ii) is more suitable than option (i). Though option (iii) is correct, it is not as close as option (ii). Options (iv) and (v) are incorrect. Hence (ii) is the answer.

1 2.1 3.1, 1.1 As the roots have been found by approximation, so while checking for the sign scheme, you should not take such values of x which are nearer to the roots.

Summary of the above discussions

23 2 23 2 1 Putting x = 1 2 2

or a1x 2 b1x c1 0 lies in the two parts of the sign scheme

The solution of the inequality ax 2 bx c 0

2

2 x 4 x 7 2 4 7 9 0 Putting x = 5, 2x2 – 4x – 7 = 2 × 25 – 20 – 7 = 23 > 0 Putting x = –2, 2x2 – 4x – 7 = 2 × 4 + 8 – 7 = 9 > 0 Thus, we find that the sign scheme is correct. Ex. 5: The inequality of b2 + 8b 9b + 14 is correct for (i) b 5, b 5 (ii) b 5, b 4 (iii) b 6, b 6

(iv) b 4, b 4

(v) b 6, b 4 Soln: The given equality is b + 8b 9b + 14 . i.e. b2 + 8b – 9b – 14 0 or, b2 – b – 14 0. For the equation b2 – b – 14 = 0, we have 2

D (–1)2 4 1 (–14) 1 56 57 0

Here, the coefficient of b 2 1 0 and the roots are

( 1) 57 1 7.6 2 1 2

[ 57 7.6 (approximately)] = 4.3 and –3.3 So, the sign scheme for the expression b2 – b – 14 is as follows: (+ve) 4.3 (-ve)

Thus for b 3.3 or, b 4.3, b 2 – b 14 0 To decide the correct option, we draw the points of the option on the real number line of the sign scheme (+ve)

(+ve)

(–ve)

– .... -7 -6 -5 -4 -3.3 -3

4

The solution (ie, the value of x) will lie either between its roots (ie, x ) or outside its roots (ie x or

x ). Now, the question is, in which case will the value of x lie between its roots and in which case outside its roots? Remember the following two points and forget all the above discussion. For the above two inequalities. where a and a1 are +ve, (i) ax 2 bx c 0 x (ii) a1x 2 b1x c1 0 x or x Note: To remember the above points, mark that when the inequality is less than zero (ie 0), the value of x is in the smaller range (ie x )and when the inequality is more than zero (ie 0) the value of x is in the larger range (ie x or x ) Now we shall discuss the above examples again. Take Ex. 4: For 2x 2 4x 7 0, find the range of x. The corresponding equation is 2x 2 4x 7 0 and 23 2 2 Here we see the inequality is less than zero so the range of x will lie between its roots. the roots are

– (-ve) -3.3

4.3

5

6

7 .....

Now, we tally each of the options one by one. Clearly, option (i) is correct. Option (ii) is also correct. Between these two options, we observe that option

So,

23 2 23 2 x 2 2

Take Ex 5: b2 8b 9b 14

b2 b 14 0 b = –3.3, 4.3 Since the inequality is greater than zero, the value of b will lie in the larger range b 3.3 or b 4.3 So, the closest answer is (ii), ie b 5, b 4

Elementary Algebra

69

Take Ex 3: x 2 3x 28 The roots of the corresponding equation are –4 and 7. Case I: When –x2 + 3x + 28 0 Change the -ve sign of x2 ie multiply by –1. Then, x 2 3x 28 0 Now, as the inequality is greater than zero, the value of x will lie in the larger range ie x 4, x 7 Case II: When –x2 + 3x + 28 0 Then, x 2 3x 28 0 As the inequality is less than zero, the value will lie in the smaller range, ie 4 x 7 . Note: In the above case, don't follow the original sign of the inequality. First make the coefficient of x2 positive and accordingly change the sign of the inequality. Ex. 6: If x + y is constant, prove that xy is maximum when x = y. 2

2

x y x y Soln: xy = , 2 2 We know that the square of any real no. 0 2

2

x y x y So, 0 and 0 2 2 As x + y is constant, xy is the maximum when 2

x y 0 2

xy = 0 or, x – y = 0 or, x = y 2 Ex. 7: Which of the following values of P satisfies the inequality = P (P – 3) < 4P – 12? 1) P > 14 or P < 13 2) 24 P 71 3) P > 13; P < 51 4) 3 < P < 4 5) P = 4, P = +3 Soln: P(P – 3) < 4 (P – 3); P(P – 3) – 4(P – 3) < 0; (P – 3) (P – 4) < 0 This means that when (P – 3) > 0 then (P – 4) < 0 ... (i) or, when (P – 3) < 0 then (P – 4) > 0 ... (ii) From (i) P > 3 and P < 4 3

4 Which is not in choices. Hence, from (i) our answer is (4). or,

Ex. 8: The inequality 3n 2 – 18n 24 0 gets satisfied for which of the following values of n? 1) n < 2 & n > 4 2) 2 < n < 4 3) n > 2 4) n > 4 5) None of these Soln: The equation for the given inequality is 3n2 – 18n + 24 = 0 3(n 2) (n – 4) 0

n 2,4 The real number line of the sign scheme is –

+ +ve

2

-ve

4

+ve

The two values of n give the two end points of the inequality. On the above axis we have three ranges in which value of n can move. (i) n < 2 (ii) 2 < n < 4 (iii) n > 4 Put a value which is less than 2. Suppose n = 0 Then 3n2 – 18n + 24 = 0 – 0 + 24 = 24 > 0 which satisfies the inequality. Hence n < 2 is a valid value. Again put a value which is between 2 and 4. Suppose n = 3. Then 3n2 – 18n + 24 = 27 – 54 + 24 = –3 < 0 which does not satisfy the inequality. Hence 2 < n < 4 is not a valid value. Again put a value which is greater than 4. Suppose n = 5. The 3n 2 – 18n + 24 = 75 – 90 + 24 = +9 > 0 which satisfies the inequality. Hence n > 4 is a valid value. Note: The inequality (3n – 6) (n – 4) > 0 or, 3(n – 2) (n – 4) > 0 is true when (n – 2) and (n – 4) both are either +ve or -ve. When both are +ve, we have (n – 2) > 0 and (n – 4) > 0 n > 2 and n > 4 n > 4 When both are -ve (n – 2) < 0 and (n – 4) < 0 n < 2 and n < 4 n < 2 Thus required value is n < 2 and n > 4. Ex. 9: Which of the following values of x satisfies the inequality. 2x(x 2) x 12 ? 1)

3 x4 2

3 2 5) None of these 3) x 4, x –

2) 3 2x 4 4) x 4, x

3 2

Quicker Maths

70 Soln: 1; 2x(x – 2) < x + 12 or, 2x 2 – 4x – x – 12 0

Soln: 10. 3; I. 2p p 3

or, 2x 2 – 5x 12 0 or, 2x 2 8x 3x 12 0 or, (x 4)(2x 3) 0 There are two cases for the above inequality. Case I: x – 4 < 0 and 2x + 3 > 0 x < 4 and x

p p p 1 1 2 3 6 p=6

11. 1; I.

3 2

II. q 2 12q 36 0 (q 6) 2 0

q=6 Therefore p = q

3 x4 2 Case II: x – 4 > 0 and 2x + 3 < 0 3 2 Which is not possible. Although it is given in choice

12. 2; I.

or, x > 4 and x < –

(3) but both inequalities (x > 0 and x < 3 2 ) are not possible at a time. So, from case I only our answer is (1). Note: If you have no idea about quadratic equation, you can verify the inequality by putting the suitable value of x from each choice. Whichever satisfies the inequality should be the answer. Directions (Ex. 10-14): In each question one or more equation(s) is/are provided. On the basis of these you have to find out the relation between p and q. Give answer (1) if p = q, Give answer (2) if p > q, Give answer (3) if q > p, Give answer (4) if p q , and Give answer (5) if q p .

5 p3 2

5 3.5 2

Therefore q > p

or,

10. I. 2p

q 1

II.

or, 2x(x 4) 3(x 4) 0

5 1 p ; 2 2

II. q –

5 1 2

p 2 10 p 5 7 7

II. q 2 2q 1 0 (q – 1) 2 0

q 1 Therefore p > q 13. 5; I. p2 9 p 3 II. 3q q 1 5 q 3 Therefore q p 14. 2; I. p

8 64 – 28 8 6 2 14 1 , ,1 14 14 14 14 7

9q 1 1 or, q 4 8 18 q

q Give answer (3) if q > p Give answer (4) if p q and

II.

11. I.

p p – 1 2 3

II. q 2 36 12q

12. I.

p 2 – 0 5 7

II. q 2 2q – 1

15.

13. I. p2 3 12

II. 3q – 5 1 q

16. (i) p – 7 = 0

(ii) 3q 2 – 10q 7 0

17. (i) 4p 2 16

(ii) q 2 – 10q 25 0

14. I. 7p – 8p 1 0

5q q 1 – II. 2 4 8

18. (i) 4p2 – 5p 1 0

(ii) q 2 – 2q 1 0

2

Give answer (5) if q p

5 9 15 13 p q 28 8 14 16

Elementary Algebra 19. (i) q 2 – 11q 30 0 Solutions: 15. 2;

71 (ii) 2p2 – 7p 6 0

5 9 15 13 p q 28 8 14 16 or,

19. 3; (i) q

11 121 – 120 11 1 5, 6 2 2

7 49 – 48 7 1 6 ,2 4 4 4 We see that p < q or q > p (ii) p

p 15 13 8 28 13 .... (*) q 14 16 9 5 3

p>q Answer = (2) Note: (*) shows that if p = 13 then q is 3. 16. 2; (i) p – 7 = 0 (ii) 3q 2 10q 7 0

Useful points to remember about the above types of question: In such questions three combinations of equations can be asked: (a) Both equations are linear (b) One equation is linear and the other quadratic (c) Both equations are quadratic

(i) p 7

(a) Both equations are linear There are different methods to solve two linear equations.

(ii) 3q 2 3q 7q 7 0

Method I: "Find the value of p in terms of q from any of the two equations and put it in the other equation to get the value of q."

3q (q – 1) – 7(q – 1) 0

(3q – 7) (q – 1) 0 q

7 or 1 3

Take Ex 15: (i) p

pq Answer (2) Note: In a quadratic equation ax 2 bx c 0 x

b b2 4ac 2a

q

–(–10) (–10) 2 – 4 3 7 10 4 7 1, 23 6 3

17. 3; (i) 4p 2 16 (ii) q 2 – 10q 25 0 (i) p 2

10 100 – 4 1 25 5 2 We see that q > p (ii) q

18. 5; (i) 4p2 – 5p 1 0 (ii) q 2 – 2q 1 0 (i) p

5 25 – 16 5 3 1 ,1 8 8 4

(ii) p

2 4–4 1 2

We see that

p q or q p

4 3q 3 2 q .... (*) 2 2

3 3 5 Put it in (ii) 2 q q 13 4 2 2

3 2

9 8

5 2

q q 13 29 29 q 8 2 q = –4 Again we put q = –4 in (*) and get p = 4. Thus p > q

Method II: "Eliminate one of the two variables (p or q) by equating their coefficients." Take Ex 15: (i) 2p + 3q + 4 = 0 3 5 (ii) p q 13 0 4 2 5 9 (i) + (ii) ×3 5p p 10 39 0 2 4 29p 29 p 4 4 Now, put p = 4 in (i) and get q = –4. Thus p > q Both the above methods are well-known to you. Adopt whichever you find easier.

Quicker Maths

72 Method III: Graph method: it is of no use to us.

which must be known to you.

Method IV: Suppose the two equations are a1x b1 y c1 0 ... (i)

Useful conclusions: In such cases, we can't reach to answer when one value of p is less than q and the other value of p is more than q (the reason is the same as discussed in (b)). So, both the values of p are either more or less than both the values of q. This further emplies that if p1, p2, q1 and q2 are the values of p and q then either p1 p2 q1 q 2

a 2 x b2 y c 2 0 ... (ii) If we perform (i) ×b2 – (ii) ×b1 b1c 2 b2 c1 Then x a b a b 1 2

y

2 1

a1c2 a 2c1 a 2 c1 a1c 2 b1a 2 b 2a1 a1b2 a 2 b1

or

Note: We see that the denominators of x and y are the same. This does not imply that x > y if b1c 2 b2 c1 a 2c1 a1c2 as it fails when a1b2 a 2 b1 is negative. (b) One equation is linear and the other is quadratic: Take Ex 16: (i) p – 7 = 0 (ii) 3q2 – 10q + 7 = 0 Equation (ii) gives two values of q. According to the given choices, both the values of q should be either more than or less than the value of p. Why? Because, if one value is more and the other is less than p, none of the given choices match our answer. Now, if both the values of q are more than p then the sum of the two values of q should be more than 2p. And if both the values of q are less than p then the sum of the two values of q should be less than 2p. In the above case; (i) p = 7 (ii) sum of two roots of q = As

( 10) 10 3 3

10 27 p > q 3

Note: When (i) p + q = 7; and (ii) q2 – q – 6 = 0 In such a case, solve equation (ii). For each value of q find the corresponding values of p from (i). Here (ii) q = –2, 3 (i) when q = –2, p = 9 and when q = 3, p = 4 p > q (c) Both equations are quadratic: Ex 17, Ex 18 and Ex 19 are the examples of such questions. You can see the most common method to solve them as given under their solutions. The other method to solve the quadratic equations is factorisation method,

p1 p2 q1 q 2

Take Ex 19: (i) q 2 11q 30 0 (ii) 2p2 7p 6 0 (i) Sum of roots (ie q1 + q2) =

( 11) 11 1

(ii) Sum of roots (ie p1 + p2) =

( 7) 7 2 2

Thus we may conclude that q > p But what will happen when one value of p is equal to one value of q? For example: 2 I. p p 6 0 2 II. q 6q 8 0

I. Sum of roots =

1 1 1

( 6) 6 1 From the above result we conclude that q > p. But our answer is not perfect because one of the roots in the two equations are common and our answer should be q p. Now, the problem is how can we confirm the case of equality without getting the roots? [I. p = 2, –3 and II q = 2, 4] See the following relation (also given in Note (vii) on page 64) If two quadratic equations ax 2 bx c 0 and II. Sum of roots =

a1x 2 b1x c1 0 have one common root, then

bc1 b1c ab1 a1b

ac1 a1c and vice-versa.

In the above example: p2 p 6 0 and q 2 6q 8 0

2

Elementary Algebra

73

1 8 6 6 1 6 (1)(1) 1 8 (1)(6)

Ex: (1) I. 4q 2 8q 4q 8

II. p2 9p 2p 12

2

(8 – 36) (–7) = (14) 196 = 196, which implies that one root is common and hence equality holds. So, our correct answer is q p . 2

(2) I. 2p2 40 18p 2 (3) I. 6q

1 7 q 2 2

II. q 2 13q 42 II. 12p 2 2 10p

The above method of checking the equality is not much time-saving. Sometimes it is easier to get the roots. Another Method to check equality Suppose the common root is x. Then I x2 x 6 0

(4) I. 4p2 5p 1 0

II. q 2 2q 1 0

(5) I. q 2 11q 30 0

II. 2p2 7p 6 0

II x 2 6x 8 0 Now, I – II gives 7x – 14 = 0 x = 2 x = 2 is the common root of the two equations. If we put p = 2 or q = 2 in the respective equations, those should be satisfied. If we perform I – II and get the value of p or q which satisfies the given equation then equality must hold.

(7) I. q 2 15q 56 0

II. 2p2 10p 12 0

(8) I. 18p 2 3p 3

II. 2p2 10p 12 0

(9) I. p2 12p 36 0

II. q 2 48 14q

(10) I. 2p2 12p 16 0

II. 2q 2 +14q + 24 = 0

(11) I. 2p2 48 20p

II. 2q 2 18 12q

(12) I. q 2 q 2

II. p2 7p 10 0

(13) I. p2 36 12p

II. 4q 2 144 48q

(14) I. p2 6p 7

II. 2q 2 13q 15 0

(15) I. 3p2 7p 2 0

II. 2q 2 11q 15 0

(16) I. 10p2 7p 1 0

II. 35q 2 12q 1 0

(17) I. 4p2 25

II. 2q 2 13q 21 0

(18) I. 3p2 7p 6

II. 6(2q 2 1) 17q

(19) I. p2 4

II. q 2 4q 4

(20) I. p2 p 56

II. q 2 17q 72 0

(21) I. 3p2 17p 10 0

II. 10q 2 9q 2 0

(22) I. p2 3p 2 0

II. 2q 2 5q

(23) I. 2p2 5p 2 0

II. 4q 2 1

(24) I. p2 2p 8 0

II. q 2 2 7

(25) I. 2p2 20p 50 0

II. q 2 25

For example:

I. p2 p 6 0 2 II. q 6q 8 0

or p2 6p 8 0 (changing q to p) Now I – II +7p – 14 = 0 p = 2 We put p = 2 in I or II. The equations hold true, which confirms that 2 is the common root of the two equations. Another example: I. 3p2 7p 2 0 II. 15q 2 8q 1 0 (Put p = q in I × 5)

15q 2 35q 10 0

27q 9 0

Now, put

q

1 3

1 in I and II. As it satisfies the equations the equality 3

holds.

( 7) 7 3 3 ( 8) 8 Sum of roots in II = 15 15

Note: For final answer, Sum of roots in I =

Therefore, our correct answer is p q . Now let us take some more examples from Previous years’ papers.

(6) I.

4p 8 0 5 15

II. 9q 2 12q 4

Solution: (You are suggested to go through the detailed discussion under the given solutions.)

Quicker Maths

74 (1) I 4q 2 4q 8 0 q 2 q 2 0; 1 1 Sum of roots = 1 ( 7) 7 II p2 7p 12 0; Sum of roots = 1 Therefore, our first conclusion is q > p. Now, check the equality: {1 × 12 – 7) (–2)} {1× 7 – 1 × 1} = {1 × 12 – 1(–2)}2 or, {26} {6} = {14}2 which is not true. Hence, our answer is q > p. Apply another method to check the equality. I

q2 q 2 0

II

q 2 7q 12 0

Apply I-II:

q2 q 2 0

S1: +2

–1

2 1 1 1 S3: –2 +1 See the solution for S2:

II. p 2 7 p 12 0 S1:

(Put p = q in II)

–6q – 14 = 0

7 3 Put this value in I or II. If we put it in I,

q= 2

7 49 7 7 20 2 0 3 3 9 3

49 21 18 0 9 10 0 which is not true. Hence our assumption 9 that p = q is wrong. Note: Such type of equation can be solved easily if we find the roots by the method of factorisation. For example:

I q q 2 0 (q + 2) (q – 1) = 0 q = –2, 1 2

II p2 7p 12 0 (p + 3) (p + 4) = 0 p = –3, –4 So, first try to find out the factors. If it seems difficult to factorise the equations only then go for the other methods. The above method can be a short cut like: STEP 1: Multiply the coefficient of q2 with the constant (the c in ax 2 bx c 0 ). Here, in I, coefficient of q2 is +1 and the constant is –2; so the product is (+1) (–2) = – 2. Now, break the coefficient of q (ie +1) in two parts so that its product becomes –2. In this case +1 = +2, –1 are two parts. STEP 2: Now divide these two parts by the coefficient of q2, ie (+1). So the two parts remain (+2) and (–1). STEP 3: Now change the sign, ie +2 becomes (–2) and (–1) becomes (+1). These are the two values or roots of the equation. See the picturised presentation of the above method:

+3

+4

3 1 S3: –3

4 1

S2:

(2) I

–4 2p 40 18p 2

p2 9p 20 0

II q 2 13q 42 q 2 13q 42 0 Which of the three methods gives the answer easily? Naturally, the method of factorisation. If we factorise, (I) (p – 4) (p – 5) = 0 p = 4, 5 (II) (q – 7) (q – 6) = 0 q = 6, 7 So, answer is q > p. See the solution by picturised presentation (I) p2 – 9p + 20 = 0 S1:

–4

4 1 S3: +4+5 S2:

–5

5 1

(II)

q2 – 13q + 42 = 0

S1:

-7

-6

S2:

-7 1

-6 1 +6

S3: +7

See the other method (Method of assumption). ( 9) 9 I sum of roots = 1 (13) 13 II sum of roots = 1 So q > p. But without checking the equality we can't confirm our answer. So, suppose p = q. Then

Elementary Algebra

75 ( 5) 5 6 6 So, p > q. But to check equality, suppose p = q. Then

p2 9p 20 0

(II) sum of roots =

p2 13p 42 0 4p 22

p

12q 2 7q 1 0

11 2 2

Put p =

11 11 11 in (I). As 9 20 0, our 2 2 2

assumption that p = q is wrong. Therefore the final answer remains the same as q > p. 2 (3) I 6q

1 7 q 2 2

12q 2 7q 1 0

II 12p 2 2 10p 6p2 5p 1 0 By factorisation Method:

1 1 3 4

(3q – 1) (4q – 1) = 0 q ,

I

II (3p – 1) (2p – 1) = 0

1 1 p , 3 2

So, the answer is p q . See the solution by picturised presentation : (I)

6q 2 5q 1 0 Now perform (I) – 2 × (II), which gives 1 3q – 1 = 0 q 3

1 7 1 in (I), we have 12 1 0 9 3 3

Putting q =

Which is true. Hence our final answer is p q. (4) I. 4p2 5p 1 0 II. q 2 2q 1 0

1 By Factorisation: I. (4p – 1) (p – 1) = 0 p , 1 4 II. (q – 1) (q – 1) = 0 q = 1 So, answer is q p . Picturised presentation: (I)

4p2 5p 1 0

S1: –4

7 1 6q 2 q 0 2 2

S2:

–1

4 4

S1:

-3

-4

-3 12 S3: + 1 4

-4 12 + 31

S2:

–1

–1

1 4

S2:

1 1

1 1

1 4

S3:

+1

+1

By Assumption: (I) sum of roots =

5 (II) sum of roots = 2 4

Therefore q > p. Now, Suppose p = q then

I

(II)

q 2 2q 1 0

S1:

S3: +1

(II)

4p2 5p 1 0

4 II 4p2 8p 4 0 S1:

-3

-2

-3 6 S3: + 1 2

-2 6 + 31

S2:

3p 3 0 p 1 Put p = 1 in I. 4 – 5 + 1 = 0, which is true, hence our final answer is q p . (5) By Factorisation (I) (q – 6) (q – 5) = 0 q = 5, 6

By Method of Assumption: (I) sum of roots =

(7) 7 12 12

3 (II) (2p – 3) (p – 2) = 0 p , 2 2 So, the answer is q > p.

Quicker Maths

76 Note: Try to solve these equations by picturised presentation. This saves time as well as space for writing. Don't write Step 1, Step 2, Step 3 in three separate lines. Change the appropriate forms in the same line to save your time and space. From the sum of roots it is clear that ( 11) ( 7) ; hence q > p. 1 2 But also suppose p = q. Now, 2p2 7p 6 0 54 18 2p2 22p 60 0 p 15 5 15p 54 0 2

18 18 Putting it in I, we get 11 30 0 5 5 or, 324 – 990 + 750 0 Hence our assumption (p = q) is wrong. So, the final answer is q > p.

4p 8 2 0 12p 8 p (6) I. 5 15 3 2 3

II. 9q 2 12q 4 0 3q 2 0 q 2

Therefore p = q.

(8) (I) 6p2 p 1 0 By Factorisation:

(II) 14q 2 9q 1 0

(II) (7q + 1) (2q + 1) = 0 q

1 1 , 7 2

Therefore, the answer is p q Note: (I) 6p2 p 1 0

(II) 14q 2 9q 1 0

S1: +3

–2

S1:

3 6

2 6

S2:

1 2

7

14p 9p 1 0

3

2

1 3

+7

S3:

7 14 1 2

+2

2 14

1 7

p

20p 10 0

1 2

2

1 1 Put it in I, Then. 6 1 0 2 2

3 1 1 0 , which is true. Hence our final answer is 2 2

or,

p q. (9) By Factorisation: (I) p2 12p 36 0

p 62 0

p=6

(II) q 2 14q 48 0 (q – 6) (q – 8) = 0 q = 6, 8 Therefore, our answer is q p. By Assumption: (I) Sum of roots

( 12) 12 1

Mark that p has two values each equal to 6

( 14) 14 1 Thus q > p. Now suppose p = q. Then (I) – (II) gives 2p – 12 = 0 or p = 6. When we put it in (I) 36 – 12 × 6 + 36 = 0. Which is true. Hence, the final answer is q p . (II) Sum of roots =

(10)

1 1 (I) (3p – 1) (2p + 1) = 0 p , 3 2

S3:

6p2 p 1 0

=

(7) I q 2 15q 56 0 II p2 5p 6 0 By Factorisation: (I) (q – 7) (q – 8) = 0 q = 7, 8 (II) (p – 3) (p – 2) = 0 p = 2, 3 Therefore, the answer is q > p. Note: Try to solve these two equations in a single-line step.

S2:

1 9 , p q. 6 14 Now suppose, p = q. Then By Assumption: As

(I) p2 6p 8 0

(II) q 2 7q 12 0 By Factorisation: (I) (p + 4) (p + 2) = 0 p = –2, –4 (II) (q + 4) (q + 3) = 0 q = –3, –4 We can't make any conclusion in such question. If we say p q, then –4 should be more than –3. Which is not true. Also, when we say q p, then –4 should be greater than –2, which is not true. Hence we can't answer this question. Note that although this question has been asked in a bank exam. You are suggested to leave such questions. (11) By Factorisation: (I) p2 10p 24 0 (p – 6) (p – 4) = 0 p = 4, 6

Elementary Algebra

77

(II) q 2 6q 9 0 (q – 3) (q – 3) = 0 q=3 Therefore our answer is p > q. By Assumption: Compare the sum of roots. As 10 > 6, p > q. Now, suppose, p = q and perform (I) – (II) then –4p +

(15) (I)

3p2 7p 2 0

Step 1: –6

–1

Step 1:

–6

–5

6 3

1 3

Step 2:

6 2

5 2

1 3

Step 3:

+3

Step 2:

2

15 = 0 p

15 15 15 . Put it in I: 10 24 4 4 4

= 225 – 600 + 384 0. Hence our final answer remains the same as p > q. (12) By Factorisation: (I) q 2 q 2 0 (q + 2) (q – 1) = 0 q = 1, – 2 (II) p2 + 7p + 10 = 0 (p + 5) (p + 2) = 0 p = –2, –5 Therefore, the answer is q p By Assumption: As –1 > –7, q > p Now, put p = q and do (I) – (II) then –6p – 12 = 0 p = –2. As it satisfies equation (I) our assumption (p = q) is true. Hence final answer is q p. (13)

(16) (I) 10p2 7p 1 0

(II)

Step 1: –5

Step 1:

Step 2:

2p2 12p 14 0

(17)

25p 29

29 , so None of the equations is satisfied with the value 25 our assumption (p = q) is wrong. Note: Now onwards, the solutions by factorisation will be presented in the picturised form; solution by assumption will not be given. You are suggested to solve the following questions by that method also.

2 10

35q 2 12q 1 0

Step 2: Step 3:

2 (I) 4p 25 p

(II)

–7

–5

7 35

1 5

5 35

1 7

25 5 4 2

2q 2 13q 21 0

Step 1: –7

–6

7 2

6 2

Step 2:

7 +3 2 Therefore q > p. Step 3: (18) (I)

29 25

5 10

1 1 2 5 Therefore, p q.

Step 1:

p

–2

Step 3: +

3 (II) (2q + 3) (q + 5) = 0 q , 5 . 2 Therefore, p > q By Assumption: We compare the sum of roots.

2p2 13p 15 0

5 2

Therefore, q > p.

(II) q 2 12q 36 0 As both are the same equations, p = q (14) By Factorisation: (I) (p – 7) (p + 1) = 0 p = 7, –1

13 p q. 2 Now, suppose p = q. Then II – 2 × I

Step 3: +2

(I) p2 12p 36 0

6

(II) 2q 2 11q 15 0

Step 2:

3p2 7p 6 0

+9

9 3

Step 3: –3

(II) 12q 2 17q 6 0

–2

Step 1:

–9

–8

2 3

Step 2:

9 12

8 12

2 3

Step 3:

3 4

2 3

Therefore, q p. (19) (I) p2 = 4 p = +2, –2

(II)

Step 1:

q 2 4q 4 0 +2

+2

Quicker Maths

78 2 1 –2

2 1 –2

Step 2: Step 3:

Therefore, p q. p2 p 56 0

(20) (I)

Step 1: +8

–7

8 7 1 1 Step 3: –8 +7 Therefore, q > p. Step 2:

(II)

q 2 17q 72 0

Step 1:

–8

–9

8 1 +8

9 1 +9

Step 2: Step 3:

(21) (I) 3p2 17p 10 0

(II)

(24) (I)

p2 2p 8 0

Step 1:

+4

4 2 1 1 Step 3: –4 +2 (II) q2 = 9 q = ±3, –3 Although this question is from the previous paper asked in BSRB Mumbai, yet no conclusions can be drawn. You are suggested to leave such questions. Step 2:

(25) (I)

Step 1:

+15

+2

Step 1:

+5

+4

Step 2:

15 3

2 3

Step 2:

5 10

4 10

Step 3:

1 2

2 5

Step 3: –5 Therefore, q > p.

p2 3p 2 0

(22) (I) Step 1:

+2

Step 3:

(II) 2q 5q 0 q(2q – 5) = 0

5 q = 0, 2

Therefore, q > p. (23) (I)

2p2 5p 2 0

Step 1:

+4

Step 2:

4 2

1 2

Step 3: –2

1 2

Therefore, q p

5 1 –5

(II) q 2 25 q 5, 5 Therefore, q p. Conditions for the Maximum and the Minimum Values of the Quadratic Expression ax2 + bx + c (= y, suppose) For the minimum value of y, the condition is a > 0. (if a < 0, y has no minimum value). The minimum value

–D b which is possible when x – 4a 2a Ex. 1: If x be real, find the maximum value of –2x2 + x + 3 and also find the corresponding value of x. Soln: The corresponding equation is –2x2 + x + 3 = 0 of y

D (1) 2 – 4 (–2) 3 1 24 25 The required maximum value of the given

+1

2 2 (II) 4q 1 q

5 1 Step 3: –5 Step 2:

+5

of y

1 1 –1

2

+5

–D b which is possible when x – 4a 2a For the maximum value of y, the condition is a < 0 (if a > 0, y has no maximum value). The maximum value

+1

2 1 –2

Step 2:

p2 10p 25 0

10q 2 9q 2 0

Step 1:

2 3

–2

–25 25 –D 3.125 and 4 (–2) 8 4a the corresponding value of expression

x

1 4

1 1 q , 2 2

b 1 1 – 0.25 2a 2 (–2) 4

Explanation: 2x 2 x 3 y .......(i) or, –2x 2 x (3 – y) 0

Elementary Algebra

79 or, 16x2 – 40x + 25 = 0 or, (4x – 5)2 = 0 or, 4x – 5 = 0

D (1) 2 – 4 (–2) (3 – y)

1 8(3 – y) 1 24 – 8y 25 – 8y Given that x is real, so D 0 or 25 – 8y 0 ,

y

25 25 i.e. the maximum value of y is 8 8

Substituting y

Ex. 3:

25 in (i), we get 8

25 8 or, 16x 2 8x 24 25 or, 16x 2 8x 1 0 2x 2 x 3

5 1.25 4 A certain number of tennis balls were purchased for 450. Five more balls could have been purchased for the same amount if each ball was cheaper by 15. Find the number of balls purchased. 1) 15 2) 20 3) 10 4) 25 5) None of these Suppose he purchased x balls. Then comparing the prices in two conditions, we x

or 25 8y

Soln:

get an equation

or, (4x – 1)2 0 or, 4x – 1 0

Ex. 2: Soln:

1 x 4 0.25 If x be real, find the minimum value of 2x2 – 5x – 3 and also find the corresponding value of x. The corresponding equation is 2x2 – 5x – 3 = 0 D = (–5)2 – 4 × 2 × (–3) = 25 + 24 = 49 The required minimum value of the given

or,

30 30 1 x x5

or,

30 30 x 5 x x 5

or, 30(x 5) x(35 x) or, 30x 150 35x x 2 or, x 2 5x – 150 0 or, x 2 15x – 10x – 150 0

D 49 49 – – 6.125 expression 4a 42 8 and the corresponding value of –

b (–5) 5 – 1.25 2a 22 4 Explanation: 2x2 – 5x – 3 = y ..... (i) or, 2x2 – 5x – (y + 3) = 0

450 450 15.......(*) x x 5

or, x(x 15) – 10(x 15) 0

–

Note:

or, (x – 10) (x 15) 0 x = 10, –15 Neglecting the -ve value we find the no. of balls = 10. (1) In equation (*)

450 x When we get 5 balls more for the same amount the

D (–5) 2 4 2 (y 3)

Cost price of a ball

25 8(y 3) 49 8y Given that x is real D 0 or 49 8 y 0 or 49 (–8) y

cost price of a ball

49 or, y (–8) 6.125 i.e. the minimum value of y is –6.125 and substituting y –

49 in (i), we get 8

49 8 or, 16x2 – 40x – 24 + 49 = 0 2x2 – 5x – 3 = –

450 x 5 We are given that price of a ball in second case is 450 450 15 x x5 Quicker Approach: Equation (*) is our first step of the solution. From this very first step we see that we are going to be traped in a quadratic equation. Naturally, it will take more time to solve it by solving the quadratic equation. So, we suggest you to stop your further calculation and look at the choices given. cheaper by 15. So,

Quicker Maths

80 Choice (1): Put x = 15. It can’t be our answer because 450 at the right-hand-side we get ; which is not 15 5 a complete number. Choice (2): Put x = 20. It also gives absurd values. Choice (3): Put x = 10. This satisfies the equation and give meaningful values. So, our answer is (3). Ex. 4: A businessman knows that the price of commodity will increase by 5 per packet. He bought some packets of this commodity for 4,500. If he bought this packet on new price then he gets 10 packets less. What is the number of packets bought by him? 1) 90 2) 100 3) 50 4) 125 5) None of these Soln: Suppose he bought x packets. 4500 Then cost price of a packet x When he gets 10 packet less then cost price of a 4500 packet x – 10

4500 4500 5 .......(*) x x – 10 The above equation is a quadratic equation so we should stop our further calculation (to save time). Put the value of x from choices given in the question. The value of x and x – 10 should be such that they divide 4500 exactly. So, our correct answer should be 100. Note: (*) should be our first and last step in examination hall.

(ii) Now, four years ago, their ages were x – 4 and y – 4 years respectively. The statement that his father was six times older than Ravi four years ago, forms another equation i.e., 6(x – 4) = y – 4 . Converted to general form of ax + by + c = 0, it looks like 6x – y – 20 = 0. (iii) The two simultaneous equations are: (1) 4x – y + 0 = 0 (2) 6x – y – 20 = 0 a1 b1 4 –1 (iv) Test for common solution a b i.e. 6 –1 2 2 Therefore, there exists an ordered pair which can satisfy both the equations. (v) Solving for x by elimination, by subtracting (2) from (1), we get –2x + 20 = 0 (by transposition). x 10; putting the value of x = 10 in (1) 4x – y + 0 = 0 40 – y = 0 y 40 The present ages of Ravi and his father are 10 and 40 years respectively.

Now, we have,

Miscellaneous Examples If x

Soln:

x2

Ex 2:

If x

Soln:

1 1 1 x 2 x – 2.x. 9 – 2 7 x x x

2

Linear Equations Equations with One Variable: A statement of equality that contains an unknown quantity or variable is called an equation. The graph of such an equation is a straight line, whose abscissae x or ordinates y satisfy the given equation. Root or Solution: Any value of the variable that makes the statement of equation true is called a root of the equation. Solved Example: Ravi’s father is four times as old as Ravi. Four years ago, his father was six times as old as he was then. Find their present ages. Solution: (i) Suppose the present age of Ravi is ‘x’ years, and the present age of his father is ‘y’ years. Ravi’s father is four times older than Ravi. This statement forms an equation, i.e., 4x = y. Converted to general form of ax + by + c = 0. It looks like 4x – y + 0 = 0.

1 1 2 the value of x 2 2 ? x x

Ex 1:

1 1 1 x – 2(x) 4 – 2 2 2 x x x

1 1 3 , the value of x 8 8 ? x x 2

2

2

x4

1 2 1 1 x 2 – 2.x 2 . 2 49 – 2 47 x4 x x

x8

1 4 1 1 x 4 – 2.x 4 . 4 (47)2 – 2 2207 x8 x x

2

Ex. 3:

If x + y = 3, xy = 2, find the value of x 3 y 3

Soln:

x 3 – y 3 ( x – y)(x 2 y 2 xy) Now, x – y (x y)2 – 4xy 9 8 1 x 2 y 2 xy (x y)2 – xy 9 – 2 7 x3 – y3 1 7 7

Elementary Algebra

Ex 4:

x

If

81 Similarly,

1 21 , x

the

value

of

3 1 ( 3 1) 2 2 3 3 1 ( 3 1) ( 3 1)

1 2 1 x 2 x is x x

x 2 y2 (2 – 3) 2 (2 3)2 2[(2)2 ( 3)2 ] 14 Ex. 9:

As (a – b) 2 (a b) 2 2(a 2 b 2 ) If y varies as x and y = 7 when x = 2, find y when x = 5.

Soln:

If y varies as x then

2

Soln:

x2

1 1 1 x 2.x. 21 2 23 2 x x x 2

2

1 1 1 x x – 4.x. 21 4 25 x x x

1 x x 25 5

1 2 1 x x x x 2 5 23 115 Ex. 5:

273n 1 .81– n 9 n 5 .33n –7

Soln:

27 33 273n 1 33(3n 1) 39n 3

y = constant x

7 y 35 y 2 5 2 3

2

Ex. 10: If x + 2 exactly divides x 6x 11x 6k; find the value of k. Soln: If x + 2 divides the given expression, then the given expression must be equal to zero when x equals –2, ie., f(–2) = 0 (–2)3 6(–2) 2 11(–2) 6k 0

81 34 (81) – n 3–4n

8 24 22 6k 0 6k 6 k 1

9 32 9 n 5 32(n 5) 32n 10 39n 3 .3–4n the given expression becomes 32 n 10 .33n –7

x y Ex. 11: What must be added to y to make it ? x The required value

y x y2 – x 2 x y xy

3(9n + 3 – 4n) – [(2n + 10) + (3n – 7)] 30 = 1

Soln:

Ex. 6:

If x = 12, y = 4; find the value of (x y) x y .

Ex. 12: What is the square root of

Soln:

(x y)

Ex. 7:

If a + b + c = 0, find the value of a 3 b3 c3 .

Soln:

a 3 b3 c 3 (a b c) 3 3(b c)(c a)(a b)

(x 2 4x 4)(x 2 6x 9) (x 2)2 (x 3) 2 the square root is: (x + 2)(x + 3) Ex. 13: If 3x – y = 27 and 3x + y = 243, find the value of x.

0 – 3(b c)(c a)(a b) –3(–a)(–b)(–c)

Soln:

x

y

(x 2 4x 4)(x 2 6x 9) ?

(16)3 4096

a 3 b3 c3 3abc

Soln:

(Remember)

35 243 3x y

3 1 3 –1 ,y Ex. 8: If x find the value of 3 –1 3 1

5 x y -------- (2) Adding (1) and (2), we get (x y) (x – y) 5 3

x 2 y2 . 2

Soln:

3 1 3 1 x y 3 – 1 3 1 Now, 2

2

2x 8 x 4.

2

3 –1 ( 3 – 1) 2 4–2 3 2– 3 3–1 3 1 ( 3 1)( 3 – 1)

33 27 3x –y x – y = 3 ------ (1)

Ex. 14: If x = 9, y 17 , the value of (x 2 – y 2 ) equal to ____? Soln:

(x 2 – y 2 )

–1

3

(81 – 17)

–1

3

3

1

3

is

1 1 1 3 . 81 – 17 64 4

Quicker Maths

82 Ex. 15: Find the value of x bc Soln:

bc

x c a

c a

x a b

On rearrangement, we obtain:

a b

x 2 x(a b – 2c) ab – bc – ca 0 . Now, the sum of roots

We have: x ( b c)(b c) x (c a )(c a ) x (a b)(a b) = x (b xb

2

2

–c 2 )

x (c

2

–a 2 )

x (a

–c2 c 2 –a 2 a 2 –b 2

2

–(a b 2c) 0 a b 2c. Product of roots = ab – bc – ca

– b2 )

x 0 1.

Ex. 16: The roots of 2kx 2 5kx 2 0 are equal if k is equal to _____ . Soln: The roots will be equal if in ax2 + bx + c = 0, b2 = 4ac. Hence, here the roots are equal if: (5k)2 4(2k)2 0

a b ab – c(b c) ab (a b) 2

Ex. 19: Find the roots of the equation, Soln:

We have, on squaring. 3y + 1 = y – 1 2y 2 y 1

k (25k 16) 0

But y = –1 means

16 25 Ex. 17: What is the condition that one of the roots of the

Ex. 20: If , are the roots of the equation

2

equation is double the other in ax bx c 0 ? Let one root be equal to Then other root will be 2 Now, we know that the sum of roots

(2 ) 2 2

Soln:

–b c and the product of roots a a (Note)

2 3

2 3a c 0 On adding, we get

c ------- (1) a

2

2 3a( ) 2c 0

3a but = sum of roots 1 3a . 5 + 3a(–3a) + 2c = 0

b2 2b2 –b 2 2 2 2 2 ------- (2) 3a 9a 9a

9a 2 5 2c a

Hence, from (1) and (2) the condition is:

1 1 1 is zero. xa xb c Find the product of the roots. Multiplying both sides of the equation by c(x a)(x b) we obtain: c(x + b) + c(x + a) = (x + a)(x + b)

2 2 x 2 3ax c 0 and if 5 , find the value of a. Since , satisfy the given equation, we must have:

2 3a c 0

–b and a

2b2 c 2b2 9ac 9a 2 a Ex. 18: The sum of the roots of the equation

y 1 2 which is not a

real number. Hence, no real root exists.

i.e. if k = 0 or if k

Soln:

3y 1 y 1

25k – 16k 0 2

Soln:

–1 2 (a b2 ) , on simplification. 2

5 2c 9

Ex. 21: If , are the roots of the equation

Soln:

2 2 x 2 3ax 2a 2 0 and if 5 , find the value of a. In the previous example, we obtained:

9a 2 5 2c Now, put, c 2a 2 We get, 9a 2 5 4a 2 5a 2 5 a 2 1 a 1

Elementary Algebra

83

Ex. 22: The area of a rectangle is the same as that of a

Soln:

35 cm. If the length of the 11 rectangle exceeds its breadth by 3 cm; find the dimensions of the rectangle. circle of radius

Soln:

Area of the circle

Thus, x 3 y 3 z 3 3xyz (From Ex. 7) Therefore,

22 35 35 7 11 11

22 35 10sq.units. 7 11

Let the breadth be x. Then the length is x + 3. Area of rectangle = Area of circle

a b 2c3 b c 2a 3 c a 2b3 = 3 a b 2c b c 2a c a 2b Ex. 25: Find the value of x 3 y 3 z 3 3xyz when Soln:

x(x 3) 10 x 2 3x 10 0. x

or, x 2 y 2 z 2 2 78 256 or, x 2 y 2 z 2 100 Now, we have,

x 3 y 3 z 3 3xyz

(x y z) x 2 y 2 z 2 (xy yz zx)

x 15x 100 x 15 x 100 0 Solving the quadratic equation using the formula, 2

b b2 4ac we obtain, 2a x = 5, –20. But radius can't be negative. Hence, we accept x = 5. Now, the volume of the pipe is x

= r 2 h = 3.14 52 5 15 = 3.14 × 25 × 20 = 1570 cub.m. Ex. 24: Show that

a b 2c3 b c 2a 3 c a 2b3 = 3 a b 2c b c 2a c a 2b

16 100 – 78 352

Ex. 26: Find the value of a 2 b 2 c 2 2ab 2ac 2bc , if a = x + y, b = x – y and c = 2x – 1 Soln: a 2 b 2 c 2 2ab 2ac 2bc a 2 (–b) 2 c 2 2(a)(–b) 2ac 2( b)c

a ( b) c (a b c) 2 2

6.28x (x 15) 628 x(x 15) 100 2

x + y + z = 16 and xy yz zx 78 x + y + z = 16 or, (x + y + z)2 = 256 or, x 2 y 2 z 2 2(xy yz xz) 256

3 32 4 1 ( 10) 2

3 49 3 7 5, 2. 2 2 But breadth cannot be negative. Hence, we discard x 5 and accept x = 2. Breadth = 2 cm, Length = 5 cm. Ex. 23: The surface area of a pipe, open at both ends, is equal to 628 sq. m. The difference between its radius and its length is 15 m; the length being the larger. If the pipe was closed at one end, what amount of water can it hold? (Use 3.14 ). Soln: Let radius = x. Then length = x + 15. Now, we have surface area = 2rh 628

Let x = a + b – 2c y = b + c – 2a z = c + a – 2b Now, we see that x + y + z = 0

(x y x y 2x 1)2 2x 2y 1

2

Ex. 27: Find the combined product of (a 4 b4 )(a 2 b 2 )(a b)(a b) Soln:

Given expression (a 4 b4 )(a 2 b2 )(a 2 b2 ) (a 4 b4 )(a 4 b4 ) a 8 b8

3 Ex. 28: Find the value of x

1 1 x a 3 when x x 3

Soln:

x

1 1 a or, x a 3 x x

Since, (a b) 3 a 3 b 3 3ab(a b) , we have 3

3

1 1 1 1 3 3 x x 3x x a x x x x

Quicker Maths

84 3 or, x

Ex. 32: If a + b + c = 0, then find the value of

1 3a a 3 x3

1 1 1 2 2 2 2 2 2 b c a c a b a b2 c2 a+b+c=0 or, a + b = –c 2

1 3 3 x x 3 a 3a Ex. 29: Find the value of Soln:

3n 4 – 6 3n 1 3n 2

3n 4 2 3n 2 3n 2 (32 2) 3n 2 3n 2

Ex. 30: If x 2 Let 2

1

3

1

3

2

13

32 2 7

, find the value of 2x 3 6x .

a , then 2

13

Now, we have x a

1 2

1

3

1 a

1 1 1 1 2 a 3 3 3 a a 3 a a a a a

1 3 1 3 = 2 a 3 2 2 2 3 a 2 2 Ex. 31: Find the value of

(a b) 2 (b c) 2 (a c) 2 . (b c)(c a) (a b)(c a) (a b)(b c) Given expression

(a b)3 (b c)3 (a c)3 (a b)(b c)(c a)

3(a b)(b c)(c a) 3 (a b)(b c)(c a) Note:

1 1 1 2bc 2ac 2ab

(a b c) 0 , since a + b + c = 0 2abc Ex. 33: If x + y + z = 0, then find the value of

(x y)(y z)(z x) . xyz

1 a

3 1 1 So, 2x 3 6x 2 a 3 a a a

Soln:

or, (a b)2 ( c) 2 or a2 + 2ab + b2 = c2 or, a 2 b2 c2 2ab Similarly, a 2 c 2 b2 2ac and b2 c2 a 2 2bc Hence, the given expression

3n 4 6 3n 1 3n 4 2 3 3n 1 3n 2 3n 2

Soln:

Soln:

Soln:

x+y+z=0 x + y = –z; y + z = –x; z + x = –y

The given expression

( z)( x)( y) 1 xyz

Ex. 34: If a + b + c = 0 then find the value of

Soln:

a 4 b4 c4 . b 2c 2 c 2 a 2 a 2 b 2 We have, a + b + c= 0 or, (a b c) 2 0 or, a 2 b 2 c 2 2(ab ac bc) Squaring both sides; (a 2 b2 c2 )2 4(ab bc ca) 2

or, a 4 b 4 c 4 2a 2 b 2 2b2 c2 2a 2c 2

4 a 2 b2 b 2c 2 a 2c 2 2abc(a b c)

We see that, a – b + b – c + c – a = 0 Therefore, by the well-known conditional identity, when x + y + z = 0

or, a 4 b4 c 4 2(a 2 b2 b2 c 2 a 2c 2 ) ( a + b + c = 0)

then x 3 y 3 z 3 3xyz we have

a b3 (b c)3 (c a)3 3(a b)(b c)(c a)

a 4 b4 c4 2 a 2 b2 b 2 c 2 a 2 c 2

Ex. 35: If x + y = 2z, then find the value of

x z . xz yz

Elementary Algebra Soln:

85

We have, x + y = 2z or, x z z y Thus, the given expression

1 y x y 1 -- - - - - (1)

x z xz zy 1 zy zy zy zy

Ex. 36: If x

Again, y

z

1 1 1 and y 1 then find the value y z

x

-------- (2)

1 1 y 1 y = z x 1 y y 1 1 y 1

1 1 y

x 1

1 1 y

From (1) & (2), the given expression

1 of z . x Soln:

1 1 1 1 y z z

Ex. 37: If a = b = c, then find the value of 1 y 1 y y

Soln:

The given expression

(a b c) 2 . a 2 b2 c2

(3a) 2 9a 2 2 3 3a 2 3a

EXERCISES 1. The values of a and b for which 3x 3 ax 2 74x b

6.

is a multiple of x 2 2x 24 are 1) a = –5, b = 24 2) a = 5, b = 24 3) a = 13, b = 16 4) a = –13, b = 16 5) None of these 6

3

2. The remainder when 4x 5x 3 is divided by

7.

x n y n is exactly divisible by 1) x – y 2) x + y 3) Both x – y and x + y 4) Neither x – y nor x +y 5) None of these 1 (a b c) (a b)2 (b c)2 (c a)2 ? 2

1) a 3 b3 c3 3abc

x 3 2 is 1) 0 4) 3

2) 1 3) 2 5) None of these

3. What should be added to 8x 3 12x 2 4x 5 to make it exactly divisible by 2x + 1? 1) 11 2) 5 3) 6 4) –11 5) None of these 4. Find the values of a and b when f (x) 2x ax 11x b is exactly divisible by (x – 2)(x + 3). 1) a = 3, b = 6 2) a = 3, b = –6 3) a = –3, b = 6 4) a = –3, b = –6 5) None of these 3

2

5. If f (x) 4x 3 2x 2 5x 8 is divided by x – 2, what will be the remainder? 1) 25 2) 42 3) 16 4) 26 5) None of these

2) a 3 b3 c3 3abc 3) a 3 b 3 c 3 3abc(a b c) 4) 3abc 5) 3 8. Find the value of a 3 b3 c3 3abc when a + b + c = 9 and a 2 b2 c 2 29. 1) 9 2) 3 3) 27 4) 81 5) None of these 9. Find the value of x 3 y 3 z 3 3xyz when x = 89, y = 87, z = 84. 1) 260 2) 19 3) 4940 4)4490 5) None of these

Quicker Maths

86 10. If x = a(b – c), y =b(c – a) and z c(a b) , then 3

3

(a 1)(1 a)(1 a a 2 )(1 a a 2 )(1 a 6 ) .

3

x y z ? a b c

3xyz 1) abc 4) 3 11. When x 1) 3 4) 27

xyz 2) 3) 3xyzabc abc 5) None of these 1 1 3 , find the value of x 2 2 . x x 2) 6 3) 9 5) None of these

1 1 5, find the value of x 2 2 . x x 1) 9 2) 27 3) 81 4) 7 5) 15 13. Find the value of a 2 b 2 c 2 2ab 2ac 2bc when a = 17, b = 15 and c = 13. 1) 111 2) 121 3) 225 4) 361 5) None of these 12. When x

1 1 3 14. When x 3 find the value of x 3 . x x 1) 9 2) 27 3) 18 4) 21 5) None of these 15. When a = –5, b = –6 and c = 10 find the value of

1) 1 a12

2) 1 a12

4) 1 a 36

5) None of these

3) 1 a 36

19. If x + y =1, find the value of x 3 y 3 3xy . 1) 1 2) 0 3) 2 4) 6 5) None of these 20. Find the value of 1) 2 4) 3 21. What

2 n 6m 1 10m n 15m n 2 . 4 m 32m n 25m 1

2 1 3) 3 3 5) None of these the value of the expression 2)

is

(a b) 3 (b c)3 (c a)3 ? (a b)(b c)(c a)

1) 1 4) 3

2) 0 3) 2 5) None of these

22. If a + b + c = 0, find the value of 1) 0 4) 3

2) 1 5) None of these

a 2 b2 c 2 . a 2 bc 3) 2

a 2 b2 c2 . c2 ab 1) 0 2) 1 3) 2 4) 3 5) None of these 24. If a + b + c = 0, then find the value of 23. If a + b + c = 0, find the value of

a 3 b3 c3 3abc (ab bc ca a 2 b2 c 2 )

1) –1 2) 1 4) –2 5) 3 16. If a = 1, find the value of

18. Find the value of

3) 2

15a 3 (3a 3 1) (4a 4 a 3 3) (a 3 1) 1) 11 2) 1 3) 10 4) 17 5) None of these 17. If a + b + c = 10 and ab + bc + ac = 31, find the value

of a 2 b2 c 2 . 1) 69 2) 162 3) 131 4) 38 5) None of these

(a 2 b2 c 2 )2 . a 2 b 2 b 2c 2 c 2a 2 1) 1 4) 4

2) 2 3) 3 5) None of these

2 2 2 25. If x y z, y z x , z x y, then the value

1 1 1 of x 1 y 1 z 1 is 1) 1 4) 4

2) –1 3) 2 5) None of these

Elementary Algebra

87

ANSWERS or, 29 2(ab bc ac) 81

1. 1; x 2 2x 24 (x 6)(x 4) Since (x + 6)(x – 4) is a multiple of the given expression, (x + 6) and (x – 4) should divide the expression exactly. That is, if we divide the given expression by (x + 6) and ( x – 4), there should be no remainder. Thus, by the remainder theorem,

or, (ab bc ac) Now, a 3 b3 c3 3abc

f ( 6) 3( 6)3 a( 6)2 74( 6) b 0

(a b c) a 2 b2 c 2 ab bc ac

and f (4) 3(4) a(4) 74(4) b 0 Solving these two equations, we get the values of a and b. So, a = –5 and b =24 3

81 29 26 2

2

= 9{29 – 26} = 27 9. 3; Use; x 3 y 3 z 3 3xyz

2. 4; Put, x 3 y then the divisor is y 2 and the given expression is 4y2 – 5y – 3. By the remainder theorem, the remainder is f (2) 4(2)2 5(2) 3 3 3. 1;By the remainder theorem, the remainder is

1 (x y z) (x y)2 (y z)2 (x z)2 2

10. 1;

x y z b c, c a, a b a b c x y z 0 a b c Now, use the well-known conditional identity

1 1 1 1 f 8 12 4 5 2 2 2 2

Now, we have

1 3 2 5 11 For the expression to be exactly divisible, the remainder should be zero. Hence, 11 should be added.

3 3 If l + m + n = 0 then l3 m n 3lmn

3

2

4. 2; f (2) 2(2)3 a(2) 2 11 2 b 0 or, 16 + 4a – 22 + b = 0 or, 4a + b = 6 ------ (1) f ( 3) 2( 3) 3 a( 3)2 11( 3) b 0 or, -54 + 9a + 33 + b = 0 or, 9a + b = 21 ------ (2) Solving (1) and (2), we get the values of a and b.

5. 4; f (2) 4(2)3 2(2) 2 5(2) 8 = 32 – 8 + 10 – 8 = 26

1 2 2 2 7. 1; (a b c) 2a 2b 2c 2ab 2bc 2ac 2

(a b c) a 2 b2 c2 ab bc ac

a 3 b3 c 3 3abc 8. 3; a + b + c = 9 or, (a b c) 2 81 or, a 2 b2 c 2 2(ab bc ac) 81

2 or, x

1 1 2x 9 2 x x

2 x

1 927 x2 2

1 12. 2; x 25 x

x2

6. 1; x n y n is exactly divisible be x – y for any +ve integer (odd or even).

2

1 2 11. 5; x 3 9 x

1 25 2 27 x2

13. 2; The given expression (a b c) 2 (17 – 15 13)2 ( 11) 2 121 3

1 14. 3; x 27 x 3 or, x

1 1 1 3 x x 27 x3 x x

Quicker Maths

88 3 or, x

22. 3; a + b + c = 0

1 3 3 27 x3

or, (a b c)2 0

1 18 x3 15. 2; The given expression = –(a + b + c) (find it) = – (–5 – 6 + 10) = – (–1) = 1. 16. 1; 15a 3 3a 3 a 3 a 3 4a 4 1 3 1

or, a 2 b 2 c 2 2(ab bc ac) 0

x3

or, a 2 b2 c 2 –2 bc a(b c) 2 bc a ( a) 2(bc a 2 ) 2(a 2 bc)

12a 3 4a 4 3 12 4 3 11 17. 4; a + b + c = 10

a 2 b2 c2 2 a 2 bc

or, (a b c)2 100

23. 3; Same as question (22).

or, a 2 b 2 c 2 2(ab bc ac) 100

24. 4; (a b c)2 0 or, a 2 b 2 c 2 2(ab bc ac)

a 2 b 2 c 2 100 2(31) 38 18. 1; The given expression

= (a 1)(1 a a ) (1 a)(1 a a ) 1 a 2

2

6

(1 a 3 )(1 a 3 )(1 a 6 ) (1 a 6 )(1 a 6 ) 1 a12

Note: We have used (a b)(a b) a 2 b2 19. 1; x + y = 1 or, (x y) 1

or, a 2 b2 c 2

or, x 3 y3 3xy 1 20. 2; Given expression =

2n 2m 1 3m 1 2m n 5m n 3m n 2 5m n 2 22m 32m n 52m 2

22m 1 32m n 1 52m 2 2 22m 32m n 52m 2 3 21. 4; Use: when x + y + z = 0, x 3 y 3 z 3 3xyz or,

x 3 y3 z 3 3 xyz

In this case, (a – b) + (b – c) + (c – a) = 0

2

4(ab bc ac) 2

or, 4 a 2 b2 b2 c 2 a 2 c 2 2abc(a b c)

or, 4 a 2 b2 b2 c2 a 2 c 2

a

2

b 2 c2

(Since a + b + c = 0)

2

a 2 b2 b 2 c 2 a 2 c 2

3

or, x 3 y 3 3xy(x y) 1

4

25. 1; x 2 y z, y 2 z x and z 2 x y or,

x x 2 x y z, y y 2 x y z

and

z z2 x y z

or, x x 2 y y 2 z z 2 x y z k (say) or, x(1 x) y(1 y) z(1 z) k 1 x 1 y 1 z 1 x k , 1 y k and 1 z k

the given expression

x y z xyz k 1 k k k k k

Chapter 12

Problems on Comparison of Quantities EXERCISES Directions (Q. 1-28): In each of the following questions, read the given statement and compare the two given quantities on its basis. 1. Given that AB = 21 cm, BM is the bisector of AC at point M. BC is the diameter of the circle, and the circumference of the circle is 132 cm.

Quantity I. Area of shaded region Quantity II. 21 cm2 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity II Quantity I 4) Quantity II = Quantity I or No relation can be established 5) Quantity II Quantity I 2. Quantity I. 21x3y Quantity II. 9x4y4 if x < 0 and y > 0 1) Quantity I > Quantity II 2) Quantity I Quantity II 3) Quantity I Quantity II 4) Quantity I < Quantity II 5) Quantity I = Quantity II or No relation can be established 3. The speed of a boat in still water and the speed of the current are in the ratio 5 : 3. The difference between the distance covered by the boat in 2 hours upstream and that in 2 hours downstream is 24 km. Quantity I. Speed of the boat in still water Quantity II. Speed of the cyclist who goes 28 km in 2 hours

1) Quantity I < Quantity II 2) Quantity I Quantity II 3) Quanttity I Quantity II 4) Quantity I > Quantity II 5) Quantity I = Quantity II 4.

Quantity I. xº Quantity II. 60º 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) No relation can be established 9

5. If 9x2 = 3x Quantity I. 1.5 Quantity II. x 1) Quantity I Quantity II 2) Quantity I < Quantity II 3) Quantity I > Quantity II 4) Quantity I Quantity II 5) No relation can be established 6. A batsman has a certain average of runs for 12 innings. In the 13th innings, he scores 96 runs, thereby increasing his average by 5 runs. Quantity I. The average after 13th innings Quantity II. 65 runs 1) Quantity I > Quantity II 2) Quantity I Quantity II 3) Quantity I Quantity II 4) No relation can be established 5) Quantity I < Quantity II

Quicker Maths

90 7. Ratio of A’s age to B’s age is 4 : 3. A will be 34 years old after 6 years. Quantity I. The present age of B Quantity II. 3 years ago A’s age 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) No relation can be established 8. A basket contains 6 red, 5 green and 8 blue balls. Four balls are picked at random. Quantity I. The probability that all four of them are red. Quantity II. The probability that two of them are green and the other two are blue. 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) No relation can be established 9. Ram’s salary, which is more than 15000, is 80% of Daya’s salary. Jay’s salary is 80% of Ram’s salary. Quantity I. Daya’s salary Quantity II. Jay’s salary 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established 10. A merchant earned a profit of 125 on the selling price of a sweater that cost the merchant 375. Quantity I. The profit expressed as a percentage of the cost to the merchant Quantity II. The profit expressed as a percentage of the selling price 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established 11. Quantity I. The perimeter of a triangle whose sides are 15 cm, 18 cm and 21 cm Quantity II. The perimeter of a square whose diagonal is 17 2 cm 1) Quantity I > Quantity II

2) Quantity I Quantity II 3) Quantity I < Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established 12.

Quantity I. Area of the shaded part Quantity II. 126 cm 2 1) Quantity I Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I > Quantity II 5) Quantity I = Quantity II or No relation can be established 13. Quantity I. The volume of a cube whose surface area is 54 cm2 Quantity II. The volume of a cuboid whose sides are 6 cm, 7 cm and 9 cm 1) Quantity I < Quantity II 2) Quantity I > Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established 14. The perimeter of a square is equal to twice the perimeter of a rectangle of length 11 cm and breadth 10 cm. Quantity I. The circumference of a semicircle whose diameter is equal to the side of the square Quantity II. The circumference of another semicircle whose radius is 14 cm 1) Quantity I Quantity II 2) Quantity I Quantity II 3) Quantity I > Quantity II 4) No relation can be established 5) Quantity I < Quantity II 15. A person bought two watches for 540. He sold one of them at a loss of 15% and the other at a gain of 19% and he found that each watch was sold at the same price.

Problems on Comparison of Quantities Quantity I. The cost price of the watch which was sold at 19% profit Quantity II. The selling price of the watch which was sold at 15% loss 1) Quantity I Quantity II 2) Quantity I < Quantity II 3) Quantity I > Quantity II 4) Quantity I Quantity II 5) No relation can be established 16. A man buys milk at 44 per litre. After adding some water he sold it at 48 per litre and thus earns a profit 1 3

of 33 % . Quantity I. Proportion of water to milk in the mixture Quantity II. 1 : 10 1) Quantity I Quantity II 2) No relation can be established 3) Quantity I < Quantity II 4) Quantity I Quantity II 5) Quantity I > Quantity II 17. A box contain 3 blue pens, 4 red pens and 5 black pens. Quantity I. If two pens are drawn at random, the probability that both the pens are either black or blue Quantity II. If three pens are drawn, the probability that all are black 1) Quantity I > Quantity II 2) Quantity I Quantity II 3) Quantity I < Quantity II 4) No relation can be established 5) Quantity I Quantity II 18. The ratio of the speeds of a car, a train and a bus is 5 : 9 : 4. The average speed of the car, the train and the bus is 72 km/h. Quantity I. The average speed of the car and the train Quantity II. The average speed of the train and the bus 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) No relation can be established 19. At the start of a seminar, the ratio of the number of male participants to that of female participants was 3 : 1. During tea-break 24 participants left in the same gender ratio as they were present there in and 14 more male and 4 more female participants registered

91 themselves. The ratio of male to female participants thus became 16 : 5. Quantity I. 1.5 times the number of male participants before tea break Quantity II. 5 times the number of female participants before tea break 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established 20. Quantity I. The average of 13 numbers is 45. If the average of the first six numbers is 49 and that of the last six numbers is 43, then the value of the seventh number Quantity II. The average of 11 numbers is 47, that of the first five is 52 and that of the last five is 46. Value of the sixth number 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established 21. Avinash, Aashish and Rahul are three typists who, working simultaneously, can type 324 pages in four hours. In one hour Avinash can type as many pages more than Aashish as Aashish can type more than Rahul. During a period of 9 hours, Avinash can type as many pages as Aashish can during eight hours. Quantity I. The total number of pages typed by Aashish and Rahul in one hour Quantity II. The total number of pages typed by Avinash and Aashish in one hour 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established 22. Quantity I. Value of x if 42x2 – 73x – 47 = 0 Quantity II. Value of y if 53y2 + 79y – 54 = 0 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established

Quicker Maths

92 23. The diameter of a solid cylinder is 5 cm and its height is 14 cm. Quantity I. Total surface area of the cylinder Quantity II. 220 cm2 1) Quantity I > Quantity II 2) Quantity I < Quantity II 3) Quantity I Quantity II 4) Quantity I Quantity II 5) Quantity I = Quantity II or No relation can be established 24. 1 > a > 0 > b Quantity I. Value of

(a b)2 – a 2 – b2 (a b ) 2 – (a – b) 2

1 Quantity II. 2(ab3 ab)

1) Quantity I < Quantity II 2) Quantity I Quantity II 3) Quantity I Quantity II 4) Quantity I > Quantity II 5) Quantity I = Quantity II 25. There are three positive numbers a, b and c. The average of a and b is less than the average of b and c by 1. Quantity I. Value of c Quanity II. Value of a 1) Quantity I = Quantity II 2) Quantity I > Quantity II 3) Quantity I < Quantity II 4) Quantity I Quantity II 5) No relation can be established 26. Three equal circles are drawn on a triangle ABC, with points A, B, C as the centres. Radius of each of the circle is equal to half of the side of the triangle ABC (Figure not to scale)

Area of shaded region 1

1 3

2 = 128 cm

Quantity I. The area of the shaded region 2 (in cm2) Quantity II. 30cm2 1) Quantity I > Quantity II 2) Quantity I Quantity II 3) Quantity I = Quantity II or No relation can be established 4) Quantity I < Quantity II 5) Quantity I Quantity II 27. Ram invested P in scheme A and 2P in scheme B for two years each. Scheme A offers simple interest p.a. Scheme B offers compound interest (compounded annually) at the rate of 10% p.a. The ratio of the interest earned from scheme A to that earned from scheme B was 8 : 21. Quantity I. Rate of interest offered by scheme A Quantity II. Rate of interest offered by scheme C (simple interest p.a.) when 1600 invested for 3 years earns an interest of 384 1) Quantity I = Quantity II 2) Quantity I > Quantity II 3) Quantity I < Quantity II 4) Quantity I Quantity II 5) No relation can be established 28. Rutuja bought two articles – article A at X and article B at X + 50. She sold article A at 20% profit and article B at 10% loss, and earned 35 as profit on the whole deal. Quantity I. Profit earned (in ) by Rutuja on selling article A Quantity II. Loss incurred (in ) when an article which costs 480 is sold at 20% loss 1) Quantity I < Quantity II 2) Quantity I Quantity II 3) Quantity I Quantity II 4) Quantity I > Quantity II 5) Quantity I = Quantity II

Problems on Comparison of Quantities

93

ANSWERS 132 132 7 = = 21 cm 2 2 22 Quantity I. Area of the shaded region 1 = 21 21 – 21 21 2 2 1 ( Ar BMC = Ar ABC) 2

1. 1; Radius of the circle =

22 21 21 – 1 441 72 2 = 693 – 220.5 = 472.5 cm2 =

21 22 = 66 cm2 7 Hence Quantity I > Quantity II 2. 4; As x < 0 and y > 0. Quantity I will always be –ve. And Quantity II will be +ve. Hence Quantity I < Quantity II. 3. 1; Quantity I. Let the speed of the boat downstream = 5x + 3x = 8x and upstream = 5x – 3x = 2x Now, 8x × 2 – 2x × 2 = 24 or, 16x – 4x = 24 Quantity II.

24 x= = 2 kmph 12 Speed of the boat = 5 × 2 = 10 kmph Quantity II.

Speed of the cyclist =

5. 3; 9x2 =

9 3x

3x3 = 1 x3 =

1 3

1

3 x = 1

3

Hence Quantity I > Quantity II 6. 5; Let the average of runs be x.

12x 96 =x+5 13

6

8. 2; Quantity I. Reqd probability =

28 = 14 2

C4 15 = C4 3876

19

Quantity II. 5

Reqd probability =

kmph Hence Quantity I < Quantity II 4. 2; Given OBC = OCB = 37º(base angles of an isosceles triangle) COB = 180º – (37º + 37º) = 106º

1 1 x = BAC = COB = × 106º = 53º 2 2 Hence Quantity I < Quantity II

or, 12x + 96 = 13(x + 5) or, x = 31 runs The average after 13th innings = 31 + 5 = 36 runs Hence Quantity I < Quantity II Note: Average increases by 5 in 13th innings He scored (13 × 5 = 65) extra runs in13th innings than previous average of 12 innings. Therefore average before 13th innings = 96 – 65 = 31 Average after 13th innings = 31 + 5 = 36 7. 2; Let the present ages of A and B be 4x and 3x respectively. 4x + 6 = 34 x = 7 years The present age of B is 21 years and 3 years ago A’s age 28 – 3 = 25 years Hence Quantity I < Quantity II

C 2 8 C 2 10 28 70 19 3876 969 C4

Hence Quantity I < Quantity II 9. 1; Daya’s salary > Ram’s salary > Jay’s salary Hence Quantity I > Quantity II. 10. 1; Quantity I. % profit on the cost price

125 100 125 100 1 1 = = 100 33 % 375 375 3 3 Quantity II. Selling price = 375 + 125 = 500 =

% profit on the selling price =

125 100 25% 500

Hence Quantity I > Quantity II Note: % profit calculated on the basis of cost price is always more than the % profit calculated on the basis of selling price. Because in case of profit, SP is always more than CP. And for smaller base (in denominator) the % profit will be higher. 11. 3. Quantity I. The perimeter of the triangle = 15 + 18 + 21 = 54 cm Quantity II. Diagonal 17 2 = = 17 cm 2 2 The perimeter of the square = 17 × 4 = 68 cm Hence Quantity I < Quantity II

Side of the square =

Quicker Maths

94 12. 2; Quantity I. Area of the shaded part = =

1 r 2 4

1 22 21 21 = 346.5 cm2 4 7

22 Quantity II. 126 cm2 = 126 = 396 cm2 7 Hence Quantity I < Quantity II 13. 1; Quantity I. Volume of the cube 3

3

Surface area 54 6 = (3)3 = 27 cm3 = 6 Quantity II. Volume of cuboid = 6 × 7 × 9 = 378 cm2 Hence Quantity I < Quantity II Note: In such case, we do not need to calcualte the exact volume. It is clear that 3 × 3 × 3 is less than 6 × 7 × 9. So QI < QII 14. 5; Quantity I. Perimeter of the square = 2 × 2(11 + 10) = 2 × 42 = 84 cm Let the side of the square be a. Then, 4a = 84 cm a = 21 cm Diameter of the circle = 21 cm

21 = 10.5 cm 2 Circumference of the semicircle

Quantity II = Selling price of the watch sold at 15%

85 loss= 315 = 267.75 100 Hence Quantity I < Quantity II Quicker Approach: CP x : y SP 0.85x = 1.19y (given that both selling prices are equal)

y QI 1 QII 0.85x 1.19 (from the above relation) QI 1 QII 1 ( is less than 1) 1.19 QI < QII 16. 5; CP of the mixture 100 300 48 48 400 = 36 = 1 100 33 3 Applying Alligation on the cost price

Radius =

22 10.5 2 10.5 = = 33 + 21 = 54 cm 7 Quantity II. Circumference of another semicircle 22 = × 14 + 2 × 14 = 44 + 28= 72 cm 7 Hence Quantity I < Quantity II Note: We can conclude that Quantity I < Quantity II on the basis of the values of the radius only. Since the first radius is less than the second. 15. 2; Quantity II Quantity I Cost price x 540–x Selling price

=x×

85 100

(540–x)

119 100

17x 119 Now, = (540 – x) × 20 100 or, 5x = (540 – x) × 7 or, 5x + 7x = 540 × 7 540 7 = 315 12 Quantity I = Cost price of the watch at 19% profit = 540 – 315 = 225 x=

Therefore water : milk = 2 : 9 = Quantity I and Quantity II = 1 : 10 Quantity I > Quantity II 3

17. 1; Quantity I.

C 2 5 C 2 3 10 13 12 66 66 C2

5

Quantity II.

C3 10 C3 220

12

Hence Quantity I > Quantity II 18. 1; The total speed of car, bus and train = 72 × 3 = 216 km/hr Speed of the car and the train

5 9 216 = 168 km/hr 59 4 Average speed of the car and the train together =

168 = 84 km/hr 2 Speed of the train and the bus =

=

9 4 216 = 156 km/h 59 4

Problems on Comparison of Quantities

95

Average speed of the train and the bus

From eqn (ii), we get 54 = x + z ... (iv) z = 54 – 20 = 30 Quantity I. The no. of pages typed by Aashish and Rahul in one hour= 27 + 30 = 57 Quantity II. The no. of pages typed by Avinash and Aashish in one hour = 24 + 27 = 51 Hence Quantity I > Quantity II

156 = 78 km/hr 2 Hence Quantity I > Quantity II Quicker Approach: We do not need to find the speeds or average speeds. We can simply conclude from the given ratio. =

QI =

59 = 7 units 2

QII =

9 4 = 6.5 units 2

QI > Q II 19. 2; Let the number of males to females be 3x and x. Total = 4x

3x – 24 3 14 4 16 Now, 1 5 x – 24 4 4 or,

3x – 18 14 16 x –64 5

or, 15x – 20 = 16x – 32 x = 12 Male participants = 3x = 3 × 12 = 36 Female participants = x = 12 Quantity I. 36 × 1.5 = 54 Quantity II. 12 × 5 = 60 Hence Quantity I < Quantity II 20. 1; Quantity I. 7th number = (13 × 45) – (6 × 49 + 6 × 43) = 585 – (294 + 258) = 585 – 552 = 33 Quantity II. 6th number = Total of 11 nos. – (Total of first five no. + total of last five no.) = 47 × 11 – (52 × 5 + 46 × 5) = 517 – (260 + 230) = 517 – 490 = 27 Hence, Quantity I > Quantity II 21. 1; Let Avinash, Aashish and Rahul type x, y and z pages respectively in one hour. Therefore they together can type 4(x + y + z) pages in 4 hours 4(x + y + z) = 324 x + y + z = 81 ... (i) Also, x – y = y – z 2y = x + z ... (ii) Again, 9x = 8y ... (iii) From equation (i) and (ii), we get 3y = 81 81 y= = 27 3 8 27 24 and x = (from (iii)) 9

22. 5;

Step II.

– 94 42

21 42

47 1 Step III. x = 21 , – 2

Step II.

106 53

– 27 53

27 53 Hence no relationship can be established. 23. 1; Quantity I. Surface area of the solid cylinder = 2r2 + 2rh Step III. y = –2,

22 2.5 2.5 2 22 2.5 14 = 2 7 7 = 39.28 + 220 = 259.28 cm2 Quantity II. 220 cm2 Hence Quantity I > Quantity II ( a b) 2 – a 2 – b 2 2ab 1 24. 4; Quantity I. = (a b) 2 – (a – b) 2 4ab 2

1 1 Quantity II. 2(ab 3 ab) 2ab( b 2 1) 1 > a > 0 > b b will be negative Quantity II will be negative. Hence QI > QII 25. 2; Average of

a b b c –1 = 2 2

ab bc–2 = 2 2 or, a + b = b + c – 2 or, a – c = – 2 or,

Quicker Maths

96 Thus, the value of c will be greater than the value of a. So, QI > QII 26. 4; Area of shaded region 1= area of circle – sector of circle

2 60º 1 = r2 1 – 360º = r2 1 – 6 = 5r 6

21% of 2P 21 2 R% of P 8 or, R = 8% Quantity II. Rate of scheme C=

2 1 385 Now, 5r = 128 = 3 3 6

385 6 7 = 49 3 22 5 r = 7 cm r2 =

= (0.16) × (7)2 = 49 × 0.16 = 7.87 cm2 Quantity II. 30 cm2 Hence Quantity I < Quantity II

10 10 Then, 10 10 = 21% for 2 years 100 Now,

= r – × r2 360º 2

Quantity I. Area of shaded region 2 = 3 –

27. 1; Quantity I. Scheme B rate of interest 10% compounded annually.

r 2 2

384 100 = 8% 1600 3

Hence Quantity I = Quantity II 28. 1; Article A Article B Cost price x (x + 50) Selling price x × 1.2 (x + 50) × 0.9 Now, profit = SP – CP or, 1.2x + 0.9x + 45 – x – x – 50 = 35 or, 0.1x = 35 + 5 = 40 x = 400 Quantity I. Profit earned on Article A = 20% of 400 = 80 Quantity II. Loss = 20% of 480 = 96 Hence Quantity I < Quantity II

Chapter 13

Surds Surds: The roots of those quantities which cannot be exactly obtained are called surds e.g. 2, 4 8 etc. Mixed Surds: A rational factor and a surd multiplied to-

1m Order of Surds: a is called a surd of the mth order.. Changing the surds into that of the same order: Ex.: Express 31/4, 21/3 and 51/6 as surds of the same order and arrange them in the ascending order of magnitude. Soln: The LCM of 4, 3 and 6 (the root indices) is 12. We then reduce them to the 12th order.

2

1 1

4

3 2

3

1

3 12

4

12

33 3

4

1 12

1 12

27

16

1 2 12

2

1 12

1

Soln:

1

1

1

1

2

1

2

1

3

44 2

3

2

6

6

1

1 42 6 16 6 4 3 4 4 2 6 8 2 In solving the examples under this chapter, the following simple results will be used:

1)

a a a

2)

a b ab

3)

a2 b a b

6) ( a b ) ( a – b ) a – b The following roots are also useful; so they should be remembered.

3 1.73205 ;

2 5, 5 5, 12 5 can be added but dissimilar surds

5 2.23607 ;

6 2.44949 ;

like 5 3,3 2, 4 7 cannot be added.

7 2.64575 ;

8 2.82842

Simplify

75 48.

Ex. 1: Find the value of

75 48

a) 1

Find the product of 4 3 , 6

1

6

4

6

6

1

6

5

3

1

6

Soln: a)

Division of Surds: Divide 12 4

3

75 147 5 5 3 7 7 3 5 3 7 3 12 3 20.7846

6

4 6 5 12000 2

75 147

b) 80 3 245 – 125

and 5 .

Soln: 4 1 3 6 1 6 5 1 2 . 2

300 .

Soln: 300 10 10 3 10 3 10 1.732 17.32 Ex. 2: Evaluate the following:

Multiplication of Surds:

Ex.:

44

2 1.41421 ;

25 3 16 3 5 3 4 3 9 3

Ex.:

5) ( a – b )2 a b – 2 ab

1 12

16 12 25 12 27 12 or, 2 3 5 6 3 4 Addition and Subtraction of Surds: similar surds like

Ex.:

3

4) ( a b )2 a b 2 ab

1 12

5 6 5 12 5 25 Hence, in order of magnitude, they should be 1

1

1

gether produce a mixed surd; e.g. 2 3, 4 5 , etc.

3

12 4 Soln: 3 2

1

6

b) 80 3 245 – 125

4 45 3 775 – 555 1

3

by 3 2 .

4 5 21 5 – 5 5 20 5 44.7214

Quicker Maths

98 Ex. 3: Evaluate the following: a)

2 3

c)

242 72

Soln: a)

b)

2 – 1 ( 2 – 1)( 2 – 1) 2 1 ( 2 1)( 2 – 1)

c)

6 150

2 3 6 2.44949 2.4495

b)

6 150 6 150 900 30

c)

121 2 11 2 11 5 242 72 1 6 6 36 2 6 2

5 1 ( 5 1)2 5 – 1 ( 5 – 1)( 5 1)

d)

Ex. 4: Find the values of a)

1 2

b)

1 3

d)

1 6

e)

1 7

Soln: a)

b)

c)

c)

1 5

1 2 2 0.7071 2 2 2 2

1 3

4 2 2 – 4 – 2 3 2 – 2 2.2426

( 3 2)2 9.8989 1 Note: In each of the above examples, we made the denominator a whole number. Suggested Quicker Method (Direct Formula) for

1 7 0.3779 7 7 Ex. 5: Evaluate the following: e)

d)

g)

Soln: a)

b)

5 1 5 –1

b)

e)

3 2 ( 3 2 )2 3 – 2 ( 3 – 2) ( 3 2 )

g)

1 5 0.4472 5 5

3 –1

(2 3)2 13.9282 1

4 2 (4 2 ) ( 2 – 1) 2 1 ( 2 1)( 2 – 1)

f)

3 3 0.5773 3 3 3

1

( 5 1)2 2.6180 4

2 3 (2 3) 2 2 – 3 (2 – 3)(2 3)

1 6 0.4082 d) 6 6

a)

e)

( 2 – 1)2 ( 2 – 1) 2 0.1716 2 –1

(i)

14 3 2

2 3 2– 3

c)

f)

2 –1 2 1

(i)

4 2 2 1

a– b a b

1 3 1 3 1 1.3660 2 3 – 1 ( 3 – 1) ( 3 1) 14 14 (3 – 2) 14(3 – 2 ) 9–2 3 2 (3 2 )(3 – 2 )

2(3 – 2 ) = 3.1716

a b a– b

a– b ( a – b )2 a b ( a )2 – ( b )2

3 2 3– 2

and (ii)

a b – 2 ab a b 2 ab – a–b a–b a–b a b ( a b )2 a – b ( a )2 ( b )2

(ii)

a b 2 ab a b 2 ab a–b a–b a–b If we combine both the results, we have

a b a b 2 ab ------- (*) a b a–b a–b

Surds

99

a b and The above formula has the first term a – b 2 ab . Both the terms are the same for a–b both the cases. The two terms are added when the numerator of the surd has a '+' sign and subtracted when the numerator of the surd has a '–' sing. For example: the second term

7 – 5 75 2 75 – 6 – 35 7–5 7 5 7–5 7 5 75 2 75 6 35 7–5 7 – 5 7–5 Note: You are suggested to remember the direct result (*). It will save you several precious seconds in your examination. Ex. 6: Find the value of the following expressions correct to 3 decimal places: 1)

1 7

1

4)

7)

7 –1

2 3 2– 3

2)

5)

1 11

2 –1 2 1

8)

5 1 5 –1

3)

6)

9)

1

5)

2 – 1 ( 2 – 1)2 0.172 2 –1 2 1

6)

5 1 ( 5 1)2 2.618 5–1 5 –1

7)

2 3 (2 3) 2 13.928 4–3 2– 3

8)

( 5 1)2 5 1 1.618 5–1 2

9)

2 3 (2 3) 2 2 3 3.732 4–3 2– 3

Ex. 7: Arrange the following in an ascending order.

7 – 5, 5 – 3, 9 – 7, Soln:

7– 5

( 7 – 5) ( 7 5) 7 5

3 –1

5 1 5 –1

5– 3

2 3

2– 3

2)

1 11 0.302 11 11

3)

1 3 1 3 – 1 ( 3 – 1)( 3 1)

4)

3 1 3 1 1.366 3–1 2

1 7 1 7 1 0.608 6 7 –1 7 –1

Similarly,

7–5 2 7 5 7 5

( 5 – 3)( 5 3) 5 3

Solution:

1 7 0.378 1) 7 7

11 – 9

9– 7

5–3 2 5 3 5 3

2 and 9 7

2 11 9 We know that when the denominator is greater, the value of a fraction is lower. This way, we may say that 11 – 9

2 2 2 2 11 9 9 7 7 5 5 3

or, 11 – 9 9 – 7 7 – 5 5 – 3 Note: The above example gives an important result. It should be remembered.

100

Quicker Maths

Chapter 14

Number System Quantitative Aptitude deals mainly with the different topics in Arithmetic, which is the science which deals with the relations of numbers to one another. It includes all the methods that are applicable to numbers. Numbers are expressed by means of figures - 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0 — called digits. Out of these, 0 is called insignificant digit whereas the others are called significant digits. Numerals: A group of figures, representing a number, is called a numeral. Numbers are divided into the following types: Natural Number: Numbers which we use for counting the objects are known as natural numbers. They are denoted by ‘N’. N= {1, 2, 3, 4, 5, ......} Whole Number: When we include ‘zero’ in the natural numbers, it is known as whole numbers. They are denoted by ‘W’. W = {0, 1, 2, 3, 4, 5, .......} Prime Number: A number other than 1 is called a prime number if it is divisible only by 1 and itself.

To test whether a given number is prime number or not If you want to test whether any number is a prime number or not, take an integer larger than the approximate square root of that number. Let it be ‘x’. Test the divisibility of the given number by every prime number less than ‘x’. If it is not divisible by any of them then it is prime number; otherwise it is a composite number (other than prime). Ex. 1: Is 349 a prime number? Soln: The square root of 349 is approximately 19. The prime numbers less than 19 are 2, 3, 5, 7, 11, 13, 17. Clearly, 349 is not divisible by any of them. Therefore, 349 is a prime number. Ex. 2: Is 881 a prime number? Soln: The approximate sq. root of 881 is 30. Prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Thus, 881 is not divisible by any of the above numbers, so it is a prime number. Ex. 3: Is 979 a prime number? Soln: The approximate sq. root of 979 is 32.

Prime numbers less than 32 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31. We observe that 979 is divisible by 11, so it is not a prime number. Ex. 4: Are 857, 509, 757, 1003, 1009, 919, 913, 647, 649, 657, 659 prime numbers? (Solve it.) Composite Numbers: A number, other than 1, which is not a prime number is called a composite number. e.g., 4, 6, 8, 9, 12, 14. Even Number: The number which is divisible by 2 is known as an even number. e.g., 2, 4, 8, 12, 24, 28.... It is also of the form 2n {where n = whole number} Odd Number: The number which is not divisible by 2 is known as an odd number. e.g., 3, 9, 11, 17, 19, ...... Consecutive Numbers: A series of numbers in which each is greater than that which precedes it by 1 is called a series of consecutive numbers. e.g., 6, 7, 8 or, 13, 14, 15, 16 or, 101, 102, 103, 104 Integers: The set of numbers which consists of whole numbers and negative numbers is known as a set of integers. It is denoted by I. e.g., I = {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ....} Rational Numbers: When the numbers are written in fractions, they are known as rational numbers. They are denoted by Q.

1 3 8 13 e.g., , , , are rational numbers. 2 4 9 15 Or, the numbers which can be written in the form a {where a and b are integers and b 0} are called rational b numbers. Irrational Numbers: The numbers which cannot be written in the form of p/q are known as irrational numbers (where p and q are integers and q 0) . For example:

3 1.732...., 2 1.414....

Quicker Maths

102 8 2.666 or 2.6 form, 3

4)

Real Numbers: Real numbers include both rational as well as irrational numbers.

5)

But recurring decimals like so they are rational numbers.

Rule of Simplification (i) In simplifying an expression, first of all vinculum or bar must be removed. For example: we know that – 8 –10 = –18 but, – 8 – 10 –(–2) 2 (ii) After removing the bar, the brackets must be removed, strictly in the order ( ), {} and [ ]. (iii) After removing the brackets, we must use the following operations strictly in the order given below. (a) of (b) division (c) multiplication (d) addition and (e) subtraction Note: The rule is also known as the rule of ‘VBODMAS’ where V, B, O, D, M, A and S stand for Vinculum, Bracket, Of, Division, Multiplication, Addition and Subtraction respectively. Ex.: Simplify: 3 1 7 3 8 1 of (6 8 3 – 2) – 7 5 25 7 14

3 1 7 14 – Soln: 1 of (6 8 1) 7 5 25 14 1

3 1 25 of (6 8) 1 7 5 7

1

3 5 2 of 14 – 1 1 6 – 7 7 7

1 2 7 – 12 5 – – 6 7 42 42

General Rules for Solving Problems in Arithmetic 1)

(a b) (a – b) a 2 b2

or,

a b ab ab

or,

a 2 b2 ab ab

2

2

2)

(a b) 2 a 2 2ab b 2

3)

(a – b) 2 a 2 2ab b 2

(a b)3 a 3 3a 2 b 3ab 2 b3 a 3 b3 3ab(a b) (a – b)3 a 3 – 3a 2 b 3ab 2 b3

= a3 – b3 – 3ab(a – b)

a 3 b3 ab a – ab b2

6)

2

or, a 3 b3 (a b) (a 2 – ab b 2 ) 7)

a 3 b3 ab a 2 ab b 2 or, a 3 b3 (a b) (a 2 ab b 2 )

8)

a 3 b3 c 3 3abc (a b c) a 2 b2 c2 ab bc ca

9) a x a y a x y 10) a x a y a x y 11) (a x ) y a xy 12) a x b x either a = b or x = 0 13) a x a y either x = y or , a = 0, 1 14) a x 1, then x is 0 for all values of a (except 0).

Ascending or Descending Rational Numbers

Orders in

Under this chapter, we shall first learn to compare two fractions. Rule 1: When the numerator and the denominator of the fractions increase by a constant value, the last fraction is the biggest. Ex. 1: Which one of the following fractions is the greatest?

3 4 5 , and 4 5 6 Soln: We see that the numerators as well as denominators of the above fractions increase by 1, so the last 5 , is the greatest fraction. 6 Ex. 2: Which one of the following fractions is the greatest? fraction, i.e.

2 4 6 , and 5 7 9 Soln: We see that the numerators as well as the denominators of the above fractions increase by 2,

Number System

103

6 , is the greatest fraction. 9 Ex. 3: Which one of the following fractions is the greatest?

Increase in Num. 2 In Ex. 5; Increase in Den. 8 is less than the

1 4 7 , , 8 9 10 Soln: We see that the, numerator increases by 3 (a constant value) and the denominator also increases by a

first fraction

so the last fraction, ie

7 constant value (1), so the last fraction, ie , is the 10 greatest fraction. Thus, a generalised form can be seen as In the group of fractions

x x a x 2a x 3a x na , , , ,....., y y b y 2b y 3b y nb x na y nb has the highest value where i) a = b

or, ii) a > b

But what happens when a < b? See in the following examples Ex. 4: Which one of the following is the greatest?

1 2 3 4 , , , 8 12 16 20 Soln: In the above example, we see that the numerators and the denominators of the fractions increase by constant values (the numerators by 1 and denominators by 4). If we change the above fractions into decimal values, we see that the fractions are in an increasing order and hence, the last fraction, i.e.

4 , is the greatest. 20 Ex. 5: Which one of the following fractions is the greatest? 2 4 6 , and 7 15 23 Soln: In the above example also, we see that the numerators and denominators increase by 2 and 8 respectively. So, the last fraction, i.e.

6 , is the least. 23

Increase in Num. 1 Note: In Ex. 4; Increase in Den. 4 is greater than the first fraction

1 . 8

2 7 In such a case, when a < b, we have the following relations: Increase in Num. 1. If Increase in Den. > First fraction, the last value is the greatest. 2. If

Increase in Num. First fraction, the last Increase in Den.

value is the least. 3. If

Increase in Num. = First fraction, all the Increase in Den.

values are equal. Rule 2:The fraction whose numerator after crossmutiplication gives the greater value is greater.

5 9 or ? 8 14 Soln: Students generally solve this question by changing the fractions into decimal values or by equating the denominators. But we suggest you a better method for getting the answer more quickly. Step I: Cross-multiply the two given fractions. 9 5 14 8 we have, 5 × 14 = 70 and 8 × 9 = 72 Step II: As 72 is greater than 70 and the numerator involved Ex. 6: Which is greater:

with the greater value is 9, the fraction

9 is the 14

greater of the two.

4 6 or ? 15 23 Soln: Step I: 4 23 15 6 Step II: As the greater value has the numerator 4 Ex. 7: Which is greater:

involved with it,

4 is greater.. 15

Ex. 8: Which is greater:

13 20 or ? 15 23

Soln: Step I: 13 23 15 20 Step II:

20 is greater.. 23

Quicker Maths

104 You can see how quickly this method works. After good practice, you won’t need to calculate before answering the question. The arrangement of fractions into the ascending or descending order becomes easier now. Choose two fractions at a time. See which one is greater. This way you may get a quick arrangement of fractions. Note: Sometimes, when the values are smaller (i.e., less than 10), the conventional method, i.e., changing the values into decimals or equating the denominators after getting LCM, will prove more convenient for some of you. Ex. 9: Arrange the following in ascending order.

Step II: Take the third fraction. Apply CM with the third fraction and the larger value obtained in Step I.

7 4 9 5 49 5 7

3 , 630 ÷7 7

4 3 3 7 and . Therefore, we apply CM with and 5 7 7 9 and see that

504 490 378 315 270 630 630 630 630 630

4 7 3 1 3 5 9 5 2 7 Method II: Change the fractions into decimals like or

3 4 7 1 3 0.428, 0.8, 0.777, 0.5, 0.6 7 5 9 2 5 Clearly,

4 7 3 1 3 5 9 5 2 7

Method III Rule of CM (cross-multiplication) Step I: Take the first two fractions. Find the greater one by the rule of CM.

4 3 5 7 3 5 7 4 4 3 57

7 3 . 9 7

4 7 3 5 9 7

Step III: Take the next fraction. Apply CM with

= 90 so, multiply 3 by 90. Thus, the fractions change to

270 504 490 315 378 , , , and . 630 630 630 630 630 The fraction which has larger numerator is naturally larger. So,

7 3 can lie after or between 9 7

Now we see that

3 4 7 1 3 , , , and 7 5 9 2 5 Soln: Method I: The LCM of 7, 5, 9, 2, 5 is 630. Now, to equate the denominators, we divide the LCM by the denominators and multiply the quotient by the respective numerators. Like, for

4 7 5 9

and see that

3 1 and 7 2

1 3 . Next, we apply CM with 2 7

7 1 7 1 and see that . Therefore , 9 2 9 2

4 7 3 1 3 5 9 5 2 7 Step IV: With similar applications, we get the final result as:

4 7 3 1 3 5 9 5 2 7 Note: This rule has some disadvantages also. But if you act fast, it gives faster results. Don’t reject this method at once. This can prove to be the better method for you.

Some Rules on Counting Numbers I.

Sum of all the first n natural numbers

n(n 1) 2

For example: 1+2+3+....+105 = II.

105(105 1) 5565 2 2

Sum of first n odd numbers = n For example: 1+3+5+7 = 42 = 16 (as there are four odd numbers).

Number System

III.

IV.

105

For example: 1+3+5+......+20th odd number (ie, 20 × 2 – 1 = 39) = 202 = 400 Sum of first n even numbers = n (n+1) For example: 2+4+6+8+.......+100 (or 50th even number) = 50 × (50 + 1) = 2550 Sum of squares of first n natural numbers

Ex. 2: Difference between (26) 2 and (25)2 51 (an odd number) Reasoning: Derived from the above rule II. 3. The difference between the squares of two consecutive numbers is the sum of the two consecutive numbers.

n(n 1)(2n 1) 6 For example:

Ex. 2: (26) 2 – (25)2 26 25 51

12 22 32 ....... 102

V.

10(10 1)(2 10 1) 6

10 11 21 385 6 Sum of cubes of first n natural numbers 2

n(n 1) 2 For example:

2

6 (6 1) 1 2 .... 6 (21) 2 441 2 3

3

3

Ex. 1: 52 4 2 = 5 + 4 = 9 Reasoning: a 2 – b2 (a – b) (a b) (a b) a – b =1

Solved Examples (1)

What is the total of all the even numbers from 1 to 400? Soln: From 1 to 400, there are 400 numbers.

400 200 even numbers. 2 Hence, sum = 200 (200+1) = 40200 [From Rule III] (2) What is the total of all the even numbers from 1 to 361? Soln: From 1 to 361, there are 361 numbers; so there are So, there are

n Note: 1. In the first n counting numbers, there are odd 2 n even numbers provided n, the number of 2 numbers, is even. If n, the number of numbers, is

(3)

and

odd, then there are

1 (n + 1) odd numbers and 2

361 – 1 360 180 even numbers. 2 2 Thus, sum = 180 (180 + 1) = 32580 What is the total of all the odd numbers from 1 to 180?

Soln: There are

range. So, the sum = (90) 2 8100 What is the total of all the odd numbers from 1 to 51?

1 (n – 1) even numbers. 2

(4)

50 25 For example, from 1 to 50, there are 2

Soln: There are

50 25 even numbers. And odd numbers and 2 from 1 to 51, there are

51 1 26 odd numbers 2

51 – 1 25 even numbers. 2 2. The difference between the squares of two consecutive numbers is always an odd number. Ex. 1: 16 and 25 are squares of 4 and 5 respectively (two consecutive numbers). 25 – 16 = 9 an odd number. and

180 90 odd numbers between the given 2

51 1 26 odd numbers between the 2

given range. So, the sum (26) 2 676 (5) Find the sum of all the odd numbers from 20 to 101. Soln: The required sum = Sum of all the odd numbers from 1 to 101 – sum of all the odd numbers from 1 to 20 = Sum of first 51 odd numbers – sum of first 10 odd numbers = (51)2 – (10)2 = 2601 – 100 = 2501

Power and Index If a number ‘p’ is multiplied by itself n times, the product is called nth power of ‘p’ and is written as p n . In p n , p is called the base and n is called the index of the power.

Quicker Maths

106

Some solved examples (1) What is the number in the unit place in (729) Soln: When 729 is multiplied twice, the number in the unit place is 1. In other words, if 729 is multiplied an even number of times, the number in the unit place will be 1. Thus, the number in the unit place in (729)58 is 1. 59

(2)

(729) 59 (729)58 (729) (....1) (729) 9 in the unit place. Find the number in the unit place in

(623) 36 , (623)38 and (623) 39 . Soln: When 623 is multiplied twice, the number in the unit place is 9. When it is multiplied 4 times, the number in the unit place is 1. Thus, we say that if 623 is multiplied 4n number of times, the number in the unit place will be 1. So, (623) (623)

49

1 in the unit place

(623) (623) the unit place

49

(623) (...1) (...9) 9

36

38

2

in

(623)39 (623) 4 9 (623) 3 (...1) (...7) 7 in the unit place Note: When you solve this type of questions (for odd numbers) try to get the last digit 1, as has been done in the above two examples. (3) Find the number in the unit place in 20

(122) , (122)

22

and (122) . 23

Soln: (...2) (...2) ...4

(...2) (...2) (...2) ...8 (...2) (...2) (...2) (2...) ...6 We know that (....6) × (....6) = (....6) Thus, when (122) is multiplied 4n times, the last digit is 6. Therefore, (122) 20 (122) 4 5 (...6) 6 in the unit place (122)22 (122) 4 5 (122)2 (...6) (...4) 4 in the unit place.

(4)

(122) 23 (122) 4 5 (122) 3 (...6) (...8) 8 in the unit place. Find the number in the unit place in (98) 40 , (98)42 and (98) 43 .

Soln: (98) (...6) 4

(98) 4n (...6)

Thus, (98) 40 (98) 4 10 (...6) 6 in the unit place

(98)42 (98)410 (98) 2 (...6) (...4) 4 in the unit place

(98)43 (98)4 10 (98)3 (...6) (...2) 2 in the unit place Note: When there is an even number in the unit place of base, try to get 6 in the unit place, as has been done in the above two questions. This chapter should be concluded with some general rules derived for this type of questions. Rule 1: For odd numbers When there is an odd digit in the unit place (except 5), multiply the number by itself until you get 1 in the unit place.

(...1) n (...1) (...3) 4 n (...1) (...7) 4 n (...1) where n=1,2,3,.... Rule 2: For even numbers When there is an even digit in the unit place, multiply the number by itself until you get 6 in the unit place. (...2)4 n (...6)

(...4) 2n (...6) (....6) n (...6) (....8) 4n (...6), where n =1, 2, 3, ..... Note: If there is 1, 5 or 6 in the unit place of the given number, then after any times of its multiplication, it will have the same digit in the unit place, i.e., (....1) n = (....1) (....5) n = (....5) (...6) n = (....6) (5) What is the number in the unit place when 781, 325, 497 and 243 are multiplied together? Soln: Multiply all the numbers in the unit place, i.e., 1 5 7 3 ; the result is a number in which 5 is in the unit place.

Rule 1 and Rule 2 can be combined to make a more general formula. See the following points: (1) If we raise any number to the power 4n, the unit's digit of the new number comes as 1, 5 or 6.

Number System (2) If we raise the new number to any power, the unit's digit of the second new number remains the same. (3) 5 as a unit's digit remains the same after multiplying any number of times (we know this well.) So, it does not create any problem. (6215)x = (... 5) where x is any positive integer. (4) 1, when multiplied any number of times, gives the same number, so the unit's digit remains the same. (5) When an even number is multiplied by 6 the unit's digit remains unchanged. 12 × 6 = 72 14 × 6 = 84 16 × 6 = 96 18 × 6 = 108 Keeping the above points in mind we may work as follows: Suppose the questions are to find the unit’s digit of 1. (623)49 2. (624)50 3. (627)52 55 59 4. (629) 5. (628) 6. (625)60 64 7. (622) To solve such questions, we will divide the powers by 4 and find the remainder, like, 49 ÷ 4 = 12 × 4 + 1 So, the unit’s digit of (623)49 = unit’s digit of (623)1 = 3 Similarly, 2. unit’s digit of (624)50 = unit’s digit of (624)2 {as 50 = 12 × 4 + 2} = 6 3. unit’s digit of (627)52 = 1 {Because 52 is divisible by 4 and we know that odd numbers give 1 as unit’s digit when they have a power of the multiple of 4} 4. unit’s digit of (629)55 = unit’s digit of (629)3 = 9 5. unit’s digit of (628)59 = unit’s digit of (628)3 = 2 6. unit’s digit of (625)60 = 5 {as 5 does not change as unit’s digit} 7. unit’s digit of (622)64 = 6 {Because 64 is divisible by 4 and we know that even numbers give 6 as unit’s digit when they have a power of the multiple of 4.}

Miscellaneous Examples In a division sum, we have four quantities – Dividend, Divisor, Quotient and Remainder. These are connected by the relation Dividend = (Divisor × Quotient) + Remainder When the division is exact, the remainder is zero (0). In this case, the above relation becomes Dividend = Divisor × Quotient

107 Ex. 1: The quotient arising from the division of 24446 by a certain number is 79 and the remainder is 35; what is the divisor? Soln: Divisor × Quotient = Dividend – Remainder 79 × Divisor = 24446 – 35 = 244111 Divisor = 24411 ÷ 79 = 309. Ex. 2: What least number must be added to 8961 to make it exactly divisible by 84? Soln: On dividing 8961 by 84, we get 57 as the remainder. the number to be added = 84 – 57 = 27 Ex. 3: What least number must be subtracted from 8961 to make it exactly divisible by 84? Soln: On dividing 8961 by 84, we get 57 as the remainder. Therefore, the number to be subtracted is 57. Note: In Ex 2, we see that the given number needs 27 to make it exactly divisible by 84. But in Ex 3, the given number exceeds by 57. Ex. 4: Find the least number of 5 digits which is exactly divisible by 89. Soln: The least number of 5 digits is 10,000. On dividing 10,000 by 89 we get 32 as remainder. if we add (89 – 32) or 57 to 10,000, the sum will be divisible by 89. the required number = 10,000 + 57 = 10,057. Ex. 5: Find the greatest number of 5 digits which is exactly divisible by 137. Soln: The greatest number of five digits is 99,999. On dividing 99,999 by 137, we get 126 as remainder. the required number = 99,999 – 126 = 99873 Note: Do you find the difference between Ex. 4 and Ex. 5? Ex. 6: Find the nearest integer to 1834 which is exactly divisible by 12. Soln: On dividing 1834 by 12, we get 10 as the remainder. Since the remainder 10 is more than the half of the divisor 12, the nearest integer will be found by adding (12 – 10) =2 Thus, the required number = 1834 + (12 – 10) =1836 Ex. 7: Find the nearest integer to 1829 which is exactly divisible by 12. Soln: On dividing 1829 by 12, we get 5 as the remainder. Since the remainder 5 is less than half of the divisor 12, the nearest integer will be found by subtracting 5 from 1829. the required number = 1829 – 5 = 1824. Note: Do you realize the difference between Ex. 6 and Ex. 7? Ex. 8: A number when divided by 899 gives a remainder 63. What remainder will be obtained by dividing the same number by 29?

108

Quicker Maths

Soln:

under 6 in the first line. Hence, we must multiply by 4 tens. So,

Number = 899 × Quotient + 63 = 29 × 31× Quotient + 2 × 29 + 5 Therefore, the remainder obtained by dividing the number by 29 is clearly 5. Ex. 9: A number when divided by 899 gives a remainder 62. What remainder will be obtained by dividing the same number by 31? Soln: Number = 899 × Quotient + 62 = 31 × 29 × Quotient + 31× 2 + 0 Therefore, the remainder obtained by dividing the number by 31 is clearly 0. Note: From Ex. (8) and (9) it is clear that the first divisor must be a multiple of the second divisor. But what happens when the first divisor is not a multiple of second divisor? See in the following examples. Ex. 10: A number when divided by 12 leaves a remainder 7. What remainder will be obtained by dividing the same number by 7? Soln: We see that in the above example, the first divisor 12 is not a multiple of the second divisor 7. Now, we take the two numbers 139 and 151, which when divided by 12, leave 7 as the remainder. But when we divide the above two numbers by 7, we get the respective remainders as 6 and 4. Thus, we conclude that the question is wrong. Ex. 11: A boy was set to multiply 432051 by 56827, but reading one of the figures in the question erroneously, he obtained 21959856177 as his answer. Which figure did he mistake? Soln: On dividing we find that 56827 does not exactly divide 21959856177. Hence the error was in reading 56827. But on dividing the number by 432051, we get 50827. Hence the boy read the figure 6 of the multiplier as 0. Ex. 12: A boy multiplied 423 by a certain number and obtained 65589 as his answer. If both the fives are wrong, but the other figures are right, find the correct answer. Soln: Step I: In the product, the figures 9,8 and 6 are correct. To get 9 in the unit place, we must multiply by 3 units. So,

4 2 3 * * 3 1 2 6 9 ..... ..... 6 * * 8 9 Step II: To get 8 in the ten’s place, we must have 2

423 *43 12 69 16 9 2 .... 6* * 8 9 Step III: To get 6 in the product we must have 4 under 1 in the second line. Hence, we must multiply by 1 hundred. So,

4

2 3 1

4 3

1 2

6 9

1 6 9 2 4 2 3 6 0 4 89 Thus, the correct answer is 60489. Ex. 13: Find the number of prime factors in N = 67 353 1110 ? Soln:

67 353 1110 (2 3)7 (5 7)3 1110

27 37 53 73 1110 Thus, there are 7 + 7 + 3 + 3 + 10 = 30 prime numbers. Ex. 14: On dividing a number by 5, 7 and 8 successively the remainders are respectively 2, 3 and 4. What will be the remainders if the order of division is reversed? Soln:

5 *** 7 ** 2 8 * 3 1 4 We have

* = 8 × 1 + 4 = 12 * * = 7 × 12 + 3 = 87 * * * = 5 × 87 + 2 = 437 Thus, the number may be 437. Now, when order of division is reversed,

8 437 7 54 5 5 7 5 1 2

Hence, the required remainders will be 5, 5 and 2.

Number System

109

Note:

We have used the words “may be” because this number is one of those many numbers which satisfy the conditions. We have used 1 as our final quotient and hence, got the number as 437. But for the other values, like 2, 3, ..... the numbers will be different. And surprisingly, all of them give the same result. You may verify it for yourself! Ex. 15: A watch ticks 90 times in 95 seconds and another watch ticks 315 times in 323 seconds. If both the watches are started together, how many times will they tick together in the first hour? Soln:

The first watch ticks every

95 seconds and the 90

323 second watch ticks every seconds. 315 They will tick together

4 3 25 9 ab 3 .....7 7 7 ....... ...... 437 Now, we see that the second digit (from right) of product is 3. This is possible only when 7 is added to 6. And for this ‘b’ must be equal to 4. Thus,

4

after

...

19 5 17 45 The number of times they will tick in the first 3600 seconds 19 5 17 3600 45 100 100 45 19 5 17 323 Once they have already ticked in the beginning; so in 1 hour they will tick 100 + 1 = 101 times Ex. 16: By what number less than 1000 must 43259 be multiplied so that the last three figures to the right of the product may be 437? Soln: It is clear that the required number is of three digits. Let that number be abc. Then we have,

4

3600

............... ............... 437 It is clear that c = 3; thus,

a 4

3

3 6

Now, for the third digit (4), we see that the third column has 1 (carried) + 7 + 3 = 11 and it needs 3 more. This is possible only when ‘a’ is equal to 7. Thus, we finally see

.............7

9

4 3 7

95 323 LCM of 95, 323 Now, LCM of 90 and 315 HCF of 90,315

ab c

2 5

7 7 7

95 323 and LCM of seconds. 90 315

4 3 25 9

3

3

2 5

9

7 4

3

..... 7 7

7

..... 3 6 ..... 3 4

3

7

Hence, the required number is 743. The above question is given in detail so that you can grasp each step, but once you understand the method, you can solve it in a single step. The only thing which should be kept in mind is that when the last three digits of the product are given, you should not calculate beyond the third column for any row. Ex.17: Find the least number by which 19404 must be multiplied or divided so as to make it a perfect square? Note:

Soln:

19404 2 2 3 3 7 7 11 22 32 72 11 Thus, if the number is multiplied or divided by 11, the resultant number will be a perfect square. Therefore, the required number is 11.

Quicker Maths

110 Ex.18: Fill in the blank indicated by a star in the number 4 * 56 so as to make it divisible by 33. Soln: The number should be divisible by 3 and 11. To make the number divisible by 3, as the digit-sum should be divisible by 3, we may put * = 0, 3, 6, or 9. We also know that a number is divisible by 11 if the sums of alternate digits differ by either 0 or a number divisible by 11. We have

S1 = 4 + 5 = 9 (sum of digits at odd places)

Now, we have, (x y) 2 (x y) 2 4xy or, (14) 2 (10) 2 4xy

(14)2 – (10) 2 96 24 4 4 Direct Formula: xy

Thus, S1 should be 9 and hence * = 3.

S2 cannot be 9 + 11x (ie, 20, 31, 42,....) because in that case * becomes a double-digit number. Theorem: When two numbers, after being divided by a third number, leave the same remainder, the difference of those two numbers must be perfectly divisible by the third number. Proof: Let two such numbers be A and B; and the divisor (the third number) and the remainder be x and y respectively. Then we have nx + y = A --------- (1) mx + y = B --------- (2) Subtracting (2) from (1), we get. x(n – m) = A – B Thus, A – B is perfectly divisible by x. Ex. 19: 24345 and 33334 are divided by a certain number of three digits and the remainder is the same in both the cases. Find the divisor and the remainder. Soln: By the above theorem, the difference of 24345 and 33334 must be perfectly divisible by the divisor. We have the difference = 33334 – 24345 = 8989 = 101 × 89 Thus, the three-digit number is 101. The remainder can be obtained by dividing one of the numbers by 101. If we divide 24345 by 101, the remainder is 4. Ex. 20: 451 and 607 are divided by a number and we get the same remainder in both the cases. Find all the possible divisors (other than 1). Soln: By the above theorem; 607 – 451 = 156 is perfectly divisible by those numbers (divisors). Now, 156 = 2 × 2 × 3 × 13 Thus, 1-digit numbers = 2, 3, 2 × 2, 2 × 3 = 2, 3, 4, 6 2-digit numbers = 12, 13, 26, 39, 52, 78 3-digit number = 156 Ex. 21: The sum of two numbers is 14 and their difference is 10. Find the product of the two numbers. Soln: Let the two numbers be x and y, then, x + y = 14 & x – y = 10 Note:

(Sum + Difference) (Sum – Difference) 4 (14 10) (14 – 10) 24 4 The numbers can be found by the direct formula

Product =

Note:

x

Sum Difference 14 10 12 2 2

Sum – Difference 14 – 10 2 2 2 Ex. 22: The sum of two numbers is twice their difference. If one of the numbers is 10, find the other number. Soln: Let the numbers be x and y. From the question, we have x + y = 2(x – y) or, x = 3y we are given, y = 10 the other number, x = 3 × 10 = 30 Ex. 23: Two numbers are said to be in the ratio 3 : 5. If 9 be subtracted from each, they are in the ratio of 12 : 23. Find the numbers. Soln: Let the numbers be 3x and 5x. Then, by the question y

3x – 9 12 5x – 9 23 or, 69x – 9 23 60x – 12 9 or, 9x = 207 – 108 = 99

x 11 or, 3x = 33 & 5x = 55 Therefore, the numbers are 33 and 55. Ex. 24: A boy was asked to find

7 of a fraction. He made 9

7 and so got 9 an answer which exceeded the correct answer by a mistake of dividing the fraction by

Soln:

8 . Find the correct answer.. 21 Let the fraction be x. Then, x

7 7 8 – x of 9 9 21

Number System

or,

9x 7x 8 – 7 9 21

or,

x(81 – 49) 8 79 21

111 54 6 ------ (2) 9 From equations (1) and (2) or, x – y =

8 7 9 3 x 21 32 4

3 7 7 The correct answer 4 9 12 Ex. 25: Four-fifths of a number is more than three-fourths of the number by 4. Find the number. 4 3 1 – 4 4 Soln: or, 5 4 20 or, 1 = 20 × 4 = 80 Therefore, the required number is 80. Ex. 26: If one-fifth of one-third of one-half of number is 15, find the number. Soln: Let the number be x. Then we have, 1 1 1 x 15 5 3 2

x = 15 × 5 × 3 × 2 = 450 Direct Method: 5 3 2 (*) The required number 15 450 1 1 1 Note:

(*) The resultant should be multiplied by the reverse of each fraction. Ex. 27: If the numerator of a fraction be increased by 12% and its denominator decreased by 2%, the value of the fraction becomes Soln:

86 7 and y = 1 2 the required number = 7 × 10 + 1 = 71 Direct Formula: The required number x

6 . Find the original fraction. 7

x Let the fraction be y . 112% of x 6 Then we have, 98% of y 7 x 98% of 6 98 6 3 y 112% of 7 112 7 4

Ex. 28: The sum of the digits of a two-digit number is 8. If the digits are reversed, the number is decreased by 54. Find the number. Soln: Let the two-digit number be 10x + y. Then, we have; x + y = 8 ------ (1) and 10y + x = 10x + y – 54

Decrease = 5 Sum of digits 9

1 Decrease Sum of digits 2 9

1 (8 – 6) = 70 + 1 = 71 2 Ex. 29: Three numbers are in the ratio of 3 : 4 : 5. The sum of the largest and the smallest is equal to the sum of the third and 52. Find the smallest number. Soln: From the question, it is clear that the sum of the largest and the smallest is 52 more than the third. Thus we have, 3 + 5 – 4 52 4 52 1 13 Therefore, the smallest number is 3 × 13 = 39 Ex. 30: If 40% of a number is 360, what will be 15% of 15% of that number? Soln: Let the number be x. Then we have 40% of x = 360 = 5 (8 + 6) +

x

360 100 900 40

Now, 15% of x

15 900 135 100

Again, 15% of 135

15 135 = 20.25 100

Direct Method: 40% = 360

360 15% of 15% = 40% 15% of 15% 360 360 15 15 15 of 15% 20.25 40 40 100 Ex. 31: The ratio of the sum and the difference of two numbers is 7 : 1. Find the ratio of those two numbers.

Quicker Maths

112 Soln:

Let the two numbers be x and y. Then we have

xy 7 . x–y 1 x y 7x – 7y or, 6x = 8y

x 8 4 y 6 34:3 Quicker Method: (Using the rule of componendodividendo)

x=3&y=6 Therefore, the number is 36. Quicker Method: For the unit’s place, we see that 50% = 3 100% = 6 For the ten’s place, we see that 100% = 3 the number is 36. Ex. 34: It is given that 232 1 is exactly divisible by a certain number. Which one of the following is also divisible by the same number?

xy 7 xy 1 (x y) (x – y) 7 1 or, (x y) – (x – y) 7 – 1 2x 8 or 2y 6

x 4 4:3 y 3

Ex. 32: The difference between a two-digit number and the number obtained by interchanging the digits is 27. What are the sum and the difference of the two digits of the number? Soln: Let the number be 10x + y. Then we have (10x + y) –(10y + x) = 27 or, 9 (x – y) = 27

xy

27 3 9

Thus, the difference is 3, but we cannot get the sum of two digits. Direct Formula: Difference of two digits =

Diff. in original & interchanged number 9

27 3 9 Ex. 33: The digit at the unit’s place of a two-digit number is increased by 50%. And the digit at the ten’s place of the same number is increased by 100%. Now, we find that the new number is 33 more than the original number. Find the original number. Soln: Let the number be 10x + y. The new number is =

200 150 10 x y 20x 1.5y 100 100 Now, by the question, (20x + 1.5y) – (10x + y) = 33 or, 10x + 0.5y = 33 = 10 × 3 + 3

Soln:

3) 216 1

1) 296 1

2) 216 1

4) 7 233

5) None of these

a b (a b)(a 2 – ab b 2 ) 3

3

296 1 (232 )3 (1)3

(232 1) (2 32 )2 – 1 232 1 32

Thus 296 1 is divisible by (2 1) and hence also divisible by that number. Ex. 35: When a certain number is divided by 223, the remainder is 79. When that number is divided by 179, the quotient is 315; then what will be the remainder? Soln: When 179 × 315 = 56385 is divided by 223, we have the remainder 189. Therefore the required remainder will be 79 + (223 –189) = 79 + 34 = 113. Detail Method: Let the number be N. Then N = 223 Q + 79. Where Q is the quotient when N is divided by 223 Again N = 179 × 315 + R = 56385 + R = 223 × 252 + 189 + R or, 223 Q + 79 = 223 × 252 + 189 + R R = 223 (Q – 252) + 79 – 189 Since R can't be more than 179, Q – 252 = 1 ie R = 223 × 1 + 79 – 189 = 113 Note: 1. When Q – 252 = 2 R = 223 × 2 + 79 – 189 = 336 which is not possible 2. When Q – 252 = 0 then R is -ve which is also not possible 3. The only possible value of Q is 253. In this case we see that N = 223 × 253 + 79 = 56498. And 56498 = 179 × 315 + 113, which confirms our answer. Ex 36: Consider a 21-digit number created by writing side by side the natural numbers as follows: N = 123456789101112 ........

Number System What will be the remainder when above number N is divided by 11? Soln: N = 1234 .........1415 Sum of digits at odd places = 1 + 3 + 5 + 7 + 9 + 0 + 1 + 2 + 3 + 4 + 5 = 40 Sum of digits at even places = 2 + 4 + 6 + 8 + 1 + 1 + 1 + 1 + 1 + 1 = 26 For the number to be divisible by 11, difference of sums of digits at odd and even places should be multiple of 11. Here the difference is 40 - 26 = 14, which should be 11. As the last digit of N is at odd place, so if we reduce the digit at 21st place by 3 the number will be divisible by 11. In other words, the number N is 3 more than the multiple of 11. That is, if we divide the number by 11 we get a remainder 3. Ex. 37: The first 100 multiples of 10, i.e., 10, 20, 30, ...... 1000 are multiplied together. How many zeroes will be there at the end of the product? Soln: We are not able to find any quicker method for this question. We suggest that you understand the working of its detailed solution. Total zeroes at the end are produced due to three factors: 1) Due to zeroes at the end of numbers. 2) Due to 5 at the ten-place of numbers followed by zero at the unit place. 3) Due to 5 at the hundred-place of numbers followed by zeroes at the end. Due to factor (1): Number of numbers with single zero = 90 Number of numbers with double zeroes = 9 Number of numbers with triple zeroes = 1 Thus, the total no. of zeroes due to this factor = 90 × 1 + 9 × 2 + 1× 3 = 111 Due to factor (2): When 5 is multiplied by any even number, a zero is produced. There are 10 numbers which have 5 at the ten-place followed by zero (like 50, 150, 250, ...., 950). Thus, 10 zeroes are produced due to factor (2). Due to factor (3): Following the same reasoning, we may say that 500 × 200 will produce an extra zero. Thus, there are total 111 + 10 + 1 = 122 zeroes. Ex. 38: The average of 7 consecutive integers is 7. Find the average of the squares of these integers. Soln: Use the formula: [for odd number of consecutive integers]

113 Average of Squares

1 = No.of integers n1 (n1 1)(2n1 1) n 2 (n 2 1)(2n 2 1) 6 6 Where, n1 = Average + and n2 = Average –

No. of integers -1 2

No. of integers +1 2

In the above case n1 = 7 +

7 1 = 10 2

n2 = 7 –

7 1 =3 2

Average of squares

1 10 11 21 3(4)(7) 7 6 6

1 371 [385 14] 53 . 7 7 Ex. 39: Find the least number which, when divided by 9, 11, and 13, leaves 1, 2 and 3 as the respective remainders. Soln: In fact, there is no possible method for this question. The only thing you can do is to apply the hit-and-trial method by taking the choices one by one. There may be such a number but it is impossible to find it. However, this is not so in all the cases. See the following example. "Find the least number which when divided by 9, 11 and 13 leaves 1, 3, 5 as the respective remainders." We see that 9 – 1 = 11 – 3 = 13 – 5 = 8. Now, we have an established method for this question. LCM of 9, 11, and 13 = 1287 the required least number = 1287 – 8 = 1279. Note: 1. Find the least number which, when divided by 13, 15, and 19, leaves the remainders 2, 4 and 8 respectively. Can we find the specific solution? Soln: YES. this question can be solved because 13 – 2 = 15 – 4 = 19 – 8 = 11 Now, LCM of 13, 15, 19 = 3705 the required least number = 3705 – 11 = 3694. 2.Find the least number which, when divided by

=

Quicker Maths

114

Soln: Ex. 40:

Soln:

Ex. 41:

Soln:

Ex. 42: Soln:

Ex. 43: Soln:

13, 15 and 19, leaves the remainders 1, 2, 3 respectively. Can we find the solution? No. But why? Because 13 – 1 15 – 2 19 – 3. 461 + 462 + 463 + 464 + 465 is divisible by 1) 3 2) 5 3) 11 4) 17 5) None of these 461[1 + 4 + 42 + 43 + 44] = 461 [1 + 4 + 16 + 64 + 256] = 341 × 461 Since 341 is divisible by 11, the given expression is also divisible by 11. The ratio between a two-digit number and the sum of the digits of that number is 4 : 1. If the digit in the unit's place is 3 more than the digit in the ten's place, what is the number? Suppose the two-digit number is = 10x + y 10x y 4 Then we have xy 1 or, 10x + y = 4x + 4y or, 6x = 3y or, 2x = y or, x = y – x = 3 (given) and y = 6 the number is 36. Find the remainder when 713 + 1 is divided by 6. See the following binomial expansion: (x + y)n = xn + nc1xn–1y + nc2xn–2y2 + nc3xn–3y3 + ... + ncn–1xyn–1 + yn We find that each of the terms except the last term (yn) contains x. It means each term except yn is perfectly divisible by x. (Note: yn may be perfectly divisible by x but we cannot say without knowing the values of x and y.) Following the same logic: 713 = (6 + 1)13 has each term except 113 exactly divisible by 6. Thus, when 713 is divided by 6 we have the remainder 113 = 1 and hence, when 713 + 1 is divided by 6 the remainder is 1 + 1 = 2. The product of two numbers is 7168 and their HCF is 16. Find the numbers. The numbers must be multiples of their HCF. So, let the numbers be 16a and 16b where a and b are two numbers prime to each other. 16a 16b 7168 ab 28

Now, the pairs of numbers whose product is 28 are (28, 1), (14, 2) and (7, 4). (14, 2), which are not prime to each other, should be rejected. Hence, the required numbers are (28 × 16, 1 × 16) and (7 × 16, 4 × 16) or, (448, 16) and (112, 64)

Note:

(1) We see that there may be more than one pair of numbers. (2) If you have understood the logic of working, you may simplify the task in the following way. Product Step I: Find the value of (HCF)2

Step II: Find the possible pairs of factors of value got in Step I Step III: Multiply the HCF with the pair of prime factors obtained in Step II For the above question: 7168 Step I: (16) 2 28

Step II: (1, 28), (2, 14), (4, 7) Step III: (1 × 16, 28 × 16) and (4 × 16, 7 × 16) or, (16, 448) and (64, 112) Ex. 44: Find the greatest number that will divide 55, 127 and 175 so as to leave the same remainder in each case. Soln: Let x be the remainder, then the numbers (55 – x), (127 – x) and (175 – x) must be exactly divisible by the required number. Now, we know that if two numbers are divisible by a certain number, then their difference is also divisible by that number. Hence, the numbers (127 – x) – (55 – x), (175 – x) – (127 – x) and (175 – x) – (55 – x) or, 72, 48 and 120 are also divisible by the required number. HCF of 72, 48 and 120 is 24. Therefore, the required number is 24. Note: If you don't want to go into the details of the method, find the HCF of the positive differences of numbers. It will serve your purpose quickly. Ex. 45: A number on being divided by 5 and 7 successively leaves the remainders 2 and 4 respectively. Find the remainder when the same number is divided by 5 × 7 = 35 Soln:

5 A 7 B 2 C 4

In the above arrangement, A is the number which, when divided by 5, gives B as a quotient and leaves 2 as a remainder. Again, when B is divided by 7, it gives C as a quotient and 4 as a remainder. For simplicity, we may take C = 1. B = 7 × 1 + 4 = 11 and A = 5 × 11 + 2 = 57

Number System

115

Now, when 57 is divided by 35, we get 22 as the remainder. Direct Formula: The required remainder = d1 × r2 + r1 Where, d1 = the first divisor = 5 r1 = the first remainder = 2 r2 = the second remainder = 4 the required remainder = 5 × 4 + 2 = 22. Ex. 46: In a long-division sum, the remainders from the first to the last are 221, 301, 334 and 280 respectively. Find the divisor and the quotient if the dividend is 987654. Soln: Step I: ___)987654(_ abc 221 (a) Since we have four remainders, our multiplier (i.e., abc) should be a three-digit number. (Why?) (b) Since we see that all the remainders are threedigit numbers, we may guess that our divisor is also a three-digit number. (It may not be true!) (c) We can find abc = 978 – 221 = 766 Now, our divisor must be a factor of 766. Thus, it is either 766 or 383 Step II: ___)987654 (__ 766 2216 defg 301 defg = 2216 – 301 = 1915 Now, we confirm our divisor by taking the HCF of 766 and 1915, which is 383. Once we get the divisor, we can complete the sum as:

383)987654(2578 766 2216 1915 3015 2681 3344 3064 280 Thus, our divisor is 383 and the quotient is 2578 Ex. 47: Find the number of zeroes at the end of the products: (a) 12 × 18 × 15 × 40 × 25 × 16 × 55 × 105 (b) 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 40 × 45 Soln: We must know that zeroes are produced only due to the following reasons: 1) If there is any zero at the end of any multiplicand.

Note:

2) If 5 or multiple of 5 are multiplied by any even number. To generalise the above two statements, we may say that: (5)n (2)m has n zeroes if n < m; or m zeroes if m< n. Thus, write the product in the form {2m × 5n × ...}. a) 12 × 18 × 15 × 40 × 25 × 16 × 55 × 105 = 12 × 18 × 16 × 40 × 15 × 25 × 55 × 105 = (22 × 3) × (2 × 9) × (2)4 × (23 × 5) × (5 × 3) × (5)2 × (5 × 11) × (5 × 21) = (2)10 × (5)6 × ... (Since numbers other than 2 and 5 are useless.) Since 6 < 10, there are 6 zeroes at the end of the product. (b) 5 × (2 × 5)(3 × 5)(22 × 5)(5)2(2 × 3 × 5)(5 × 7) (23 × 5)(23 × 5)(5 × 9) = (2)10 × (5)11 × ... Since 10 < 11, there are 10 zeroes at the end of the product. This is the easiest way to count the number of zeroes in the chain of products. By this method, you can easily find that the product of 1 × 2 × 3 ... × 100 contains 24 zeroes. Try it.

To find the number of different divisors of a composite number Rule: Find the prime factors of the number and increase the index of each factor by 1. The continued product of increased indices will give the result including unity and the number itself. For example Ex. 1: Find the number of different divisors of 50, besides unity and the number itself. Soln: If you solve this problem without knowing the rule, you will take the numbers in succession and check the divisibility. In doing so, you may miss some numbers. It will also take more time. Different divisors of 50 are: 1, 2, 5, 10, 25, 50 If we exclude 1 and 50, the number of divisors will be 4. By rule: 50 = 2 × 5 × 5 = 21 × 52 the number of total divisors = (1 + 1) × (2 + 1) =2×3=6 or, the number of divisors excluding 1 and 50 =6–2=4 Ex. 2: The number of divisors of 40, except unity, is Soln: 40 = 2 × 2 × 2 × 5 = 23 × 51 Total number of divisors = (3 + 1)(1 + 1) = 8 number of divisors excluding unity = 8 – 1 = 7

Quicker Maths

116 Ex. 3: Find the different divisors of 37800, excluding unity. Soln: 37800 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 7 = 23 × 3 3 × 5 2 × 7 1 Total number of divisors = (3 + 1)(3 + 1)(2 + 1) (1 + 1) = 96 number of divisors excluding unity = 96 – 1 = 95

To find the number of numbers divisible by a certain integer Ex. 1: How many numbers up to 100 are divisible by 6? Soln: Divide 100 by 6. The quotient obtained is the required number of numbers. 100 = 16 × 6 + 4 Thus, there are 16 numbers. Ex. 2: How many numbers up to 200 are divisible by 4 and 3 together? Soln: LCM of 4 and 3 = 12 Now, divide 200 by 12 and the quotient obtained is the required number of numbers. 200 = 16 × 12 + 8 Thus, there are 16 numbers. Ex. 3: How many numbers between 100 and 300 are divisible by 7? Soln: Up to 100, there are 14 numbers which are divisible by 7 (since 100 = 14 × 7 + 2 ). Up to 300, there are 42 numbers which are divisible by 7 (since 300 = 42 × 7 + 6 ) Hence, there are 42 – 14 =28 numbers.

Other method: Step I: First, find the range of the limit. In this case, range = 300 – 100 = 200. Step II: Divide the range by 7 and get the quotient as your required answer. Here, 200 ÷ 7 =28 as quotient. There are 28 numbers. Note: The above method is not applicable in all cases. Sometimes it fails to give the correct answer, as in the following example. Ex. 4: How many numbers between 100 and 300 are divisible by 13? Soln: Method I: Up to 100, there are 7 numbers divisible by 13 because 100 = 7 × 13 + 9 Upto 300, there are 23 numbers divisible by 13 because 300 = 23 × 13 + 1 there are 23 – 7 =16 numbers. Method II: Range = 300 – 100 = 200 and 200 = 15 × 13 + 5 there are 15 numbers, which is not true. Note: You can see how the two methods give different answers. We suggest that you solve these questions by Method I only.

Ex. 5: By what number less than 1000 must 43521 be multiplied so that the last three figures at the right end of the product may be 791? Soln: The last digit of the product is 1, so the multiplier’s last digit should be 1. For the second digit in the product follow the following: let the second digit in multiplier be x. Then, by the rule of multiplication (cross-multiplication of last two digits of multiplicand and multiplies): 4 3521

x1 7 91 1×2+1×x=9x=7 Now, for the third digit in multiplier: 4 3521

x 71 7 91 5 × 1 + 2 × 7 + 1 × x = _7 or 19 + x = _ 7 x = 8 Thus, we see that the three-digit number is 871. Ex. 6: What would be the maximum value of Q in the following equation? 5P9 + 3R7 + 2Q8 = 1114 1) 8 2) 7 3) 5 4) 4 5) None of the above Soln: 5; 5P9 + 3R7 + 2Q8 = 1114 For the maximum value of Q, the values of P and R should be the minimum, i.e. zero each Now, 509 + 307 + 2Q8 = 1114 or, 816 + 2Q8 = 1114 or, 2Q8 = (1114 – 816 =) 298 So, the reqd. value of Q is 9. Ex. 7: If the places of last two digits of a three-digit number are interchanged, a new number greater than the original number by 54 is obtained. What is the difference between the last two digits of that number? 1) 9 2) 12 3) 6 4) Data inadequate 5) None of these Soln: 3; Let the three-digit number be 100x + 10y + z. According to the question, (100x + 10z + y) – (100x + 10y + z) = 54 or, 9z – 9y = 54 or, z – y = 6 Note: Remember that the difference between last two digits in such case is

Difference in two values 54 6 9 9

Number System

117

EXERCISES 1. How many numbers up to 120 are divisible by 8? 2. How many numbers between 200 and 500 are divisible by 13? 3. How many numbers between 100 and 300 are multiples of 13? 4. If we write the numbers from 1 to 201, what is the sum of all the odd numbers? 5. If we write the numbers from 101 to 309, what is the sum of all the even numbers? 6. If we write the numbers from 50 to 151, what is the difference between the sum of all the odd and even numbers? 7. How many different numbers of 7 digits are there? 8. How many numbers up to 200 are divisible by 5 and 7 together? 9. How many numbers between 200 and 400 are divisible by 3, 4 and 5 together? 10. The sum of two numbers is 22 and their difference is 14. Find the product of the numbers. 11. The sum of two numbers is 30 and their difference is 6. Find the difference of their squares. 12. The product of two terms is 39 and their difference is 28. Find the difference of their reciprocals. 13. What is the number just greater than 9680 exactly divisible by 71? 14. What is the difference between the largest and the smallest numbers written with all the four digits 7, 3, 1 and 4? 15. What is the least number which is a perfect square and contains 3675 as its factor? 16. What is the least number which must be subtracted from 9600 so that the remaining number becomes divisible by 78? 17. Find the least number which when added to 3000 becomes a multiple of 57. 18. In a division sum, the quotient is 105, the remainder is 195, and the divisor is the sum of the quotient and the remainder. What is the dividend? 19. Find the sum of all the numbers between 200 and 600 which are divisible by 16. 20. Is 1001 a prime number? 21. Is 401 a prime number? 22. What is the sum of all the prime numbers between 60 and 80?

23. When a certain number is multiplied by 13, the product consists entirely of sevens. Find the smallest such number. 24. Find the number which when multiplied by 16 is increased by 225. 25. How many times shall the keys of a typewriter have to be pressed in order to write first 200 counting numbers, i.e., to write 1,2,3,.... up to 200? 26. In a division sum, the divisor is 4 times the quotient and 3 times the remainder. What is the dividend if the remainder is 4? 27. How many times must 79 be subtracted from 10,000 in order to leave remainder 6445? 28. Find the total number of prime numbers which are contained in (30) 6 . 29. What is the number that added to itself 20 times, gives 861 as result? 30. If 97 be multiplied by a certain number, that number is increased by 7584. Find that number. 31. A certain number when successively divided by 3 and 5 leaves remainder 1 and 2. What is the remainder if the same number be divided by 15? 32. A certain number when successively divided by 7 and 9 leaves remainder 3 and 5 respectively. Find the smallest value of such a number. 33. A certain number when divided by 36 leaves a remainder 21. What is the remainder when the same number be divided by 12? 34. When a certain number is multiplied by 13, the product consists entirely of fives. What is the smallest such number? 35. If a = 16 and b = 15, then what is the value of

a 2 b2 ab ? a 3 – b3 36. What is the largest natural number by which the product of three consecutive even natural numbers is always divisible?

x 3 6 y–x 37. If y 4 , then the value of 7 y x is ______. 38. If

27 x 1 1 , then the value of x is 169 13

______.

Quicker Maths

118 39. What least value must be given to * so that the number 6135*5 is exactly divisible by 9? 40. What least value must be given to * so that the number 97215*6 is divisible by 11? 41. What least value must be given to * so that the number 91876*2 is divisible by 8? 42. What is the largest number of four digits which is exactly divisible by 88? 43. Write down the first prime number. 44. If the number (10n – 1) is divisible by 11 then n is 1) odd number 2) even number 3) any number 4) multiple of 11. 45. If

a 4 3a 2b , then ? b 3 3a 2b

55.

56.

57.

1 1 1 1 46. 1 – 1 – 1 ..... 1 – ? 3 4 5 n 1 3 5 997 ? 47. 2 – 2 – 2 – ..... 2 – 3 5 7 999

137 137 137 133 133 133 ? 48. 137 137 137 133 133 133 49.

? 96 54 ?

50. What is the number of prime factors in the expression (6)10 (7)17 (11)27 ?

51. The sum of two even numbers is 6 more than twice of the smaller number. If the difference between these two numbers is 6, find the smaller number? 52. The sum of three consecutive odd numbers and three consecutive even numbers together is 231. Also, the smallest odd number is 11 less than the smallest even number. What is the sum of the largest odd number and the largest even number? 53. The sum of five consecutive even numbers of set A is 220. What is the sum of a different set of five consecutive numbers whose second lowest number is 37 less than double of the lowest number of set A? 54. The sum of 5 consecutive odd numbers of Set A is 125. What will be the sum of Set B containing 4

58.

consecutive odd numbers, if the smallest odd number of Set B is 16 more than the highest odd number of Set A? The sum of a series of 5 consecutive odd numbers is 195. The second lowest number of this series is 5 less than the second highest number of another series of 5 consecutive even numbers. What is 40% of the second lowest number of the series of consecutive even numbers? There are four consecutive positive odd numbers and four consecutive positive even numbers. The sum of the highest even number and the highest odd number is 33. What is the sum of all the four consecutive odd and even numbers? The tens digit of a three-digit number is 3. If the digits at units and hundreds places are interchanged then the number thus formed is 396 more than the previous one. Also the sum of the units digit and hundreds digit is 14. Then what is the number? S1 is a series of 4 consecutive even numbers. If the sum of the reciprocal of the first two numbers of S1 is

11 , then what is the reciprocal of the third highest 60

number of S1? 59. A number is such that when it is multiplied by ‘8’, it gives another number which is as much more than 153 as the original number itself is less than 153. What is 25% of the original number? 60. On the annual day, sweets were to be distributed equally amongst 600 children of the school. But on that particular day, 120 children remained absent. Thus, each child got 2 extra sweets. How many sweets was each child originally supposed to get ? 61. Deepak has some hens and some goats. If the total number of animal heads is 90 and the total number of animal feet is 248, what is the total number of goats Deepak has? 62. Rachita enters a shop to buy ice-creams, cookies and pastries. She has to buy at least 9 units of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items. How many cookies does she buy?

Number System

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SOLUTIONS (Hints) 120 15 1. 8 2. Number of numbers up to 200 which are divisible by 200 5 15 , i.e., 15 13 13 Number of numbers up to 500 which are divisible by 13 =

500 9 38 , i.e., 38 13 13 the required numbers = 38 – 15 = 23 3. Number of numbers up to 100 which are multiples of 13

100 9 7 , i.e., 7 13 13 Number of numbers up to 300 which are multiples of 13

300 1 23 ,i.e., 23 the required numbers 13 13 = 23 – 7 = 16 4. The number of numbers (201) is odd, hence there are 13

1 (201 1) 101 odd numbers. 2

the sum of the first 101 odd numbers = (101)2 = 10,201 5. Number of even numbers up to 309 =

1 309 1 2

= 154 the sum of first 154 even numbers = 154 (154 + 1) = 23,870

1 101 – 1 50 2 the sum of first 50 even numbers = 50 (50 + 1) = 2,550 the required sum = 23,870 – 2,550 = 21,320 6. If our series starts with an even number and ends with an odd number, then the required difference Number of even numbers up to 101 =

151 – 50 1 102 51 2 2 Note:1) If our series starts with an odd number and ends with an odd number, then such difference =

last number first number 2

2) If our series starts with an even number and ends with an odd number, then such difference

last number – first number 1 2 3) If our series starts with an even number and ends with an even number, then such difference

last number first number 2 7. Least number of 7 digits = 10,00,000 Highest number of 7 digits = 99,99,999 Then, the number of 7-digit numbers = 99,99,999 – 10,00,000 + 1 = 9 106 8. Numbers which are divisible by 5 and 7 together are also divisible by their LCM. LCM of 5 and 7 = 35 Therefore, the required number of numbers

200 25 5 , i.e., 5 35 35 9. LCM of 3, 4 and 5 = 60 Number of numbers up to 200 which are divisible by

200 1 3 , i.e., 3 60 3 Number of numbers up to 400 which are divisible by 60

400 2 6 , i.e., 6 60 3 the required number = 6 – 3 = 3 10. Let the numbers be x and y. Then x + y = 22 --- (1) x – y = 14 ---(2) Squaring (1) and (2), we get 60

x 2 y 2 2 xy 484

---(3)

x 2 y 2 2xy 196

---(4)

Subtracting (4) from (3), we get 4xy = 288 288 72 xy 4 Direct formula: I: Product of two numbers

(Sum)2 – (Difference)2 4

(22) 2 – (14) 2 36 8 72 4 4

Quicker Maths

120 22 14 18 2 22 14 y 4 2 xy = 18 × 4 = 72 11. Let the two numbers be x and y. Then x + y = 30 and x – y = 6 or, (x + y) (x – y) = 180 II: x

or, x 2 y 2 180 12. Let the two terms be x and y. We are given x – y = 28 ---------(1) and xy = 39 ---------(2) x y 28 Dividing (1) by (2), we get xy xy 39

1 1 28 or, y x 39 Ans. 13. On dividing 9680 by 71, we get a remainder of 24. Now, 9680 needs (71 – 24 =) 47 more to be divisible by 71. the required number = 9680 + 47 = 9727 14. The largest number = 7431 and the smallest number = 1347 the required difference = 7431 – 1347 = 6084 15. 3675 3 5 5 7 7 3 5 2 7 2 . See that all the factors except 3 are squares. So, if we multiply 3675 by 3, the obtained number will be a perfect square and also have 3675 as its factor. Thus, the required least number = 3675 × 3 = 11025 16. On dividing 9600 by 78, we get 6 as remainder. If we subtract 6 from 9600, the obtained number will have no remainder. Thus, the required least number = 6 17. When we divide 3000 by 57, we get 36 as remainder. Then, the required least number = 57 – 36 = 21, which when added to 3000, the obtained number becomes a multiple of 57, i.e., that number is perfectly divisible by 57. 18. Q = 105, R = 195, D = Q + R = 105 +195 = 300 Dividend = D × Q + R = 31695 19. The least such number = 16 × 13 = 208 The highest such number = 16 × 37 = 592 the required sum = 16 × 13 + 16 × 14 +......+16 × 37 = 16 (13 + 14 +.....+ 37) = 16[(1 + 2 + ..... + 37) – (1 + 2 + .... + 12)] 37 38 12 13 16 – 16[703 78] 10,000 2 2

20. The approximate square root of 1001 is 32. The prime numbers which are less than 32 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31. We see that 1001 is divisible by 7; so it is not a prime number. 21. The approximate square root of 401 is 20. The prime numbers which are less than 20 are 2, 3, 5, 7, 11, 13, 17, 19. We see that 401 is not divisible by any of the above prime numbers. So it is a prime number. 22. Sum = 61 + 67 + 71 + 73 + 79 = 351 23. Write a number which consists of two sevens only (77). Divide the number by 13. If it is perfectly divisible, then the quotient obtained by division is the required number. If the number is not divisible, then add one more seven and check its divisibility by 13. If it is perfectly divisible, then the quotient is the required number. If it is not divisible, then add one more seven and .... Moving the same way, we see that 777777 ÷ 13 = 59829 24. Let that number be x. Then 16x – x = 225

225 15 15 Note:Thus, in this case, the required number

x

Increased value Multiplier – 1

25. Up to 200, the number of one-, two- and three-digit numbers are 9, 90 and 101 respectively. the number of times the keys of the typewriter to be pressed = 9 × 1 + 90 × 2 + 101 × 3 = 9 + 180 + 303 = 492

D 12 3 4 4 Dividend = DQ + R = 12 × 3 + 4 = 40

26. R = 4, D = 3 × R = 12, Q =

27. The required number of times

10,000 – 6,445 45 79

28. (30) 6 (2 3 5) 6 2 6 36 56 2, 3 and 5 are repeated 6 times each, so there are 6 + 6 + 6 = 18 prime numbers. 861 861 41 29. The required number = 20 1 21 7584 79 30. The required number 97 1 31. By Quicker Method: The required remainder = d1 r2 r1 3 2 1 7 Note: It is a very important method. It should be remembered.

Number System

121

32. Start with the last quotient, i.e., 1

3a 2b 3 4 2 3 18 3 3a 2b 3 4 2 3 6

7 * * * 9 * * 3 1 5 ** = 9 × 1 + 5 = 14 *** = 7 × 14 + 3 = 101 33. The number = 36x + 21 = 36x + 12 + 9 = (36x + 12) + 9 As (36x + 12) is divisible by 12, the remainder will be 9. Note: When the first divisor is divisible by the second, the required remainder will be obtained by dividing the first remainder by second divisor. 34. Same as in Ex. 23.

a 2 b2 ab 1 1 1 3 3 a b a b 16 15 36. The largest such number = (2 × 4 × 6) = 48 35.

x 3 37. y 4 Using the rule of componendo-dividendo: yx 43 1 yx 43 7

6 yx 6 1 7 Then 7 y x 7 7 7 1 27 196 14 1 1 169 169 13 13 x=1 39. A number is divisible by 9, when its digit-sum is divisible by 9. Digit-sum of the given number (excluding *) is 17. If we put * = 1, the number will be perfectly divisible by 9. 40. 97215 * 6 Digit-sum of odd positions = 9 + 2 + 5 + 6 = 22 Digit-sum of even positions (excluding *) = 7 + 1 = 8 The difference of the two should be either 0 or divisible by 11. So * = 3 41. The last three digits should be divisible by 8. So * = 3 42. The largest number of 4 digits = 9999 On dividing 9999 by 88, we get a remainder of 55. Now, if this remainder is subtracted from 9999, the remaining number will be exactly divisible by 88. the required number = 9999 – 55 = 9944. 43. 2 44. 2; n should be even number. 38. 1

45.

a 4 , then by the rule of componendo – dividendo, b 3

Other way:

a 4 3a 4 3 2 , b 3 2b 3 2 1

3a 2b 2 1 3a 2b 2 1 3 46.

2 3 4 n 1 2 .... 3 4 5 n n

47.

5 7 9 1001 1001 .... 3 5 7 999 3

48. Use the formula a 3 b3 (a – b)(a 2 ab b 2 ) Then the given expression =

1 1 137 133 4

49. (?)2 54 96

? 9 6 6 16 3 6 4 72 50. (6)10 (7)17 (11) 27 (2)10 (3)10 (7)17 (11) 27 total number of prime factors = 10 + 10 + 17 + 27 = 64 51. Let the bigger even number be x. And smaller even number be y. Then, x + y = 2y + 6 or, x – y = 6 ... (i) and x – y = 6 ... (ii) We can’t determine the exact value of x and y from both the equations as they are identical. So, the data provided are not adequate to answer the question. 52. Let the smallest even number be x. Then the smallest odd number will be x - 11. Also, (x + x + 2 + x + 4) + (x – 11 + x – 9 + x – 7) = 231 252 42 6 Largest even number = 42 + 4 = 46 Largest odd number = 42 – 7 = 35 Sum = 46 + 35 = 81 Quicker Method (Direct Formula): When sum of three consecutive odd numbers and three consecutive even numbers together is 'X', then Sum of the smallest odd and the smallest even number

6x – 21 = 231;

x

X 12 3 And sum of the largest odd and the largest even number =

=

X 12 3

Quicker Maths

122 Here, sum of the largest even and the largest odd

231 12 81 3 Note: You can verify the above thorem by taking another example!! 53. Let the first number be x. x + x + 2 + x + 4 + x + 6 + x + 8 = 220 5x = 220 – 20 = 200 x = 40 Second lowest number of set B = 40 × 2 – 37 = 43 Required sum = 42 + 43 + 44 + 45 + 46 = 220 Method II: We know that in an AP series the middle number is the average of the series. Here, in set A, the middle number number is

220 44 5 Lowest number of set A = 40 Second lowest number of set B = 40 × 2 – 37 = 43 Middle numbers of set B = 43 + 1 = 44 Sum of number of set B = 44 × 5 = 220 54. Let the five consecutive odd numbers of set A be x, x + 2, x + 4, x + 6 and x + 8 Then x + x + 2 + x + 4 + x + 6 + x + 8 = 125 or, 5x + 20 = 125 or, 5x = 105 x = 21 Highest odd number of Set A = 21 + 8 = 29 Smallest odd number of Set B = 29 + 16 = 45 Sum of four consecutive odd numbers of Set B = 45 + 47 + 49 + 51 = 192 or 3rd number =

125 Method II. Middle number (or average) = = 25 5 So, the five consecutive odd nos. are 21, 23, 25, 27, 29 Reqd sum = 45 + 47 + 49 + 51 = 192 55. The middle no. of the series is the average of five 195 = 39 5 Series is 35, 37, 39, 41, 43. Second highest number of consecutive even nos. = 37 + 5 = 42 Series of consecutive even numbers is 36, 38, 40, 42, 44. Now, required 40% of second lowest even number = 40% of 38 = 15.2 56. Quicker Approach: Sum of the respective even and odd numbers from the highest to the lowest will be: 33, (33 – 4), (33 – 8) and (33 – 12), or, 33, 29, 25, 21 Reqd sum of both odd and even numbers = 33 + 29 + 25 + 21 = 108 consecutive numbers, ie

Note: There are so many possible groups of such even or odd numbers like.

And so on. But the required sum will always be the same. 57. Let the units digit be x. Tens digit = y Hundreds digit = z Then, the number = 100z + 10y + x Given that, y = 3 And z + x = 14 ... (i) Now, according to the question, or, (100x + 10y + z) – (100z + 10y + x) = 396 or, 99x – 99z = 396 x–z=

396 4 99

...(ii)

14 4 14 4 = 9 and z = =5 2 2 number = 100 × 5 + 10 × 3 + 9 = 539 Quicker Method: If a three-digit number (say abc) is changed to another three-digit number by interchanging units and hundreds digits (say cba), then difference of units digit and hundreds digit From (i) and (ii), x =

=

Difference in numbers 99

So, in the given question, c – a = Also given that c + a = 14

396 4 ...(1) 99 ...(2)

4 14 9 4 14 5 and a = 2 2 required number = 539 58. Let the four consecutive even numbers be 2x, 2x + 2, 2x + 4, 2x + 6 So, c =

Then,

1 1 11 2x 2x 2 60

2x 2 2x 11 2 x ( 2x 2) 60 4x 2 11 2x ( 2x 2) 60 120x + 60 = 22x2 + 22x 22x2 + 22x – 120x – 60 = 0 11x2 – 49x – 30 = 0 11x2 – 55x + 6x – 30 = 0 11x(x – 5) + 6(x – 5) = 0

Number System

123

(11x + 6) (x – 5) = 0 x – 5 = 0 (Neglect negative value.) x=5 Reciprocal of the third highest number

60. Number of extra sweets = 2 × 480 = 960 These sweets were to be distributed among 120 children. Number of sweets to be given to each child originally

1 1 1 2x 4 2 5 4 14 Quicker (Logical) approach:

960 8 120 61. Let the no. of hens = h and the no. of goats = g Then, h + g = 90 ... (i) 2h + 4g = 248 ... (ii) Solving these, we get, h = 56 and g = 34 Another Method (Alligation Method): Apply alligation on number of legs per head.

=

=

11 5 6 5 6 1 1 60 60 60 60 12 10 First two consecutive even numbers are 10 and 12. Third number of S1 = 14 1 14 Note that as the numbers are consecutive we should break 11 in two closer parts - 5 and 6 (and not 2 + 9, 3 + 8 or 4 + 7). 59. Let the original number be x. Then, according to the question, 8x – 153 = 153 – x 9x = 306 x = 34 Reqd reciprocal =

34 8.5 4 Note: If we read the question carefully, we see that 153 is exactly in the middle of x and 8x. And 25% of x =

153 = average of x and 8x = x=

2 153 34 9

x 8x 9 x 2 2

= 112 : 68 = 28 : 17

17 No. of goats = 90 28 17 = 34 62. Total number of items = 32 Maximum number of icecreams = 9 pastries > cookies > icecream So, 13 10 9 12 11 9 Hence number of cookies is either 10 or 11. Number of pastries is either 13 or 12.

124

Quicker Maths

Chapter 15

Binary System Number System is a system which represents different numbers in different ways. There are many number systems. All the number systems have different bases. Base denotes the number of symbols in the number system. For example, in the decimal system, the base is 10 and it has 10 numbersymbols (0, 1, 2, 3, ..... 9). Some more examples of number systems are given below in tabular form. Number System Base R e p r e s e n t a t i o n o f Symbols Quinary No. System 5 0, 1, 2, 3 and 4 Octal No. System 8 0, 1, 2, 3, .... 7 Hexadecimal No. System 16 0, 1, 2, ... 9, A(10), B(11), C(12), ... F(15) Binary No. System 2 0 and 1 Binary No. System was introduced by JV Newman in 1946. Conversion of Decimal Number into its Binary equivalent: To find the binary equivalent of a decimal number, we go on dividing the decimal number by the constant divisor 2 till the last quotient is obtained. For example we convert 89 into its binary equivalent. 2

89 2 × q1 (i.e. 44) = 88; 89 - 88 = 1 (r1) 44 2 × q2 (i.e. 22) = 44; 44 - 44 = 0 (r2) 22 2 × q3 (i.e. 11) = 22; 22 - 22 = 0 (r3) 11 2 × q4 (i.e. 5) = 10; 11 - 10 = 1 (r4) 5 2 × q5 (i.e. 2) = 4; 5 - 4 = 1 (r5) 2 2 × q6 (i.e. 1) = 2; 2 - 2 = 0 (r6) 1 (last quotient)

we finally note down the remainders (including the last quotient, 1) strictly in accordance with the arrow mark, i.e. we note down finally in the way: First of all the last quotient, then the last remainder, then the second last remainder, ....., third remainder, then second remainder, and finally the first remainder. That is, the binary number equivalent to 89 is 1011001. Or, by notation, 89 10 (1011001) 2 The base 10 stands for decimal system and the base 2 stands for binary system.

Conversion of Binary Number to its Decimal equivalent: In binary system the value of 1 doubles itself every time it shifts one place to the left, and wherever '0' occurs its value becomes zero. Let us convert 1011001 into its decimal equivalent. To understand easily, we write each digit of 1011001 inside each box in the following way and the value of each box is written above it.

26 1

25 0

24 1

23 1

22 0

20 1

Now 10110012 1 26 0 25 1 2 4 1 23 0 22 0 21 1 20

64 0 16 8 0 0 1 ( 20 1) = 64 + 16 + 8 + 1 = 89 Some solved example: Ex. 1: Convert 90 into its Binary equivalent.

q1 ,q 2 ..... are the first/second/.... quotients and r1 , r2 ..... are the first/second/ .... remainders. For all the stages of division the common divisor 2 remains unchanged and the quotient obtained becomes the next dividend. The process continues till the last quotient (1) is obtained. Here, after dividing the real dividend 89 by 2, the first quotient q1 ( 44) becomes our next dividend. Now after dividing 44 by 2, the second quotient q 2 ( 22) becomes our next dividend. And so on. Every time the remainder is noted down carefully. When the last quotient as 1 is obtained

21 0

Here we get (90)10 (1011010 ) 2

Quicker Maths

126 Verification:

26 1

25 0

24 1

23 1

22 0

21 1

20

1 Verification: (i) (10)2 1 21 0 20 2 0 2 (ii) (11) 2 1 21 1 20 2 1 3 (iii) (100) 2 1 22 0 21 0 2 0 4 0 0 4 (iv) (101)2 1 22 0 21 1 2 0 4 0 1 5 (v) (110) 2 1 22 1 21 0 20 4 2 0 6 (vi) (111)2 1 22 1 21 1 2 0 4 2 1 7

Verification:

23 1

22 0

21 0

20 0

(vii) (1000) 2 1 2 3 0 22 0 21 0 20 = 8 + 0 +0+0=8 Ex. 4: Directions: In a certain code, the symbol of 0 is . There are no other symbols for and 1 is other numbers greater than 1 and are written using these two symbols only. The value of the symbol

(1011000) 2 26 24 23 64 16 8 88 Ex. 3: Convert the following numbers into their binary equivalents: (i) 2 (ii) 3 (iii) 4 (iv) 5 (v) 6 (vi) 7 (vii) 8 Also verify your answers.

Soln: (i)

2 2 0

doubles itself every time it shifts one place to the left. 0 is written as 1 is written as

i.e., ( 2)10 (10) 2

2 is written as

1 (ii)

2 3 1

2 0

4 is written as Now answer the following questions.

i.e., (3)10 (11) 2 i.e., ( 4)10 (100) 2

(i)

2 0

i.e., (5)10 (101) 2 (ii)

1 (v)

2 6 0

3 1 1

3 2 What is the value of 2 2 1.5

a) c) e) None of these

1 (iv) 2 5 1

3 is written as

1 (iii) 2 4 0

i.e., (8)10 (1000) 2

4 0 2 0

2 88 0 44 0 22 0 11 1 5 1 2 0 1 Here we get (88)10 (1011000) 2

24 1

(vii) 2 8 0

0

1 23 0 22 1 21 0 20 = 64 + 0 + 16 + 8 + 0 + 2 + 0 = 64 + 16 + 8 + 2 = 90 Ex. 2: Convert 88 into its Binary equivalent.

25 0

i.e., (7)10 (111) 2

3 1 1

(1011010) 2 1 26 0 25 1 2 4

26 1

(vi) 2 7 1

i.e., (6)10 (110 ) 2 (iii)

b) d)

45 3 3 1 ? 54 5

If be multiplied by , what will be the result? a) 165 b) 180 c) 175 d) 200 e) None of these Which of the following will represent the value of 37.5% of 56?

Binary System

a)

(iv)

b)

2 b) 4 ,

c) 23 32 , 2

(1111) 2

2

c) e) None of these

HCF of

a)

= (1101) 2

1 23 1 22 0 21 1 20 8 4 0 1 13 Now, 15 × 13 = 195 e) answer

Now the binary equivalent of 21 is 2 21 10 5 2

be divided by

b)

d)

and

i.e. (21)10 (10101)2 = answer is

– × =? a) 20 b) 21 c) -19 d) 22 e) None of these Soln: Obviously, these questions are based on Binary Number System. (viii)

(iv) 2 42 21 10 5 2 1

i.e. (42)10 (101010)2 =

d) answer (v) (a)

9 1 7 3 10 18 Now we have to find the binary equivalent of 10. 2 10 0

5 2

1 0

i.e., (10)10 (1010) 2

b)

(10011)2

16 2 1 19 27( 33 )

(b)

(10010) 2 24 0 0 21 0 16 2 18 16 ( 4 2 )

(c)

(1001000) 2 26 0 0 23 0 0 0 64 8 72 8 9 23 32 c) answer

1 Now, using the given symbols for 0 and 1, we get

(1010) 2

24 0 0 21 20

2

8 4 1.5

0 1 0 1 0

45 3 3 –1 (i) 2 2 1.5 54 5 3

1 0 1 0

1

b) d)

c) e) None of these

, the result is

a)

and

3 8

If

1 23 1 2 2 1 21 1 20 8 4 2 1 15

(iii) 37.5% of 56 56 21

d) 2 3 , e) None of these

(vii)

(ii) Using the given symbols we get,

c) d) e) None of these Which of the following pairs have the same numbers? a) 33 ,

(vi)

b)

c) d) e) None of these Which of the following will represent 42? a)

(v)

127

c) answer

(d)

(100110) 2 5 2 0 0 22 21 0 32 4 2 38 36 ( 22 32 )

Quicker Maths

128 (vi)

(10010000) 2

(vii)

(11001) 2

2 4 23 0 0 20 16 8 1 25

27 0 0 24 0 0 0 0

(1111)2

128 16 144

(10010)2 24 0 0 21 0 16 2 18 Now 144 ÷ 18 = 8 = (1000) 2 d) answer

23 22 21 20 8 4 2 1 15 HCF of 25 and 15 = 5 = (101) 2 = Ans = (c) (viii)

= 5; = 4; 5 – 4 × 6 = –19 Ans = (c)

=6

Chapter 16

Permutation and Combination To understand permutation and combination, let us take two examples: Ex. 1: How many triangles can be formed with four points (A, B, C & D) in a plane? It is given that no three points are collinear. From the three points A, B and C, have only one triangle with these points. It is irrespective of the fact where he starts. Although the arrangement of points may be in different orders like ABC, ACB, BAC, BCA, CAB and CBA, but in all these cases the triangles formed ABC , ACB , BAC , BCA , CAB and CBA are exactly the same triangle. With the 4 points A, B, C and D we can form maximum 4 triangles namely ABC , ABD , ACD and BCD . Ex. 2: How many number plates of 3 digits can be formed with four digits 1, 2, 3 and 4? Here, the order of arrangement of digits does matter. For the digits 1, 2 and 3 the different arrangements are: 123, 132, 213, 231, 312 and 321. Here, the vehicles having the number plates 123, 132, 213, 231, 312 and 321 are 6 different vehicles but in Ex 1 the six triangles were the same. The total no. of 3-digits number plates will be = 4 × 3 × 2 = 24. (The 3-digit number plates will bear the number: 123, 132, 124, 142, 134, 143, 213, 231, 214, 241, 234, 243, 312, 321, 314, 341, 324, 342, 412, 421, 413, 431, 423 and 432). Note: Ex.1 is the case of combination and Ex. 2 is the case of permutation. In Ex. 1, total number of triangles 4! 4 C3 4 3! 4 – 3 ! Factorial notation: The product of n consecutive positive integers beginning with 1 is denoted by n! or n and is read as factorial n. (5! or 5 is read as factorial five, 13! as factorial thirteen, etc.) n! 1 2 ..... (n 2) (n 1) n n (n 1) (n 2) .... 2 1 .

For example, 5! = 5 × 4 × 3 × 2 × 1 = 120

4! 4 3 2 1 24, 3! = 3 × 2 ×1 = 6, 2! 2 1 2

9! 9 8 7 6 5 4 3 2 1 4! 4 3 2 1 9 8 7 6 5 4 3 2 1 4 3 2 1 = 9 × 8 × 7 × 6 × 5 = 15120

3! 3! 1 1 7! 7 6 5 4 3! 7 6 5 4 840 If r and x be two positive integers such that r < n then n! n (n 1) (n 2) ..... 2 1 r! r (r 1) ... 2 1 = n × (n – 1) × ... × (r + 1)

n! Similarly, (n – r)! n (n 1) .... (n r 1) Arrangement: Suppose we have four different object A, B, C and D. We have to form a group of two objects out of these four objects. In other words, we have to form a group of four different objects taken two at a time. Clearly, we will have six such groups: (i) A and B, (ii) A and C, (iii) A and D, (iv) B and C, (v) B and D, (vi) C and D. By the notational representation, the total no. of such groups

4 C2

4! 4! 6 2!(4 2)! 2! 2!

Now, the two objects in each of these groups can be arranged in two different ways namely (i) A and B & B and A, (ii) A and C & C and A, and so on. Thus, there are a total of twelve such arrangements. Total no. of arrangements = total no. of groups × r! Where r is the no. of objects in each group. In the above example, total no. of arrangements

6 2! 6 (2 1) 12.

Quicker Maths

130 Definition of permutation: Each of the different arrangements which can be made by taking some or all of the given things or objects at a time is called a permutation. The symbol n Pr denotes the no. of permutations of n different things taken r at a time. The letter P stands for permutation.

n! n Also, Pr (n r)! Thus, the symbol 9 P4 denotes the no. of permutations or arrangements of 9 different things taken 4 at a time and 9

P4

9! 9! (9 4)! 5!

9 (9 1) ...(5 1) 9 8 7 6 3024 Definition of combination: Each of the different selections or groups which can be made by taking some or all of a no. of given things or objects at a time is called a combination. The symbol

n

Cr

denotes the no. of combinations of n different things taken r at a time. The letter C stands for combination.

n! n Also, C r r!(n r)! Thus, the symbol 9 C 4 denotes the no. of selections, or groups of 9 different things taken 4 at a time and 9! 9! 9C 4 4!(9 4)! 4! 5!

9 8 7 6 5! 126 4 3 2 1 5! Note: In the topic Arrangement we have, Total no. of arrangements = total no. of groups or selection r! where r is the no. of objects in each group or selection.

So, n Pr n C r r! For example, 9 P4 4! 9 C 4

Some Fundamental Principles of Counting I.

Multiplication Rule: Suppose one starts his journey from place X and has to reach place Z via a different place Y.

For Y, there are three means of transport – bus, train and aeroplane – from X. From Y, the aeroplane service is not available for Z. Only either by a bus or by a train can one reach Z from Y. Also, there is no direct bus or train service for Z from X. We want to know the maximum possible no. of ways by which one can reach Z from X. For each means of transport from X to Y there are two means of transport for going from Y to Z. Thus, for going from X to Z via Y there will be 2 (firstly, by bus to Y and again by bus to Z; secondly, by bus to Y and thereafter by train to Z. +2 (firstly, by train to Y and thereafter by bus to Z; secondly, by train to Y and thereafter by train to Z.) +2 (firstly by aeroplane to Y and thereafter by bus to Z, secondly by aeroplane to Y and thereafter by train to Z.) = 3 × 2 = 6 possible ways.

We conclude: If a work A can be done in m ways and another work B can be done in n ways and C is the final work which is done only when both A and B are done, then the no. of ways of doing the final work, C = m × n. In the above example, suppose the work to reach Y from X = the work A in m i.e. 3 ways. The work to reach Z from Y = the work B in n i.e. 2 ways. Then the final work to reach Z from X = the final work C in m × n, i.e. 3 × 2 = 6 ways. II. Addition rule: Suppose there are 42 men and 16 women in a party. Each man shakes his hand only with all the men and each woman shakes her hand only with all the women. We have to find the maximum no. of handshakes that have taken place at the party. From each group of two persons we have one handshake. Case 1: Total no. of handshakes among the group of 42 men

42 C2

42! 42! 42 41 40! 2!(42 – 2)! 2! 40! 2 1 40!

= 21 × 41 = 861 Case 2: Total no. of handshakes among the group of 16 women

16

C2

16! 16 15 14! 2!(16 – 2)! 2 1 14!

8 15 120 Maximum no. of handshakes = 861 + 120 = 981.

Permutation and Combination To find the no. of permutations or arrangements of n different things, all the n things taken at a time. Suppose a student has 3 books (B1, B2 and B3) and his book-rack has 3 shelves. He has to arrange the books in the shelves. Case I: He puts one book in each shelf: He can put anyone of the 3 books in the first shelf. He is left with 2 books and he can put anyone of the remaining 2 books in 2 ways in the second shelf. Now, he is left with a single book which can be put in 1 way in the third shelf. Total no. of ways in which he can put the books = 3 × 2 × 1 = (3! =) 6 Case II: He puts 2 books together in one shelf and the remaining 1 book in another shelf. He can put 2 books together out of the 3 different books in 3P2 ways in one shelf. The remaining one books can be put in 1 way in anyone of the remaining two shelves. Total no. of ways of putting the books 3

3! 3! P2 1 P2 3! 6 (3 – 2)! 1! 3

Case III: He puts all the 3 books together in one shelf. He can put all the 3 books in any one of the shelves in any one of the following sequences: When the book B1 is at the top, B2 and B3 can be arranged in two ways: B2 in the middle and B3 at the bottom, and B3 in the middle and B2 at the bottom. So, we see if the book B1 is at the top in anyone of the shelves there are 2 ways of arrangement. Similarly, when B2 is at the top, there are 2 ways of arrangement and when B3 is at the top there are 2 ways of arrangement. Total no. of ways of putting the books =2+2+2 = 3 × 2 (= 3!) = 6 We see in all the above three cases, total no. of ways of putting all the 3 books = 3! Thus, we conclude that total no. of arrangements of n different things, all (the n) things taken at a time = nPn = n! ..... (i) n We have Pr

n! (n r)!

n! n! n Pn (n n)! 0! ...... (ii)

131 So, n!

n! [equating (i) and (ii)] 0!

n! 1 n! Thus, we get 0! = 1 To find the no. of permutations or arrangements of n different things taken r at a time when each thing can be repeated any no. of times. or, 0! =

Note:

n

Pr , the no. of permutations or arrangements of n

n! different things taken r at a time (n r)! , when repetition is not allowed. Now, suppose a painter has to paint a 4-digit number on a number plate of vehicles using the digits 1, 2, ...., 9 and repetition of digits is allowed (i.e. he can paint the numbers 1111, 1112, 1211, 1121, 1221, 2121, etc.).

thousands place

hundreds place

tens place

units place

any one of the 9 digits 1, 2, 3, ....9 in 9 ways He can mark any one digit out of the 9 digits 1, 2, 3, .... 9 at thousands place on the number plate in 9 ways. After marking at the thousands place he has again 1, 2, ...., 9 (total 9) digits (as repetition of digits is allowed). So, he can mark the hundreds place in 9 ways. Similarly, each of tens place and units place can be marked in 9 ways. Thus, he can mark a total of 9× 9 × 9 × 9 (= 9 × 9 .... 4 times = 94) = 6561, number plates Now, we conclude the no. of permutations or arrangements of n different things taken r at a time, when repetition is allowed = n × n × n ..... r times = nr ways. Now, suppose the painter has to paint 4-digit numbers on the number plates using all the ten digits (0, 1, 2, .... 9) and repetition of digits is allowed.

thousands hundreds tens units place place place place any one of the any one of the Similarly 9 digits 1, 2, 10 digits 0, Similarly in ..... 9 1, 2, ....9 in in 10ways 10 ways in 9 ways 10 ways Note: If he puts 0 at thousands place, the 4-digit no. will reduce to a 3-digit no. Thus he cannot do so. Reqd. total no. = 9 × 10 × 10 × 10 = 9000

Quicker Maths

132

(m n 3)

From the examination point of view the following few results are useful. Without going into details you should simply remember the following results: I.

= 3024 = 9 × 8 × 7 × 6 .... (i) m+n=9 Again,

n n If C x C y then either x = y or x + y = n

mn

II. No. of permutations of n things out of which P are alike and are of one type, q are alike and are of the

(m n 3)

n! other type, and the remaining all are different = p! q! III. No. of selections of r things (r n) out of n identical things is 1. IV. Total no. of selections of zero or more things from n identical things = n + 1. V. Total no. of selections of zero or more things from n different things

= 120 5 4 3 2

mn5

Ex. 4: If n C2 n C5 , find n. Soln:

or, 5 × 4 × 3 × 2 × (n – 5)! = 2 × (n – 2) × (n – 3) × (n – 4) × (n – 5)! or, 5 × 4 × 3 = (n – 2) × (n – 3) × (n – 4) n – 2 = 5 or, n = 7

n r 1

C r 1. Note:

Solved Examples

Soln:

Here 2 5

n! 210 (n 3)!

n

7! 210 (7 – r)!

or, 7 × 6 × .... × (7 – r + 1) = 7 × 6 × 5 7–r+1=5 or, 8 – r = 5 r=8–5=3 Ex. 3: If m+nP4 = 3024 and m-nP4 = 120, find m and n. (m n)! mn P4 Soln: (m n – 4)!

(m n) (m n 1) (m n 2)

Whenever n C x n C y and x y , then n must be equal to x + y.

Ex. 1: If nP3 = 210, find n.

or, n × (n – 1) × (n – 2) = 7 × 6 × 5 =7 Ex. 2: If 7Pr = 210, find r.

n! n! 2!(n 2)! 5!(n 5)! or, 5!(n 5)! 2!( n 2)!

VI. No. of ways to distribute (or divide) n identical things among r persons where any person may get any no.

Soln:

.... (ii)

From the equations (i) and (ii), we get m = 7 and n = 2

n C0 n C1 n C2 ..... n Cn 2n

of things =

P4 (m n) (m n 1) (m n 2)

n=2+5=7 Ex. 5: How many quadrilaterals can be formed by joining the vertices of an octagon? Soln: A quadrilateral has 4 sides or 4 vertices whereas an octagon has 8 sides or 8 vertices. Reqd no. of quadrilaterals 8765 70 24 Ex. 6: How many numbers of five digits can be formed with the digits 1, 3, 5, 7 and 9, no digit being repeated? Soln: The given no. of digits = 5 8! 8 = C4 4!(8 – 4)!

Note:

Reqd no. = 5 P5 5! 120 If repetition of digits be allowed, then reqd no. 55 3125.

Permutation and Combination

133

Ex. 7: How many numbers of five digits can be formed with the digits 0, 2, 4, 6 and 8? Soln: ten thousands place thousands place hundreds place tens place units place

in 4P1 i.e. 4 ways (any one of 2/4/6/8)

After filling up ten thousands place we are left with 4 digits, including 0, and the number of blank place s is 4. So, in 4P4 = 4! = 24 ways

Required number = 4 × 24 = 96 Note: If repetition of digits be allowed then reqd. no. = 4 × 54 = 2500 (For ten thousands place, we can’t consider 0. Ex. 8: How many numbers of five digits can be formed with the digits 0, 1, 2, 3, 4, 6 and 8? Soln: Here nothing has been said about the repetition of digits. So, it is understood that repetition of digits is not allowed. ten thousands place thousands place P1 = 6 ways (exclude 0)

6

hundreds place

tens place

units place

After filling up ten thousands place we are left with 6 digits, (including 0), and the blank places ar 4. So, in 6P4 = 6×5×4×3 = 360 ways

Reqd no. = 6 × 360 = 2160 Ex. 9: How many even numbers of three digits can be formed with the digits 0, 1, 2, 3, 4, 5 and 6? Soln: Case (i): When 0 occurs at units place: hundreds place tens place units place 6 P2 = 6 × 5 = 30 ways Only 0, i.e. in 1 way Total of such numbers = 30 × 1 = 30 Case (ii): When 0 does not occur at units place:

hundreds place After filling of units place we are left with 6 digits but 0 cannot occur at hundreds place. We are finally left with 5 digits, so in 5P1=5 ways.

tens place After filling up units place and hundreds place we are left with 5 digits (including 0) so in 5 ways.

units place any one of 2/4/6 in 3 ways

Total of such numbers = 5 × 5 × 3 = 75 Reqd no. = 30 + 75 = 105 Ex. 10: How many nos. greater than 800 and less than 4000 can be made with the digits 0, 1, 2, 4, 5, 7, 8, 9, no number (digit) occurring more than once in the same number? Soln: Case 1: 3-digit numbers:

hundreds place either 8 or 9, i.e. in 2 ways Total no. = 2 × 42 = 84 Case 2: 4-digit numbers:

tens place units place 7 P2 = 7 × 6 = 42 ways

thousands place hundreds place tens place units place 7 Either 1 or 2, i.e. in 2 ways P3 = 7 × 6 × 5= 210 ways Total no. = 2 × 210 = 420 Reqd no. = 84 + 420 = 504

Quicker Maths

134 Ex. 11: Find the number of words formed with the letters of the word ‘DELHI’ which (i) begins with D, (ii) ends with I, (iii) has the letter L always in the middle, and (iv) begins with D & ends with I Soln: There are 5 letters in the word ‘DELHI’. Case (i): D 4 1 way P4 = 24 ways Reqd no. = 1 × 24 = 24 Case (ii): I P4 = 24 ways 1 way Reqd no. = 1 × 24 = 24 Case (iii): L 1 ways Remaining 4 places will be filled in = 4P4 = 24 ways Reqd no. = 24 Case (iv): 4

D I 3 1 way P3 = 6 ways 1 way Reqd no. = 6 × 1 × 1 = 6 Ex. 12: How many words can be formed with the letters of the word ‘EQUATION’? Soln: No. of permutations or arrangements of n different things, taken all at a time, i.e. nPn = n! Here, there are 8 letters in the word EQUATION. Reqd no. of words = 8! = 40320 Ex. 13: How many words beginning with vowels can be formed with the letters of the word EQUATION? Soln: There are 8 letters in the word EQUATION.

A/E/I/O/U I in 5 ways in 7P7 = 7! = 5040 Reqd. no. = 5 × 5040 = 25200 Ex. 14: How many words can be formed with the letters of the word INTERNATIONAL? Soln: There are 13 letters in the word INTERNATIONAL, of which N occurs thrice, each of I, T and A occurs twice, and the rest are different. 13! reqd. no. 3! 2! 2! 2! 13 12 11 10 9 8 7 6 5 4 3 2 6 2 2 2 = 13 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 3 × 2 = 129729600

Ex. 15: In how many ways can 4 boys and 5 girls be seated in a row so that they are alternate? Soln:

G B G B G B G B G The diagram shows the possible arrangement of sitting of boys and girls. Now, 4 boys can be seated in 4 places in 4P4 = 4! and 5 girls in 5 places in 5! ways. Reqd no. of ways = 4! × 5! = 24 × 120 = 2880 Ex. 16: There are 4 boys and 4 girls. In how many ways can they be seated in a row so that all the girls do not sit together? Soln: Total no. of persons = 4 + 4 = 8 When there is no restriction they can be seated in a row in 8! ways. But when all the 4 girls sit together, we can consider the group of 4 girls as one person. Therefore, we have only 4 (no. of boys) + 1 = 5 persons, who can be arranged in a row in 5! ways. But the 4 girls can be arranged among themselves in 4P4 = 4! ways. No. of ways when all the 4 girls are together = 5! × 4! Reqd no. of ways in which all the 4 girls do not sit together = 8! – 5! × 4! = 8 × 7 × 6 × 5! – 5! × 24 = 5! (336 – 24) = 120 × 312 = 37440 Ex. 17: How many different words can be formed with the letters of the word EQUATION without changing the relative order of the vowels and consonants? Soln: In the word EQUATION, the 5 vowels E, U, A, I and O occupy the 5 places 1, 3, 4, 6 and 7 respectively whereas the 3 consonants Q, T and N occupy the 3 places 2, 5, and 8 respectively. All the letters of the word are different i.e. there is no repetition of any letter. The 5 vowels can be arranged in the 5 places in 5 P5 = 5! = 120 ways whereas the 3 consonants can be arranged in the 3 places in 3P3 = 3! = 6 ways. Reqd no. = 120 × 6 = 720 Ex. 18: How many words can be formed out of the letters of the word BANANA so that the consonants occupy the even places? Soln:

1 2 3 4 5 6 The word BANANA contains 6 letters out of which A occurs thrice and N occurs twice. The 3 consonants B and N (which occurs twice)

Permutation and Combination

135

can be arranged at the 3 even places 2, 4 and 6 in 3! = 3 ways 2! The remaining 3 odd places can be arranged with 3! triple A in = 1 way 3! Reqd no. of words = 3 × 1 = 3 Ex. 19: Find the no. of ways in which 4 identical balls can be distributed among 6 identical boxes, if not more than one ball goes into a box? Soln: No. of identical balls = 4 and no. of identical boxes =6 Now, distributing 4 identical balls among 6 identical boxes when not more than one ball goes into a box, implies to select 4 boxes from among 6! the 6 boxes, which can be done in 6C4 = 4! 2! = 15 ways. Ex. 20: Find the no. of triangles formed by joining the vertices of a polygon of 12 sides. Soln: A polygon of m sides will have m vertices. A triangle will be formed by joining any three vertices of the polygon.

Ex. 22: In a party every person shakes hand with every other person. If there was a total of 210 handshakes in the party, find the no. of persons who were present in the party. Soln: For each selection of two persons there will be one handshake. So, no. of handshakes in the party = nC2, where n = no. of persons. Now, nC2 = 210 (given)

m! 3! (m 3)! m (m 1) (m 2) (m 3)! 6 (m 3)!

1 9 8 7 6 126 24 (ii) When one particular person has to be always excluded from the 5-member delegation, we are left with 10 – 1 = 9 persons. So selection can be done in 9C5 ways. Reqd no. = 9C5 = 126 Ex. 24: Find the no. of triangles formed by the 11 points (out of which 5 are collinear) in a plane. Soln: Let us suppose that the 11 points are such that no three of them are collinear. Now a triangle can be formed by joining any three of these 11 points. So, selection of any 3 out of the 11 points can be done in 11C3 ways. No. of triangles formed by 5 points when none of the 3 or more points are collinear = 5C3 But from 3 or more than 3 collinear points no triangle can be formed. Reqd. no. of triangles = 11C3 – 5C3

m No. of triangles formed = C3

m (m 1) (m 2) 6

Putting m = 12, we get 12 11 10 220 6 Ex. 21: Find the no. of diagonals of a polygon of 12 sides. Soln: A polygon of m sides will have m vertices. A diagonal or a side of the polygon will be formed by joining any two vertices of the polygon. No. of diagonals of the polygon + no. of sides of the polygon (= m) = mC2 No. of diagonals of the polygon = mC2 – m m! m (m 1) m m 2!(m 2)! 2 m(m – 1) – 2m m(m – 3) 2 2 Putting m = 12, we get the reqd. no. of diagonals

Reqd. no. of triangles =

12 9 54 . 2

or,

n (n 1) 210 2

or, n × (n – 1) = 2 × (2 × 3 × 5 × 7) = 21 × 20 n = 21 Ex. 23: There are 5 members in a delegation which is to be sent abroad. The total no. of members is 10. In how many ways can the selection be made so that a particular member is always (i) included (ii) excluded? Soln: (i) Selection of one particular member can be done in = 1C1 = 1 way. After the selection of the particular member, we are left with 9 members and for the delegation, we need 4 members more. So selection can be done in 9C4 ways. Reqd no. of ways of selection = 1C1× 9C4

11 10 9 5 4 165 – 10 155 6 2 Ex. 25: A person has 12 friends out of which 7 are relatives. In how many ways can he invite 6 friends such that at least 4 of them are relatives?

Quicker Maths

136 Soln:

No. of non-relative friends = 12 – 7 = 5 He may invite 6 friends in following ways: I: 4 relatives + 2 non-relatives 7 C 4 5 C 2 II: 5 relatives + 1 non-relative 7 C 5 5 C1

III: 6 relatives + 0 non-relative 7 C 6 Reqd. no. of ways = 7 C 4 5 C2 7 C5 5C1 7 C6 = 35 × 10 + 21 × 5 + 7 = 462 Ex. 26: In an examination, a minimum of marks is to be scored in each 6 subjects to pass. In how many ways can a student fail? Soln: The student will fail if he fails in one or more subjects out of 6 different subjects, i.e. 6C1 + 6C2 + 6 C3 + 6 C4 + 6 C5 + 6 C6 = (6C0 + 6C1 + ... + 6C6) – 6C0 = 26 – 1 = 64 – 1 = 63 ways Other approach: He fails if he fails in any of the 6 subjects. With each paper there two possibilities: either fail or pass. This way, for 6 subjects there are 2 × 2 × 2 × 2 × 2 × 2 =26 = 64 possible cases. This also includes the case when he passes in all the 6 subjects. Thus, he can fail in 64 – 1 = 63 ways. Note: There are 6 questions in a question paper. In how many ways can a student solve one or more questions? This is the same equation. So, answer is 63. Ex. 27: In how many ways can 12 different books be divided equally among (a) 4 persons (b) 3 persons? Soln: (a) Each person will get 12 ÷ 4 = 3 books. Now, first person can be given 3 books out of 12 different books in 12C3 ways. Second person can be given 3 books out of the rest (12 – 3 =) 9 books in 9C3 ways. Similarly, third person in 6 C3 and the fourth person in 3C3 ways. Reqd. no. of ways = 12C3 × 9C3 × 6C3 × 3C3 12! 9! 6! 3! 12! 3! 9! 3! 6! 3! 3! 3! 0! (3!)4 12 11 10 9 8 7 6 5 4 3! 369600 3! 6 6 6 (b) Now each person will get 12 ÷ 3 = 4 books. Similarly, required no. of ways = 12C4 × 8C4 × 4C4 12! 8! 4! 12! 4! 8! 4! 4! 4! 0! (4!)3

12 11 10 9 8 7 6 5 4! 34650 4! 24 24

Ex. 28: In how many ways can 12 different books be divided equally among (a) 4 sets or groups; (b) 3 sets or groups? Soln: (a) Reqd. no. of ways

C3 9 C3 6 C3 3C3 12! 15400 = 4! (3!)4 4! (b) Required no. of ways 12

12

=

C4 8C4 4 C4 12! 5775 3! 3!(4!)3

Ex. 29: How many different letter arrangements can be made from the letters of the word EXTRA in such a way that the vowels are always together? Soln: Considering the two vowels E and A as one letter, the total no. of letters in the word ‘EXTRA’ is 4 which can be arranged in 4P4, i.e. 4! ways and the two vowels can be arranged among themselves in 2! ways. reqd no. = 4! × 2! = 4 × 3 × 2 × 1 × 2 × 1 = 48 Ex. 30: Letters of the word DIRECTOR are arranged in such a way that all the vowels come together. Find out the total number of ways for making such arrangement. 1) 4320 2) 2720 3) 2160 4) 1120 5) None of these Soln: 3;Taking all vowels (IEO) as a single letter (since they come together) there are six letters among which there are two R. 6! 3! 2160 2! Three vowels can be arranged in 3! ways among themselves, hence multiplied with 3!. Hence, answer is (3). Ex. 31: How many different letter arrangements can be made from the letters of the word RECOVER? 1) 1210 2) 5040 3) 1260 4) 1200 5) None of these

Hence no. of arrangements =

Soln: 3; Possible arrangements are:

7! 1260 2! 2!

[division by 2 times 2! is because of the repetition of E and R] Ex. 32: 4 boys and 2 girls are to be seated in a row in such a way that the two girls are always together. In how many different ways can they be seated? 1) 1200 2) 7200 3) 148 4) 240 5) None of these

Permutation and Combination Soln: 4; Assume the 2 given students to be together (i.e one). Now there are five students. Possible ways of arranging them are = 5! = 120 Now they (two girls) can arrange themselves in 2! ways. Hence total ways = 120 × 2 = 240 Ex. 33: On the occasion of a certain meeting each member gave shakehand to the remaining members. If the total shakehands were 28, how many members were present for the meeting? 1) 14 2) 7 3) 9 4) 8 5) None of these Soln: 4; A combination of 2 persons gives a result of one handshake. If we suppose that there are x persons then there are total x C2 28 x(x – 1) 28 2! or, x(x – 1) = 56 = 7 × 8 x=8 Ex. 34: How many different numbers of six digits (without repetition of digits) can be formed from the digits 3, 1, 7, 0, 9, 5? (i) How many of them will have 0 in the unit place? (ii) How many of them are divisible by 5? (iii) How many of them are not divisible by 5? Soln: The total numbers of 6 digit numbers = 6 ! – 5! = 600. [Note that 5! numbers are for those having 0 in first place which will be excluded.] (i) 5! = 120 (ii) Numbers are divisible by 5 if (a) they will have zero in the unit place and hence the remaining 5 can be arranged in 5! = 120 ways. (b) they will have 5 in the last place and as above we will have 5! = 120 ways. These will also include numbers which will have zero in the first place (ie, number of 5 digits). Therefore the numbers having zero in 1st and 5 in unit place will be 4!. Therefore 6 digit numbers having 5 in the end will be 5! – 4! = 120 – 24 = 96. Therefore the total number of 6 digits numbers divisible by 5 is 120 + 96 = 216. (iii) Not divisible by 5. Total – (divisible by 5) = 600 – 216 = 384. Ex. 35: Find the total number of 9 digits numbers which have all different digits.

or,

137 Soln:

Total No. of 9 digit numbers 10! 9! – 1! 1! = 9! (10 – 1) = 9 (9!) = 9 × (9 × 8 × 7 × 6!)

=

10

C 9 – 9 C8

81 56 720 3265920. Alternative. The number is to be of 9 digits. The first place (from left) can be filled in 9 ways only (as zero can not be in the first place). Having filled up the first place the remaining 8 places can be filled up by the remaining 9 digits in 9

P8 9! ways. Hence the total is 9 9! .

Ex. 36: (a) How many different arrangements can be made by using all the letters in the word MATHEMATICS? How many of them begin with C? How many of them begin with T? (b) How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. Soln: (a) There are 11 letters Two M, Two A, Two T, H, E, I, C, S. (ii) Hence the number of words by taking all at a time

11! 11 10 9 8 7 6! 2! 2! 2! 222

= 990 7 720 990 5040 4989600 .

To Begin with C. Having fixed C at first place we have 10 letters in which 2 are M, 2 are A and 2 are T and the rest 4 are different. Hence the number of words will be

10! 10 9 8 7 6! 90 7 720 2! 2! 2! 22 2 = 630 × 720 = 453600. To Begin with T. Having fixed T in first place we will have only 10 letters out of which 2 are M s and 2 are A s and rest six are H, E, I, C, S and T. Hence the number of words is

10! 907200 = (i.e. double of part (ii)) 2! 2! (b) We can choose 4 letters from the 11 listed in part (a) as under. All the four different

Quicker Maths

138 We have 8 different types of letters and out of these 4 can be chosen in

alike, can be arranged amongst themselves in

= 12 ways. Hence the total number of words is 63 ×12 = 756. Two alike of one kind and two alike of other kind. Out of 3 pairs of like letters we can choose 2

8! 8 7 6 5 1680 4! Two different and two alike. We have 3 pairs of like letters out of which one 8

P4

pair can be chosen in 3 C1 = 3 ways. Now we have to choose two out of the remaining 7 different types of letters which can be done in

7 C2

4! 2!

pairs in 3 C 2 ways 3 ways. One such group is MM AA. These four letters out of which 2 are alike of one kind and 2 are alike of other kind, can be arranged

7! 76 21 ways 5! 2! 2

Hence the total number of groups, of 4 letters in which 2 are different and 2 are alike is 3 21 63 Let one such group be M, H, M, I. Each such group has 4 letters out of which 2 are

in

4! 6 ways. 2! 2!

Hence the total number of words of this type is 3 × 6 = 18. Therefore total number of 4 letter words is 1680 + 756 + 18 = 2454.

EXERCISES 1. If

15

C

r –1

:

15

C 5 : 11 , find r.. r

2. If n C n – 2 462 , find n. 3. How many numbers of five digits can be formed with the digits 0, 1, 2, 4, 6 and 8? 4. How many odd numbers of three digits can be formed with the digits 0, 1, 2, 3, 4, 5 and 6? 5. How many numbers of 4 digits, divisible by 5, can be formed with the digits 0, 2, 5, 6 and 9? 6. How many words of 4 letters beginning with either A or E can be formed with the letters of the word EQUATION? 7. In how many ways can be the letters of the word INTERMEDIATE be arranged? 8. How many words can be formed out of the letters of the word ARTICLE so that the vowels occupy the even places? 9. How many words can be formed with the letters used in EQUATION when any letter may be repeated any no. of times? 10. How many different words of 5 letters can be formed with the letters of the word EQUATION so that the vowels occupy odd places? 11. If 7 parallel lines are intersected by another 7 parallel lines, find the no. of parallelograms thus formed.

12. There are five students A, B, C, D and E. (i) In how many ways can they sit so that B and C do not sit together? (ii) In how many ways can a committee of 3 members be formed so that A is always included and E is always excluded? 13. There are 12 points in a plane out of which 5 are collinear. Find the no. of straight lines formed by joining them. 14. A candidate is required to answer 6 out of 10 questions which are divided into groups, each containing five questions. In how may ways can he answer the questions, if he is not allowed to attempt more than 4 questions from a group? 15. A committee of 8 students is to be formed out of 5 boys and 8 girls. In how many ways can it be done so that the no. of girls is not less than the no. of boys? 16. From 6 gentlemen and 4 ladies a committee of 5 is to be formed. In how many ways can this be done if the committee is to include at least one lady? 17. A candidate is required to answer 6 out of 10 questions which are divided into two groups each containing 5 questions and he is not permitted to attempt more than 4 from each group. In how many ways can he make up his choice?

Permutation and Combination

139

18. How many different groups can be selected for playing tennis out of 4 ladies and 3 gentlemen there being one lady and one gentleman on each side? 19. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women? 20. A team of 5 children is to be selected out of 4 girls and 5 boys such that it contains at least 2 girls. In how many different ways can the selection be made? 21. On a shelf there are four books on Economics, three on Management and four on Statistics. In how many different ways can the books be arranged so that the books on Economics are kept together? Directions (Q. 22-23): Study the following information to answer the given questions: A Committee of some members is to be made from a group of 8 men and 6 women. In how many different ways can it be made if the Committee is to be made according to the following stipulation? 22. The Committee of 4 must include at least 1 woman. 23. The Committee of 6 must have exactly 3 men and 3 women.

Directions (Q. 24-25): Answer the questions on the basis of the following data. A committee of 5 members is to be formed by selecting out of 4 men and 5 women. 24. In how many different ways can the committee be formed if it should have at least 1 man? 25. In how many different ways can the committee be formed if it should have 2 men and 3 women? Directions (26-28): Study the following information to answer the given questions. A Committee is to be formed from a Group of 6 women and 5 men. Out of the 6 women 2 are Teachers, 2 Social Workers and 2 Doctors. Out of the 5 men 3 are Teachers and 2 Doctors. In how many different ways can it be done? 26. Committee of 6 persons in which at least 2 are Doctors. 27. Committee of 2 Teachers, 2 Doctors and 1 Social Worker. 28. Committee of 5 with 3 Females and 2 Males and out of which having 2 Social Workers and at least 1 Female Doctor and at least 1 Male Doctor.

Solutions 1. Answer = 5 2. Answer = 11 3. Required no. of numbers = 5 P4 5 5! 5 120 600 4. Required no. of numbers = 5 × 5 × 3 = 75 5. For a digit to be divisible by 5, its unit digit must be either 0 or 5. When there is 0 at the unit place, the number of 5

numbers P3 1 24 When there is 5 at the unit place, the number of 4

numbers

total required numbers = 24 + 18 = 42 6. Required no. of such words

three vowels can be put in 3 even places in 3P3 = 6 ways. And the four consonants can be arranged in 4 odd places in 4 P4 24 ways.

total no. of words = 6 × 24 = 144. 9. Required no. of words = 8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 10. The three odd places can be occupied by 5 vowels in 5

P2 = 5 × 4 × 3 = 60 ways. Whereas the two even places can be occupied by 3 consonants in 3 P2 = 3 × 2 = 6 ways. Required no. of words = 60 × 6 = 360. 11. No. of such parallelograms

12! total no. of such words = 3! 2! 2! 19958400

= 7 C2 7 C2 21 21 441 12. (i) No. of ways in which A and B sit together = 2 × 4! = 48 No. of ways in which A and B do not sit together = 5! – 48 = 120 – 48 = 72 (ii) After selection of A, we are left with 3 persons (excluding E) out of which 2 are to be selected.

8. There are three vowels and four consonants. So, the

total no. of required ways = 1 × 3 C2 = 1 × 3 = 3.

= 2 P1 7 P3 2 (7 6 5) 420 7. There are 12 letters in the given word, out of which E occurs thrice, each of I and T occurs twice, and the rest occur only once.

Quicker Maths

140 13. Required no. of straight lines =

12

C2 – 5C2 1 57

14. Total no. of ways 5 5 = ( C 2 C4 )

5

C3 5 C3

5

C 4 5 C 2 200

15. Total no. of ways 8 C8 5 C1 8 C7 5 C2 8 C6 5 C3 8 C5 5 C4 8 C4

= 1230 16. 6 Gentlemen, 4 Ladies; Committee of 5. At least one lady to be included; the combinations are: (1L, 4G), or (2L, 3G), or (3L, 2G), or (4L, 1G) 4

6

4

6

4

6

4

6

C C C C C C C C 1

4

2

3

3

2

4

1

= 60 + 120 + 60 + 6 = 246. Quicker Method: [Total no. of committees (from 6 men & 4 ladies)] [No. of committee without lady (only men)] 10 9 8 7 6 6 256 – 6 246 1 2 3 4 5 1 17. Group A and group B consists of 5 questions each out of which 6 are to be attempted but not more than 4 from any group (4A, 2B), (3A, 3B), (2A, 4B) 10 C5 – 6 C5

5

C4 5C 2 5C3 5 C3 5 C 2 5 C4

50 100 50 200. 18. 4 Ladies 3 Gentlemen A B 1L, 1G 1L, 1G 4

3

women, ie 3C2 . Hence, required number of ways = 7C2 3C2

[

4 3 5 4 3 4 3 2 5 4 1 5 1 2 1 2 3 1 2 3 1 2 = 60 + 40 + 5 = 105 21. The required number of ways = 8! × 4! = 967680 22. Total number of persons = 8 + 6 = 14 Total number of selections of 4 members out of 14

=

persons =

14

C4 Total number of selections of 4 members when no woman is included = 8 C4 Required number of ways = Total number of selections – Total number of selections without women = =

14

C 4 8 C 4

14 13 12 11 – 8 7 6 5 1 2 3 4 1 2 3 4

23. The required number of ways

Selection of side B = 3 C1 2 C1 3 2 6 Hence the number of ways of selection for the team = 12 × 6 = 72. 19. The group consists of 7 men and 3 women. We have to select 5 out of 7 men, ie 7C2 , and 2 out of 3

7 6 3 2 = 63 2 1 2 1

= 4 C2 5 C3 4 C3 5 C2 4 C4 5 C1

= 1001 – 70 = 931

Selection of side A = C1 C1 4 3 12 After selecting side A, we are left with 3L and 2G from which one each is to be chosen for side B.

=

I. Selecting 2 girls out of 4 and 3 boys out of 5. This can be done in 4 C2 5 C3 ways. II. Selecting 3 girls out of 4 and 2 boys out of 5. This can be done in 4 C3 5 C2 ways. III. Selecting 4 girls out of 4 and 1 boy out of 5. This can be done in 4 C4 5 C1 ways. Since the team is formed in each case, therefore the total number of ways of forming the team.

n

Cr n Cn r ]

20. A team of five children consisting of at least two girls can be formed in the following ways:

8

c3 6 c3

8 7 6 65 4 1 2 3 1 2 3 = 56 × 20 = 1120 24. The total number of ways of forming the committee

=

= 9 C5 = 126 ways The total number of ways of forming the committee when the committee consists of no male member = 5 C5 = 1 way.. Hence, the required number of ways = 126 – 1 = 125 25. Here, two men out of 4 men can be selected in 4 C2 ways. Also, three women out of 5 women can be selected in 5 C3 ways. Hence, the total number of different ways of selection = =

4

C2 5 C3

43 5 43 = 6 × 10 = 60 2 1 3 2 1

Permutation and Combination

141

26. Total number of committee of 6 persons =

11

C6 Total number of committee without any doctor = 7 C6 Total number of committee with 1 doctor = 7 C 5 4 C1 The required number of ways 11

C 6 7 C 5 4 C1 7 C 6 = 462 – 84 – 7 = 371 27. There are total 5 teachers, 4 doctors and 2 social workers. Therefore, =

The required number of ways =

5

C2 4 C2 2 C1 =

10 × 6 × 2 = 120 28. Case I: If the group of five persons (the committee) includes two female social workers, one female doctor, one male doctor and one male teacher, then the possible number of ways = 2 C2 2 C1 2 C1 3 C1 = 12 ways Case II: If the group of five persons includes two female social workers, one female doctor and two male doctors, then the possible number of ways

C2 2 C1 2 C2 = 2 ways Total number of ways = 12 + 2 = 14 ways =

2

142

Quicker Maths

Chapter 17

Probability Probability is a measurement of uncertainty. In this chapter chances of the happening of events are considered.

Terminology Random Experiment. It is an experiment which if conducted repeatedly under homogeneous conditions does not give the same result. The result may be any one of the various possible ‘outcomes’. Here the result is not unique (or the same every time). For example, if an unbiased dice is thrown it will not always fall with any particular number up. Any of the six numbers on the dice can come up. Trial and Event. The performance of a random experiment is called a trial and the outcome an event. Thus, throwing of a dice would be called a trial and the result (falling of any one of the six numbers 1, 2, 3, 4, 5, 6) an event. Events could be either simple or compound (also called composite). An event is called simple if it corresponds to a single possible outcome. Thus, in tossing a dice, the chance of getting 3 is a simple event (because 3 occurs in the dice only once). However, the chance of getting an odd number is a compound event (because odd numbers are more than one, i.e. 1, 3 and 5). Exhaustive Cases. All possible outcomes of an event are known as exhaustive cases. In the throw of a single dice the exhaustive cases are 6 as the dice has only six faces each marked with a different number. However, if 2 dice are thrown the exhaustive cases would be 36 (6 × 6) as there are 36 ways in which two dice can fall. Similarly, the number of exhaustive cases in the throw of 2 coins would be four (2 × 2), i.e. HH, TT, HT and TH (where H stands for head and T for tail). Favourable Cases. The number of outcomes which result in the happening of a desired event are called favourable cases. Thus in a single throw of a dice the number of favourable cases of getting an odd number is three, i.e. 1,3 and 5. Similarly, in drawing a card from a pack, the cases favourable to getting a spade are 13 (as there are 13 spade cards in the pack).

Mutually Exclusive Events. Two or more events are said to be mutually exclusive if the happening of any one of them excludes the happening of all others in a single (i.e. same) experiment. Thus, in the throw of a single dice the events 5 and 6 are mutually exclusive because if the event 5 happens no other event is possible in the same experiment. Here, one and only one of the events can take place at a time, excluding all others. Equally Likely Cases. Two or more events are said to be equally likely if the chances of their happening are equal, i.e. there is no preference of any one event to the other. Thus, in a throw of an unbiased dice, the coming up of 1, 2, 3, 4, 5, or 6 is equally likely. In the throw of an unbiased coin, the coming up of head or tail is equally likely. Independent and Dependent Events. An event is said to be independent if its happening is not affected by the happening of other events and if it does not affect the happening of other events. Thus, in the throw of a dice repeatedly, coming up of 5 on the first throw is independent of coming up of 5 again in the second throw. However, if we are successively drawing cards from a pack (without replacement) the events would be dependent. The chance of getting a King on the first draw is 4/52 (as there are 4 Kings in a pack). If this card is not replaced before the second draw, the chance of getting a King again is 3/51 as there are now only 51 cards left and they contain only 3 Kings. If, however, the card is replaced after the first draw, i.e. before the second draw, the events would remain independent. In each of the two successive draws the chance of getting a king would be 4/52. While tossing a coin you are not at all sure that Head will come. Tail may also come. However, you are sure that whatever will come, will be any one of the two: either Head or Tail. Let us see the following trials: i) A coin is tossed. The outcomes (results) may be {H, T} where H is Head (of the coin) and T is Tail (of the coin).

Quicker Maths

144 ii) Two coins are tossed. The outcomes may be {(H, H), (H, T), (T, H), (T, T)}. iii) A dice is thrown. The outcomes may be {1, 2, 3, 4, 5, 6}. iv) A person is selected randomly and is asked the day of the week on which he was born. The outcomes may be {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}. v) A candidate appears at a certain examination. The outcomes may be {Pass, Fail}. vi) One person is to form triangles from the 4 noncollinear points A, B, C and D. The outcomes may be {ABC, ABD, ACD, BCD}. vii) One person is to form 3-digit numbers from the given 4 digits 1, 2, 3, 4. The outcomes may be {123, 132, 124,....}. A set containing all possible outcomes of a random experiment is known as Sample Space. For the above-mentioned trial (i), number of Sample Space n(S) = l (for H) + l (for T) = 2 Similarly, for trial (ii), n(S) = 1 + 1 + 1 + 1 = 4; for (iii), n(s) = 6; for (iv), it is 7; for (v), it is 2; for (vi) it is 4 C3= 4; for (vii), it is 4P3 = 4 × 3 × 2 = 24 Each outcome of a Sample Space is called an Event. Thus, in the experiment (iv), {Sunday}, {Monday},.....{Saturday} are events. We also see that total no. of events = n(S). In all the above mentioned experiments, it is reasonable to assume that each outcome is as likely to occur as any other outcome. While tossing a coin, the chance of Head to come is the same as the chance of Tail. Now, Probability of an event (E) is denoted by P(E) and is defined as

P(E)

n(E) n(S)

no. of desired events total no. of events (i.e. no.of sample space)

When a coin is tossed, as for example, probability of Head coming, P(H) =

1 = P(T), probability of Tail 2

coming. When two coins are tossed, probability for Heads coming on both the coins =

1 4

Probability of at least one Tail coming =

111 3 4 4

Solved Examples: Ex. 1: A dice is thrown. What is the probability that the number shown on the dice is (i) an even no.; (ii) on odd no.; (iii) a no. divisible by 2; (iv) a no. divisible by 3; (v) a no. less than 4; (vi) a no. less than or equal to 4; (vii) a no. greater than 6; (viii) a no. less than or equal to 6. Soln: In all the above cases, S = {1, 2, 3, 4, 5, 6}, n(S) = 6. (i) E (an even no.) = {2, 4, 6}, n (E) = 3

P(E)

n(E) 3 1 n (S) 6 2

(ii) E (an odd no.) = 1, 3, 5, n ( E ) 3

P(E)

n(E) 3 1 n(S) 6 2

(iii) E (a no. divisible by 2) 2, 4, 6 , n(E) 3 P(E)

3 1 6 2

(iv) E (a no. divisible by 3) = 3,6 , n(E) 2 P(E)

2 1 6 3

(v) E (a no. less than 4) 1, 2, 3 , n(E) 3 3 1 6 2 (vi) E (a no. less than or equal to 4) = {1, 2, 3, 4} n(E) = 4 P(E)

2 6 3 (vii)E (a no. greater than 6) = {}, i.e. there is no number greater than 6 in the Sample Space, P(E)

P(E)

0 0 6

Probability of an impossible event = 0 (viii) E (a no. less than or equal to 6) = {1, 2, 3, 4, 5, 6}, n(E) = 6 6 1 6 Probability of a certain event = 1. Note: 0 Probability of an event 1. Ex. 2: Two coins are tossed. What is the probability of the appearing of (i) at most one head (ii) at most two heads? P(E)

Probability Soln:

145

n(S) = 4 = {(T, T), (H, T), (T, H), (H, H)} For (i), E (of appearing at most one head) = {HT, TH, TT}, n(E) = 3 3 4 For (ii), E (of appearing atmost two heads) = {HH, HT, TH, TT}, n(E) = 4 P(E)

4 1 4 Ex. 3: A positive integer is selected at random and is divided by 7. What is the probability that the remainder is (i) 1; (ii) not 1? Soln: When a positive integer is divided by 7, the remainder may be 0 or 1 or 2 or 3 or 4 or 5 or 6; n(S) = 7 For (i), E(l) = {l}, n(E) = l. P (E)

1 7 For (ii), E (not 1) = {0, 2, 3, 4, 5, 6}, n(E) = 6 P(E)

P(E)

Note:

6 7

1 1 = l. If we 7 6 represent an event by A then the event “not A” is

We see that E(l) + E (not 1) =

represented by A and A or Ac is known as complement of an event A. P ( A) P( A) 1 , or P ( A) 1 – P( A ) . Ex. 4: A dice is thrown. What is the probability that the number shown on the dice is not divisible by 3? Soln: S = {1, 2, 3, 4, 5, 6}; n(S) = 6 E (not divisible by 3) ={1,2, 4, 5}, n(E) = 4

4 2 P(not divisible by 3) = 6 3 Other Method: E (divisible by 3) = {3, 6}, n(E) = 2 2 1 P (divisible by 3) 6 3 P (not divisible by 3) = 1 – P (divisible by 3) 1 2 3 3 Ex. 5: (i) What is the chance that a leap year selected randomly will have 53 Sundays? (ii) What is the chance, if the year selected in not a leap year? Soln: (i) A leap year has 366 days so it has 52 complete weeks and 2 more days. The two days can be {Sunday and Monday, Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday}, i.e. n(E) = 7. Out of these 7 cases, cases favorable for more Sundays are {Sunday and Monday, Saturday and Sunday}, i.e., n(E) = 2

= 1

2 7 (ii) When the year is not a leap year, it has 52 complete weeks and 1 more day that can be {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}, n(S) = 7 Out of these 7 cases, cases favorable for one more Sunday is {Sunday}, n(E) = 1. P(E)

P(E)

1 7

Chart I: When two dices are thrown: S = {(1, 1}, (1, 2),...., (1, 6), (2, 1), (2, 2),...., (2, 6), (3, 1), (3, 2), ....., (3, 6), (4, 1),....., (4, 6), (5, 1), ...., (5, 6), (6, 1),....., (6, 6)} n(S) = 6 × 6 = 36

Sum of the numbers Events n(S) of the two dice (i) (ii) (i) (ii) 2 12 1 {1,1} {6,6} 3 11 2 {1, 2}, {2,1} {6,5}{5,6} 4 10 3 {1,3}, {3, 1}, {2, 2} {6, 4}, {4, 6}, {5, 5} 5 9 4 {1,4},{4,1},{2,3},{3,2} {6,3},{3,6},{5,4},{4,5} 6 8 5 {1,5}{5,1}, {2,4},{4,2}{3,3} {6,2},{2,6},{5,3},{3,5},{4,4} 7 6 {1,6},{6,1},{2,5}, {5,2},{4,3},{3,4}

Quicker Maths

146

Chart II: A pack of cards has a total of 52 cards: Red suit (26) Black suit (26) Diamond (13) Heart (13) Spade (13) Club (13)

The numbers in the brackets show the respective no. of cards in that category. Each of Diamond, Heart, Spade and Club contains nine digit-cards 2, 3, 4, 5, 6, 7, 8, 9 and 10 (a total of 9 × 4 = 36 digit-cards) along with four Honour cards Ace, King, Queen and Jack (a total of 4 × 4 = 16 Honour cards). Ex. 6: When two dice are thrown, what is the probability that (i) sum of numbers appeared is 6 and 7? (ii) sum of numbers appeared 8? (iii) sum of numbers is an odd no.? (iv) sum of numbers is a multiple of 3? (v) numbers shown are equal? (vi) the difference of the numbers is 2?

Ex. 7: A card is drawn from a pack of cards. What is the probability that it is (i) a card of black suit? (ii) a spade card? (iii) an honours card of red suit? (iv) an honours card of club? (v) a card having the number less than 7? (vi) a card having the number a multiple of 3? (vii) a king or a queen? (viii) a digit-card of heart? (ix) a jack of black suit?

Soln: (Hint - Use Chart II) For all the above cases n(S) 52 C1 52 (i)

26 1 or, 52 2

(ii)

13 1 52 4

(iii)

4 2 2 52 13

Soln: Hint - use Chart I (i)

n(E) 5 For 6, reqd probability n(S) 36 6 1 36 6 Desired sums of the numbers are 2, 3, 4, 5, 6, 7 and 8; n(E) = 1+2 + 3 + 4 + 5 + 6 + 5 = 26 26 13 reqd probability = 36 18 Desired sums of the numbers are 3, 5, 7, 9 and 11; n(E) = 2 + 4 + 6 + 4 + 2 = 18 18 1 reqd probability = 36 2 Desired sums of the numbers are 3, 6, 9 and 12; n(E) = 2 + 5 + 4 + 1 = 12

(iv)

For 7, reqd probability (ii)

(iii)

(iv)

(v)

(vi)

12 1 reqd probability = 36 3 Events = {1, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}, {6, 6}; n(E) = 6 6 1 P(E) 36 6 Events = {3, 1}, {4, 2}, {5, 3}, {6, 4}, {4, 6}, {3, 5}, {2, 4}, {1, 3} or, n(E) = 8 P(E)

8 2 36 9

(v) (vi) (vii)

26 52

C1 26 ( n C1 n) C2 52

4 1 52 13 5 4 5 52 13 3 4 4 52 13

P (a King)

4 1 4 1 ; P(a Queen) 52 13 52 13

1 1 2 P(a King or a Queen) 13 13 13

(viii)

9 52

2 1 52 26 Ex. 8: From a pack of 52 cards, 2 cards are drawn. What is the probability that it has (i) both the Aces? (ii) exactly one Queen? (iii) no honours card? (iv) no digit-card? (v) One King and one Queen? Soln: For all the above cases,

(ix)

n(S) 52 C 2

52 51 26 51 2

Probability (i)

147

Total no. of Aces = 4 43 4 n(E) C 2 2 6

(iii) n(E) 4 C1 4 C1 4 C1 4 4 4

in 4 C1 = 4 ways. He can select the remaining 1 card from the remaining (52 – 4 =) 48 cards.

(iv) n(E) 4 C1 4 C2 4 6

Now, cards in

48

C1 48 ways.

n(E) 4 48 4 48 32 26 51 221 Total no. of honour cards = 16 To have no honour card, he has to select two cards out of the remaining 52 –16 = 36 cards which he can do in

P(E)

(iii)

8 47 46 4324 26 17 50 5525

P(E)

6 1 26 51 21 Total no. of Queens = 4 Selection of 1 Queen card out of 4 can be done

P(E)

(ii)

(ii) n(E) 48 C3 8 47 46

36

C2

36 35 18 35 ways 2

P(E)

18 35 105 26 51 221

C2 8 15 20 26 51 26 51 221 16

(iv)

P(E)

(v)

n(E) 4 C1 4 C1 4 4 16

16 8 26 51 663 Ex. 9: From a pack of 52 cards, 3 cards are drawn. What is the probability that it has (i) all three Aces? (ii) no Queen? (iii) one Ace, one King and one Queen? (iv) one Ace and two Jacks? (v) two digit-cards and one honour card of black suit? Soln: For all the above cases, n(S) P(E)

52 C3

52 51 50 26 17 50 3 2

(i) n(E) 4 C3 4 P(E)

4 1 26 17 50 5525

P(E)

P(E)

44 4 16 26 17 50 5525

46 6 26 17 50 5525

(v) n(E) 36 C2 8 C1 18 35 8 18 35 8 252 26 17 50 1105 Ex. 10: A bag contains 3 red, 5 yellow and 4 green balls. 3 balls are drawn randomly. What is the probability that the balls drawn contain (i) balls of different colours? (ii) exactly two green balls? (iii) no yellow ball? Soln: Total no. of balls = 3 + 5 + 4 = 12; P(E)

12 11 10 220 3 2 (i) In order to have 3 different-coloured balls, the selection of one ball of each colour is to be made. n(S)

12

C3

n(E) 3 C1 5 C1 4 C1 3 5 4 3 5 4 3 220 11 (ii) 2 green balls can be selected from 4 green

P(E)

balls in 4 C 2 ways and the rest one ball can be selected from the remaining (12 – 4 =) 8 balls in 8 C1 ways.

n(E) 4 C2 8 C1 6 8 48 48 12 220 55 (iii) 3 balls can be selected from 3 (red) + 4 P(E)

(green) = 7 balls in 7 C3 ways. n(E) 7 C3 P(E)

765 35 3 2

35 7 220 44

Quicker Maths

148 Ex. 11: If the letters of the word EQUATION be arranged at random, what is the probability that (i) there are exactly six letters between N and E? (ii) all vowels are together? (iii) all vowels are not together? Soln: There are eight different letters in the given word

2 43 2 5 43 5 (ii) P (not divisible by 2) = 1 – P (divisible by 2) P(E)

2 3 5 5 (iii) A number is divisible by 5, when its units digit is either 0 or 5. We have not been provided the digit 0. So, the units place can be filled up with only 5, i.e. in 1 way. The rest two places can be filled up with the

= 1

8 Total no. of arrangements, n(S) = P8 8! (i) If N occupies first place, E must occupy last place and vice versa so that there are exactly six letters in between the letters N and E. N and E

remaining 4 digits in 4 P2 ways.

can be arranged in 2 P2 2 ways and the rest six places can be filled by the remaining six letters (Q, U, A, T, I and O) in 6 P6 6! ways.

n(E) 2 6! 2 6! 2 1 8! 8 7 28 Considering all the five vowels as one letter, we have a total of 3 (consonants) + 1 = 4 letters P(E)

(ii)

which can be arranged in 4 P4 = 4! ways. But the five vowels can also be arranged in 5! ways among themselves. So, the letters can be arranged in 4! × 5! ways so that all the vowels are together. i.e., n(E) = 4! × 5!

P(E) (iii)

4! 5! 4 3 2 1 8! 8 7 6 14

P (All vowels are not together) = 1 – P(All vowels

1 13 14 14 Ex. 12: A three-digit number is formed with the digits 1, 3, 6, 4 and 5 at random. What is the chance that the number formed is (i) divisible by 2? (ii) not divisible by 2? (iii) divisible by 5? Soln: A three-digit number can be formed with the given

are together) = 1

five digits in 5 P3 ways, i.e. n(S) 5 P3 5 4 3 (i) Any one of the two digits 4 and 6 should come at units place, which can be done in 2 ways. After filling up the units place, the remaining two places can be filled up with the remaining four digits in 4 P2 ways;

n(E) 2 4 P2 2 4 3

n(E) 1 4 P2 4 3 43 1 5 43 5 Ex. 13: There are 4 boys and 4 girls. They sit in a row randomly. What is the chance that all the girls do not sit together? P(E)

13 . 14 Ex. 14: The letters of the word ‘ARTICLE’ are arranged in different ways randomly. What is the chance that the vowels occupy the even places? Soln: The 7 different letters of the word ARTICLE can be arranged in 7! ways, i.e., n(S) = 7!

Soln:

Try yourself. Answer is

n(E) 3 P3 4 P4 3! 4! 6 24 6 24 1 . 7! 35 Ex. 15: A committee of 4 is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have number of boys less than number of girls? Soln: Selection of 1 boy and 3 girls in P(E)

C1 4 C3 5 4 20 ways Selection of 4 girls and no boy in 5

C0 4 C4 1 1 1 way n(E) = total no. of ways = 21 Without any restriction, a committee of 4 can be formed from among 4 girls and 5 boys in 5

9

C4

9 8 7 6 9 7 2 ways 4 3 2

P(E)

n(E) 21 1 . n(S) 9 7 2 6

Probability

149

Ex. 16: A box contains 4 black balls, 3 red balls and 5 green balls. 2 balls are drawn from the box at random. What is the probability that both the balls are of the same colour? 1)

47 68

2)

1 6

3)

19 66

1)

Soln:

12 11 66 2

Ex. 17: In a box carrying one dozen of oranges, onethird have become bad. If 3 oranges are taken out from the box at random, what is the probability that at least one orange out of the three oranges picked up is good? 3)

45 55

15 205 41 220 220 44 Ex. 19: Out of 15 students studying in a class, 7 are from Maharashtra, 5 from Karnataka and 3 from Goa. Four students are to be selected at random. What are the chances that at least one is from Karnataka?

= 1–

3 4) 5) None of these 55 Soln: 2;P(At least one good) = 1 – P(All bad) ..... (*)

1 –

4 12

152 5) None of these 55 2; Total no. of balls = 5 + 4 + 3 = 12

10 4 1 15 ways Now, P(all the 3 marbles of the same colour) +P(all the 3 marbles are not of the same colour) =1 P(all the 3 marbles are not of the same colour)

n(E) 19 Required probability, P(E) n(S) 66

54 55

13 55

can be done in 5 C3 4 C3 3C3

4 3 3 2 5 4 2 2 2 = 6 + 3 + 10 = 19

2)

3)

n(S) 12 C3

n(E) 4 C2 3C 2 5 C 2

1 55

41 44

12 11 10 220 1 2 3 i.e., 3 marbles out of 12 marbles can be drawn in 220 ways. If all the three marbles are of the same colour, it

2 5) None of these 11 Soln: 3;Total no. of balls = 4 + 3 + 5 = 12

1)

2)

4)

4)

n(S) 12 C 2

13 44

C3 4 1 54 1– 1– C3 220 55 55

Note: (*) See the following combinations of selection of 3 oranges out of 8 good and 4 bad oranges. (i) All 3 are bad and 0 good. (ii) 1 bad 2 good (iii) 2 bad 1 good (iv) 0 bad 3 good Combination of (ii), (iii) and (iv) can be said to be “At least one good”. We have, P(i) + {P(ii) + P(iii) + P(iv)} = 1 or, P(i) + P{At least one good} = 1 P{At least one good} = 1 – P(All 3 bad) Ex. 18: A box contains 5 green, 4 yellow and 3 white marbles. 3 marbles are drawn at random. What is the probability that they are not of the same colour?

1)

12 13

2)

11 13

3)

100 15

51 5) None of these 15 2; P(At least one from Karnataka) = 1 – P(No one from Karnataka)

4) Soln:

1

10 15

C4 10 9 8 7 2 11 1– 1– C4 15 14 13 12 13 13

Ex. 20: A bag contains 5 red and 8 black balls. Two draws of three balls each are made, the ball being replaced after the first draw. What is the chance that the balls were red in the first draw and black in the second? Soln:

Required probability

5

C3 8 C3 140 13 13 C3 C3 20449

Ex. 21: A bag contains 5 black and 7 white balls. A ball is drawn out of it and replaced in the bag. Then a ball is drawn again. What is the probability that (i) both the balls drawn were black; (ii) both were white; (iii) the first ball was white and the second

Quicker Maths

150

Soln:

black; (iv) the first ball was black and the second white? The events are independent and capable of simultaneous occurrence. The rule of multiplication would be applied. The probability that 5 5 25 (i) both the balls were black 12 12 144 7 7 49 12 12 144 (iii) the first was white and second black

(ii) both the balls were white

7 5 35 12 12 144 (iv) the first was black and second white

5 7 35 12 12 144 Ex. 22: A bag contains 6 red and 3 white balls. Four balls are drawn out one by one and not replaced. What is the probability that they are alternately of different colours? Soln: Balls can be drawn alternately in the following order: Red, White, Red, White OR White, Red, White, Red If Red ball is drawn first, the probability of drawing the balls alternately

6 3 5 2 ...(I) 9 8 7 6 If White ball is drawn first the probability of drawing the balls alternately

3 6 2 2 ... (II) 9 8 7 6 Required probability = (I) + (II) ....... (*)

6 3 5 2 3 6 2 5 5 5 5 . 9 8 7 6 9 8 7 6 84 84 42 Important Note: In Ex. 20 & 21, the two events are independent and can occur simultaneously. So, we used “multiplication”. In other words, since the word AND was used between two events, we used multiplication. Mark that both also means first and second. In Ex. 22, in (*) we used addition because the two events are joined with OR. Ex. 23: A bag contains 4 white and 6 red balls. Two draws of one ball each are made without replacement.

Soln:

What is the probability that one is red and other white? Such problems can be very easily solved with the help of the rules of permutation and combination. Two balls can be drawn out of 10 balls in

10

C 2 or

10! 10 9 2! 8! or 2 or 45 ways. One white ball can be drawn out of 4 white balls 4

in

4! C1 or 1! 3! or 4 ways.

One red ball can be drawn out of 6 red balls in 6

C1 or 6 ways. The total number of ways of drawing a white and a red ball are 4 C1 6 C1 or 4 6 24. (See Important Note given above) The required probability would be

No. of cases favourable to the event Total No. of ways in which the event can happen

24 8 . 25 15 Ex. 24: A bag contains 7 white and 9 black balls. Two balls are drawn in succession at random. What is the probability that one of them is white and the other black? Soln: Total number of ways of drawing 2 balls from (7 + 9) or 16 balls is

16

C2

16! 16 15 120 2! 14! 2

Number of ways of drawing a white ball out of 7

C1

7! 7 1! 6!

Number of ways of drawing a black ball out of 9

9! 9 is C1 1! 8! 9 Number of ways of drawing a white and a black 7 9 ball would be C 1 C 1 7 9 63

The required probability

7

C1 9 C1 7 9 21 . 16 C2 120 40

Probability

151

EXERCISES 1. Out of five girls and three boys, four children are to be randomly selected for a quiz contest. What is the probability that all the selected children are girls? 1)

1 14

2)

1 7

3)

5 17

2 5) None of these 17 2. A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour?

4)

1)

41 190

2)

21 190

3)

59 190

99 77 5) 190 190 3. From a well-shuffled pack of 52 playing cards, one card is drawn at random. What is the probability that the card drawn will be a black king?

4)

1) 4)

1 26

2)

9 13

5)

7 13

3)

3 13

1 13

4. A bag contains 7 blue balls and 5 yellow balls. If two balls are selected at random, what is the probability that none is yellow? 1)

5 33

2)

5 22

3)

7 22

7 7 5) 33 66 5. A die is thrown twice. What is the probability of getting a sum 7 from both the throws?

4)

1)

5 18

2)

1 18

3)

1 9

1 5 5) 6 36 6. A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one ball is drawn from each bag, find the probability that both are green.

4)

13 1) 20

4)

8 35

1 2) 4

5) None of these

6 3) 35

7. A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red? 1)

1 3

2)

2 3

3)

11 15

1 8 5) 15 15 8. A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag, one after another, what is the probability that the first ball is red and the second ball is yellow?

4)

1)

1 14

2)

2 7

3)

5 7

3 5) None of these 14 9. A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random. What is the probability that none of the balls drawn are yellow in colour ?

4)

1)

1 7

2)

3 7

3)

2 7

5 9 5) 14 14 10. A bag contains 5 red balls, 6 yellow balls and 3 green balls. If two balls are picked at random, what is the probability that either both are red or both are green in colour?

4)

1)

3 7

2)

5 14

3)

1 7

2 3 5) 7 14 11. A bag contains 4 red balls, 6 green balls and 5 blue balls. If three balls are picked at random, what is the probability that two of them are green and one of them is blue?

4)

1)

20 91

2)

10 91

3)

15 91

5 25 5) 91 91 12. There are 3 red balls, 4 blue balls and 5 white balls. 2 balls are chosen randomly. Find the probability that one is red and the other is white.

4)

Quicker Maths

152

1)

5 22

2)

5 23

3)

7 22

4 5) None of these 9 13. In a bag there are 7 red balls and 5 green balls. Three balls are picked at random. What is the probability that two balls are red and one ball is green in colour?

4)

29 1) 44

21 2) 44

27 3) 44

23 19 5) 44 44 14. Abhay rolled a pair of dice together. What is the probability that one dice showed a multiple of 2 and the second dice showed neither a multiple of 3 nor of 2?

4)

1)

1 3

2)

1 9

3)

1 6

2 5) 3 6 15. A bag contains 24 eggs, out of which 8 are rotten. The remaining eggs are not rotten. Two eggs are selected at random. What is the probability that one of the eggs is rotten?

4)

11 1) 23

17 2) 23

13 3) 23

32 62 5) 69 69 16. A bag contains 63 cards (numbered 1, 2, 3, ... 63). Two cards are picked at random from the bag (one after another and without replacement). What is the probability that the sum of the numbers of both the cards drawn is even?

4)

1)

11 21

2)

34 63

3)

7 11

11 5) Other than those given as options 63 Directions (Q. 17-21): Study the given information carefully and answer the questions that follow: An urn contains 3 red, 6 blue, 2 green and 4 yellow marbles. 17. If two marbles are picked at random, what is the probability that both are green ?

4)

2 1 2 2) 3) 15 15 7 4) 1 5) None of these 18. If three marbles are picked at random, what is the probability that two are blue and one is yellow ?

1)

1)

2 15

4) 1

4

2)

6 91

3)

12 91

5) None of these

19. If four marbles are picked at random, what is the probability that at least one is yellow ? 91 69 125 2) 3) 123 91 143 1 4) 5) None of these 4 20. If two marbles are picked at random, what is the probability that either both are red or both are green ?

1)

3 4 2 2) 3) 5 105 7 5 4) 5) None of these 91 21. If four marbles are picked at random, what is the probability that one is green, two are blue and one is red ?

1)

4 17 6 2) 3) 15 280 91 11 4) 5) None of these 15 Directions (Q. 22-24): Study the information to answer the questions: A bag contains 6 red shirts, 6 green shirts and 8 blue shirts. 22. Two shirts are drawn randomly. What is the probability that at most one shirt is red?

1)

1)

27 38

2)

29 38

3)

25 38

33 35 5) 38 38 23. One shirt is drawn randomly and then another shirt is drawn randomly. What is the probability that the first shirt is red and the other shirt is blue?

4)

1)

14 95

2)

24 95

4)

12 95

5)

18 95

3)

15 95

Probability

153

24. One shirt is drawn randomly. What is the probability that it is not green? 7 7 9 1) 2) 3) 10 20 10 3 2 4) 5) 5 5 Directions (Q. 25-27): Study the information and answer the given questions. A bag contains four blue shirts, five red shirts and six yellow shirts 25. Three shirts are drawn randomly. What is the probability that exactly one of them is blue? 36 40 44 1) 2) 3) 91 91 91 48 31 4) 5) 91 91

26. One shirt is drawn randomly. What is the probability that it is either red or yellow? 1)

4 15

2)

7 15

11 15

3)

8 13 5) 15 15 27. Two shirts are drawn randomly. What is the probability that both of them are blue?

4)

1)

3 35

2)

1 35

4)

6 35

5)

4 35

2 35

3)

Solutions 1. 1; Total number of ways of selecting 4 children out

8 7 6 5 70 1 2 3 4 Number of ways of selecting 4 girls out of 5 = 5 C4 = 5

of 8 = 8C4 =

5 1 70 14 2. 4; Total number of balls = 13 + 7 = 20 Number of sample space = n(S) = 20C2 = 190 Number of events = n(E) = Both are white or both are black = 13C2 + 7C2 = 78 + 21 = 99 Required probability =

n(E) 99 P(E) = n(S) 190 3. 1; Total possible outcomes = 52C1 = 52 Favourable outcomes = 2 2 1 52 26 4. 3; Total number of balls = 7 + 5 = 12

Required probability

12 11 66 1 2 Number of events = n(E) = Both balls are blue Sample space = n(S) =

=

7

C2

=

12

C2 =

6. 3;

Probability that ball from bag A is green =

4 10

Probability that ball from bag B is green =

3 7

4 3 12 6 10 7 70 35 7. 2; Total no. of balls = 3 + 5 + 7 = 15 Now, n(S) = 15C1 = 15 n(E) = 7C1 + 3C1 = 7 + 3 = 10 Total probability =

Reqd probability =

6

Probability that first ball is red =

C1 11 3 C1 21 11

22

Now, there are 21 balls left in the bag. Probability that second ball is yellow =

21 7 66 22

n(E) 10 2 n(S) 15 3

8. 5; Mark that ball is selected one by one.

11

7 6 21 1 2

required probability =

5. 4; Sample space n(S) = 6 × 6 = 36 For a sum of 7, events = (1, 6), (6, 1), (5, 2), (2, 5), (4, 3), (3, 4) n(E) = 6 n(E) 6 1 Probability P(E) = n(S) 36 6

21

C1 11 C1 21

reqd probability =

3 11 1 11 21 7

Quicker Maths

154 9. 2; Total number of balls in the bag = 4 + 5 + 6 = 15 Total possible outcomes = Selection of 2 balls 15 14 105 1 2 Total favourable outcomes = Selection of 2 balls out of 4 orange and 6 pink balls 10 9 45 = 10C2 = 1 2

out of 15 balls = 15C2 =

45 3 105 7 10. 3; Total no. of balls = 5 + 6 + 3 = 14

Required probability

n(S) =

14

C2 =

13 14 = 91 2

Both are red = 5 C2 =

45 = 10 2

Both are green = 3 C2 = 3 = 3 n(E) = 10 + 3 = 13 Reqd probability =

13 1 91 7

11. 3; Number of sample space n(S) = 15 C3 = 455 Now, two of them are green = 6C2 And one of them is blue = 5 C1 n(E) = 6 C2 5 C1 = 75 75 15 455 91 12. 1; Total number of balls = 3 + 4 + 5 = 12 Now, two balls are chosen randomly. The number of sample space = n(S) = 12C2 Number of favourable events n(E) = 3C1 × 5C1 = 3 × 5 = 15

Reqd probability =

P(E) =

n(E) 15 15 5 11 12 n(S) 66 22 2

13. 2; Total number of balls = 7 + 5 = 12 Now, three balls are picked randomly Then, the number of sample space n(S) 10 11 12 12 = C3 = 220 1 2 3 The number of events n(E) = 7 C2 5 C1 = P(E) =

67 5 = 21 × 5 = 105 2

n(E) 105 21 n(S) 220 44

14. 3; Multiples of two = 2, 4, 6 3 1 P(E) = 6 2 Multiples of neither 2 nor 3 = 1, 5 2 1 P(E) = 6 3 1 1 1 Now, required probability P(E) = 2 3 6 15. 4; Total number of eggs = 24 8 eggs are rotten Remaining eggs = (24 – 8 =) 16 is not rotten Now, 2 eggs are randomly selected. Then n(S) = 24C2 = 23 × 12 n(E) = 16C1 × 8C1 = 16 × 8 16 8 32 23 12 69 16. 5; There are 63 cards, numbered 1 to 63. So, there will be two cases of getting sum as even number. Case (i): Odd + Odd = Even Case (ii): Even + Even = Even Now, Total no. of even cards = 31 And total no. of odd cards = 32

P(E) =

32 31 16 63 62 63 31 30 15 Case II. P2 = 63 62 63 Reqd probability = P1+ P2 Case I. P1=

16 15 16 15 31 = 63 63 63 63 63 17. 5; Total number of marbles in the urn = 15 n(S) = Total possible outcomes = Selection of two marbles at random out of 15 marbles =

15 14 = 105 1 2 P(E) = Favourable outcomes = Selection of 2 marbles out of 2 green marbles = 2 C2 = 1 = 15C2 =

n(E) 1 Required probability = n(S) 105

15 14 13 = 455 1 2 3 n(E) = Selection of 2 marbles out of 6 blue marbles and that of one marble out of 4 yellow marbles =

18. 3; n(S) = 15C3 =

6

C2 × 4 C1 =

65 4 = 60 1 2

Probability

155

P(E) 60 12 Required probability = P(S) 455 91

15 14 13 12 19. 2; n(S) = 15C4 = = 1365 1 2 3 4 Let no yellow marble is selected. n(E) = Selection of 4 marbles out of 11 other marbles 11 10 9 8 = 330 1 2 3 4 Required probability = 1 – P (no yellow marble) = 11C4 =

23. 4; Reqd probability = P(first shirt is red) × P (other shirt is blue) 6

8 C1 C1 6 8 12 = 20 19 C1 C1 20 19 95

24. 2; P(not green) = P (shirt is from 6 red or 8 blue) 14

=

n(S) = 15 C3 = 455 Now, for exactly one of them to be blue,

330 22 91 22 69 1 1365 91 91 91 20. 2; n(S) = 15C2 = 105 3 2 1 = 4 1 2

Required probability =

4 105

n(E) = 4 C1 11 C2 = 220 Reqd probability =

red or yellow in

= 1

6

C2 56 = 1 C2 20 19

20

3 35 = 1 = 38 38

11

C1 = 11 ways.

n(E) = 11

21. 3; n(S) = C4 =1365 n(E) = 2C1 × 6C2 × 3C1 = 2 × 15 × 3 = 90

22. 5; Total no. of balls = 6 + 8 + 6 = 20 Required probability = 1 – P (both are red)

220 44 455 91

26. 3; n(S) = 15 C1 = 15 Now, when one shirt is drawn it might be either

15

n(E) 90 6 Required probability = n(S) 1365 91

C1 14 7 C1 20 10

25. 3; Total no. of shirts = 4 + 5 + 6 = 15

= 1

n(E) = 3C2 + 2C2 =

20

P(E) = 27. 3;

n ( E ) 11 n ( S ) 15

When two shirts are drawn n(S) =

15

C2 =

15 14 = 105 1 2

n(E) = Both shirts are blue = 4 C2 = Reqd probability =

6 2 105 35

3 4 =6 2

156

Quicker Maths

Chapter 18

Ratio and Proportion The number of times one quantity contains another quantity of the same kind is called the ratio of the two quantities. Clearly, the ratio of two quantities is equivalent to the fraction that one quantity is of the other. Observe carefully that the two quantities must be of the same kind. There can be a ratio between 20 and 30, but there can be no ratio between 20 and 30 mangoes. The ratio 2 to 3 is written as 2 : 3 or

2 . 2 and 3 are 3

called the terms of the ratio. 2 is the first term and 3 is the second term. The first term of a ratio is called the antecedent and the second the consequent. In the ratio 2 : 3, 2 is the antecedent and 3 is the consequent. Note:1. The word ‘antecedent’ literally means ‘that which goes before’. The word ‘consequent’ literally means ‘that which goes after’. 2. Since the quotient obtained on dividing one concrete quantity by another of the same kind is an abstract number, the ratio between two concrete quantities of the same kind is an abstract number. Thus, the ratio between 5 and 7 is 5 : 7. Since a fraction is not altered by multiplying or dividing both its numerator and denominator by the same number, a ratio which is also a fraction is not altered by multiplying or dividing both its terms by the same number. Thus 3 : 5 is the same as 6 : 10, and 15 : 20 is the same as 3 : 4.

Note:

resulting ratio is 4 2 : 32 . It is called the duplicate ratio of 4 : 3. Similarly, 43 : 33 is the triplicate ratio of 4 : 3.

4 : 3 is called the subduplicate ratio of 4 : 3. a

Soln:

The required ratio

4 9 26 2 16 3 13 5 15 25

1

3

:b

1

3

is subtriplicate ratio of a and b.

Inverse Ratio If 2 : 3 be the given ratio, then

1 1 : or 3 : 2 is called 2 3

its inverse or reciprocal ratio. If the antecedent = the consequent, the ratio is called the ratio of equality, such as 3 : 3. If the antecedent > the consequent, the ratio is called the ratio of greater inequality, as 4 : 3. If the antecedent < the consequent, the ratio is called the ratio of less inequality, as 3 : 4. Ex. 1: Divide 1458 into two parts such that one may be to the other as 2 : 7. Soln:

1st part 2

1458 1458 2 324 27 9

1458 1134 9 Find three numbers in the ratio of 1 : 2 : 3, so that the sum of their squares is equal to 504. Let the numbers be x, 2x and 3x. Then we have, 2nd part = 7

Ex. 2: Soln:

x 2 (2x) 2 (3x) 2 504

Compound Ratio Ratios are compounded by multiplying together the antecedents for a new antecedent, and the consequents for a new consequent. Ex.: Find the ratio compounded of the four ratios: 4 : 3, 9 : 13, 26 : 5 and 2 : 15

When the ratio 4 : 3 is compounded with itself, the

Ex. 3:

or, 14x 2 504 x=6 Hence, the required numbers are 6, 12 and 18. A, B, C and D are four quantities of the same kind such that A : B = 3 : 4, B : C = 8 : 9, C : D = 15 : 16. (i) Find the ratio A : D; (ii) Find A : B : C; and (iii) Find A : B : C : D.

Quicker Maths

158

Soln:

(i)

A 3 B 8 C 15 , , B 4 C 9 D 16

A A B C 3 8 15 5 D B C D 4 9 16 8 A:D = 5 : 8 (ii) A : B = 3 : 4 = 6 : 8 B:C=8:9 A:B:C= 6:8:9 (iii) We put down the first ratio in its original form and change the terms of the other ratios so as to make each antecedent equal to the preceding consequent. A:B = 3 : 4 ----------- (1) B:C = 8 : 9 ----------- (2) = 1:

9 (divided by 8) 8

= 4:

9 × 4 (multiplied by 4) 8

9 ----------- 2(i) 2 = 15 : 16 ----------- (3) = 4:

C:D

= 1:

16 (divided by 15) 15

=

9 16 9 9 : (multiplied by ) 2 15 2 2

=

9 24 : -----------3(i) 2 5

9 24 : = 30 : 40 : 45 : 48 2 5 (1) In the Equation (2), B = 8. To make the ratios equivalent, the ‘8’ in (2) should be reduced to ‘4’ (equivalent to B in (1)). (2)In the Equation (3), C = 15. To make the ratios equivalent, the ‘15’ in (3) should be reduced to

A: B : C : D= 3 : 4 :

Note:

9

2 (equivalent to C in 2 (i)). Ex. 4: A : B = 1 : 2 , B : C = 3 : 4 , C : D = 6 : 9 and D : E = 12 : 16. Find A : B : C : D : E Soln: A : B = 1:2= 3:6 B:C = 3:4= 6:8 C:D = 6 : 9 = 8 : 12 D:E = 12 : 16 A : B : C : D : E = 3 : 6 : 8 : 12 : 16

Note:

In the above example, we moved from below, because it made the calculations easier.

Theorem: If the ratio between the first and the second quantities is a : b and the ratio between the second and the third quantities is c : d, then the ratio among first, second and third quantities is given by ac : bc : bd The above ratio can be represented diagramatically as

Proof: We have First : Second = a : b; Second : Third = c : d To equate the two ratios, we need to equate the consequent (b) of the first ratio and antecedent (c) of the second ratio. So, we multiply the first ratio by c and the second ratio by b. Therefore, First : Second = ac : bc Second : Third = bc : bd Then, First : Second : Third = ac : bc : bd Ex. 5: The sum of three numbers is 98. If the ratio between the first and second be 2 : 3 and that between the second and third be 5 : 8, then find the second number. Soln: The theorem does not give the direct value of the second number, but we can find the combined ratio of all the three numbers by using the above theorem. The ratio among the three numbers is 2 : 3 5 : 8 10 : 15 : 24

98 15 30 10 15 24 Ex. 6: The ratio of the money with Rita and Sita is 7 : 15 and that with Sita and Kavita is 7 : 16. If Rita has 490, how much money does Kavita have? Soln: Rita : Sita : Kavita 7 : 15 7 : 16 49 : 105 : 240 The ratio of money with Rita, Sita and Kavita is 49 : 105 : 240 We see that 49 490 240 2400

The second number =

Theorem: If the ratio between the first and the second quantities is a : b; the ratio between the second and the third quantities is c : d and the ratio between the third

Ratio and Proportion

159

and the fourth quantities is e : f then the ratio among the first, second, third and fourth quantities is given by

1st : 2nd

=

a :

b

2nd : 3rd =

c

:

Ex 9:

d

3rd : 4th = e : f 1st : 2nd : 3rd : 4th = ace : bce : bde : bdf Proof: It is easy to prove this by the same method used in the previous theorem. Ex. 7: If A : B = 3 : 4, B : C = 8 : 10 and C : D = 15 : 17. Then find A : B : C : D. Soln: A : B = 3:4 B:C = 8 : 10 C : D= 15 : 17 A : B : C : D = 3 × 8 × 15 : 4 × 8 × 15 : 4 × 10 × 15 : 4 × 10 × 17 = 9 : 12 : 15 : 17 Note: 1. Ex. 3 can be solved with the help of above theorems. Try it. 2. If A : B = 1 : 2, B : C = 3 : 4, C : D = 2 : 3 and D : E =3:4 Then find A : B : C : D : E. Soln: A :

B = 1

B

:

C =

C

:

D =

D

:

E =

:

2 3

:

4 2

:

3 3

Soln:

:

4

A hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 4 leaps of the hound are equal to 5 leaps of the hare. Compare the rates of the hound and the hare. 4 leaps of hound = 5 leaps of hare

5 leaps of hound =

25 leaps of hare 4

the rate of hound : rate of hare =

Or, Ratio of

12 15 1 5 7 days. 8 2 2 Or, By the rule of fraction: As B is more efficient, it is clear that ‘B’ will complete the work in less days. So, the number of days (12) should be multiplied by a less-than-one fraction

and that fraction is

25 : 6 = 25 : 24 4 Hound

Hare

Leap frequency

5

6

Leap length

4

5

100 100 i.e., . Therefore, our 100 60 160

100 12 5 15 1 7 days. 160 8 2 2 Ex 10: One man adds 3 litres of water to 12 litres of milk and another 4 litres of water to 10 litres of milk. What is the ratio of the strengths of milk in the two mixtures? required answer is 12

Soln:

A:B :C:D:E=1×3×2×3:2×3×2×3:2×4×2×3 :2×4×3×3:2×4×3×4 = 3 : 6 : 8 : 12 : 16

Ex. 8:

Soln:

Then the required ratio of speed is the ratio of the cross-product. That is, speed of hound : speed of hare = 5 × 5 : 6 × 4 = 25 : 24 A can do a piece of work in 12 days. B is 60% more efficient than A. Find the number of days it takes B to do the same piece of work. A : B Efficiency 100 : 160 Days 160 : 100 or, 8 : 5 the number of days taken by B

12 12 12 3 15 Strength of milk in the second mixture Strength of milk in the first mixture =

=

10 10 10 4 14

12 10 the ratio of their strengths = 15 : 14 = 12 × 14 : 15 × 10 = 28 : 25 Ex 11: 425 is divided among 4 men, 5 women and 6 boys such that the share of a man, a woman and a boy may be in the ratio of 9 : 8 : 4. What is the share of a woman? Soln: The ratio of shares of group of men, women and boys = 9 × 4 : 8 × 5 : 4 × 6 = 36 : 40 : 24 Share of 5 women =

425 40 = 170 36 40 24

the share of 1 woman =

170 = 34 5

Quicker Maths

160 Ex 12: If a carton containing a dozen mirrors is dropped, which of the following cannot be the ratio of broken mirrors to unbroken mirrors? 1) 2 : 1 2) 3 : 1 3) 3 : 2 4) 1 : 1 5) 7 : 5 Soln: 3;There are 12 mirrors in the carton. So, the sum of terms in the ratio must divide 12 exactly. We see that 2 + 1 = 3 divides 12 exactly. 3 + 1 = 4 also divides exactly. 3 + 2 = 5 doesn’t divide 12 exactly. Thus, our answer is (3).

Hence, 6 is the mean proportional between 9 and 4, and 4 is the third proportional to 9 and 6. Ex. 1: Find the fourth proportional to the numbers 6, 8 and 15. Soln: If x be the fourth proportional, then 6 : 8 = 15 : x

Ex. 2: Soln:

PROPORTION Consider the two ratios: 1st ratio 2nd ratio 6 : 18 8 : 24 Since 6 is one-third of 18, and 8 is one-third of 24, the two ratios are equal. The equality of ratios is called proportion. The numbers 6, 18, 8 and 24 are said to be in proportion. The proportion may be written as 6 : 18 : : 8 : 24 ( 6 is to 18 as 8 is to 24) or, 6 : 18 = 8 : 24 or,

6 8 . 18 24

The numbers 6, 18, 8 and 24 are called the terms. 6 is the first term, 18 the second, 8 the third, and 24 the fourth. The first and fourth terms, i.e., 6 and 24 are called the extremes (end terms), and the second and the third terms, i.e., 18 and 8 are called the means (middle terms). 24 is called the fourth proportional. 1. If four quantities be in proportion, the product of the extremes is equal to the product of the means. Let the four quantities 3, 4, 9 and 12 be in proportion. We have

3 9 , [Multiply each ratio by 4 × 12] 4 12

3 9 4 4 12 12 4 12 3 × 12 = 4 × 9 2. Three quantities of the same kind are said to be in continued proportion when the ratio of the first to the second is equal to the ratio of the second to the third. The second quantity is called the mean proportional between the first and the third; and the third quantity is called the third proportional to the first and second. Thus, 9, 6 and 4 are in continued proportion for 9 : 6 : : 6 : 4.

8 15 20 6 Find the third proportional to 15 and 20. Here, we have to find a fourth proportional to 15, 20 and 20. If x be the fourth proportional, we have 15 : 20 = 20 : x x

20 20 80 2 26 15 3 3 Find the mean proportional between 3 and 75. If x be the required mean proportional, we have 3 : x : : x : 75 x

Ex. 3: Soln:

Note:

x 3 75 15 It is evident that the mean proportional between two numbers is equal to the square root of their product. (Remember) Consider the proportion 5 : 15 : : 8 : x. Here, the 1st, 2nd and 3rd terms are given, and the 4th term is unknown. The unknown term is denoted by x. We want to find x. Now, the product of the means is equal to the product of the extremes. 5 x 15 8

15 8 24 5 Hence, the 4th term can be found by the following rule Rule: Multiply the 2nd and 3rd terms together, and divide the product by the 1st term. We shall now take examples concerning concrete quantities. Direct Proportion: Consider the following example. Ex.: If 5 balls cost 8, what do 15 balls cost? Soln: It will be seen at once that if the number of balls be increased 2, 3, 4 .... times, the price will also be increased 2, 3, 4.... times. Therefore, 5 balls is the same fraction of 15 balls that the cost of 5 balls is of the cost of 15 balls. 5 balls : 15 balls : : 8 : required cost or, x

the required cost =

15 8 = 24 5

Ratio and Proportion This example is an illustration of what is called direct proportion. In this case, the two given quantities are so related to each other that if one of them is multiplied (or divided) by any number, the other is also multiplied (or divided) by the same number. Inverse Proportion: Consider the following example. Ex.: If 15 men can reap a field in 28 days, in how many days will 10 men reap it? Soln: Here it will be seen that if the number of men be increased 2, 3, 4, .... times, the number of days will be decreased 2, 3, 4.... times. Therefore, the inverse ratio of the number of men is equal to the ratio of the corresponding number of days.

1 1 : : : 28 : the required number of days 15 10

or, 10 : 15 : : 28 : the required number of days

15 28 42 10 The above example is an illustration of what is called inverse proportion. In this case, the two quantities are so related that if one of them is multiplied by any number, the other is divided by the same number, and vice versa. Note: The arrangement of figures may create a problem. To overcome this, we give you a general rule known as the RULE OF THREE. The Rule of Three : The method of finding the 4th term of a proportion when the other three are given is called Simple Proportion or the Rule of Three. In every question of simple proportion, two of the given terms are of the same kind, and the third term is of the same kind as the required fourth term. Now, we give the rule of arranging the terms in a question of simple proportion. Rule: I: Denote the quantity to be found by the letter ‘x’, and set it down as the 4th term. II: Of the three given quantities, set down that for the third term which is of the same kind as the quantity to be found. III: Now, consider carefully whether the quantity to be found will be greater or less than the third term; if greater, make the greater of the two remaining quantities the 2nd term, and the other 1st term, but if less, make the less quantity the second term, and the greater the 1st term. IV: Now, the required value

the required number of days =

=

Multiplication of means 1st term

161 After having the detailed knowledge about proportions and the Rule of Three, we now solve some of the examples which are usually solved by the unitary method. Ex. 1: If 15 books cost 35, what do 21 books cost? Soln: This is an example of direct proportion. Because if the number of books is increased, their cost also increases. By the Rule of Three: Step I : ... : ... = ... : Required cost Step II: ... : ... = 35 : Required cost Step III:The required cost will be greater than the given cost; so the greater quantity will come as the 2nd term. Therefore, 15 books : 21 books = 35 : Required cost 21 35 = 49 15 In a given time, 12 persons make 111 toys. In the same time, 148 toys are to be made. How many persons should be employed? Step I: ... : ... = ... : Number of persons Step II: ... : ... = 12 : x Step III: The required number of persons is more. Hence, 111 : 148 = 12 : x

Step IV: the required cost = Ex 2:

Soln:

148 12 16 111 If 192 mangoes can be bought for 15, how many can be bought for 5? Step I: ... : ... = ... : the required number of mangoes Step II: ... = ... = 192 : x Step III: As the required quantity would be less, 15 : 5 = 192 : x Step IV: x

Ex. 3: Soln:

5 192 64 15 If 15 men can reap a field in 28 days, in how many days will 5 men reap it? Step I : ... : ... = ... : Required number of days Step II : ... : ... = 28 : x Step III : The required number of days will be more, since 5 men will take more time than 15 men. Therefore, 5 : 15 = 28 : x Step IV: x

Ex. 4: Soln:

Step IV : x

15 28 84 days 5

Ex. 5: A fort had provisions for 150 men for 45 days. After 10 days, 25 men left the fort. How long will the food last at the same rate for the remaining men?

Quicker Maths

162 Soln:

The remaining food would last for 150 men for (45 –10 =) 35 days. But as 25 men have gone out, the remaining food would last for a longer period. Hence, by the Rule of Three, we have the following relationship. 125 men : 150 men = 35 days : the required no. of days.

150 35 the required no. of days = 125 42 days

Compound Proportion or Double Rule of Three Ex. 6: Soln:

If 8 men can reap 80 hectares in 24 days, how many hectares can 36 men reap in 30 days? We can resolve this problem into two questions. 1st: If 8 men can reap 80 hectares, how many hectares can 36 men reap? 8 men : 36 men = 80 hectares : the required no. of hectares the required no. of hectares

Ex 7:

If 30 men working 7 hrs a day can do a piece of work in 18 days, in how many days will 21 men working 8 hrs a day do the same piece of work?

Soln:

21 men : 30men :: 18 days : the reqd. no. days 8 hrs : 7hrs

the reqd. no. of days = Note:

Ex. 8:

Soln:

36 80 360 hectares 8 2nd: If 360 hectares can be reaped in 24 days, how many hectares can be reaped in 30 days?

30 360 450 24 We observe that the original number of hectares, namely 80, has been changed in the ratio formed

15 12 8 2.25 18 30 6 1 Now, we have, 24 women = 15 men 12 women = 7.5 men And also, 36 boys = 15 men

the reqd. no. of men =

36 30 and . The above 8 24

question can be solved in a single step. We arrange the figures in the following form :

6 boys

Ex. 9:

Multiplication of means = Multiplication of 1st terms

80 36 30 450 8 24

30 days working 6 hrs a day? Useful reasoning (1) More days : less men. (2) Less working hrs : more men. (3) More work : more men. Therefore, by the Rule of Three, 30 days: 12 days 6 hrs : 8 hrs

the reqd. no. of hectares =

The reqd. no. of hectares

1 times as great in 4

: : 15men : the reqd. no. of men 1 1 work : 2 works 4

By the Rule of Three 24 days : 30 days = 360 hectares : the reqd. no. of hectares.

8men : 36 men :: 80 hect : the reqd. no. of hectares 24days : 30days

Two lines of reasoning are used in the above case: (1) Less men : more days. (2) More working hrs : less days. If 15 men or 24 women or 36 boys do a piece of work in 12 days, working 8 hrs a day, how many men must be associated with 12 women and 6 boys to do another piece of work 2

by compounding the ratio

18 30 7 1 22 days 21 8 2

Soln:

15 5 2.5 men 6 2

12 women + 6 boys = 7.5 + 2.5 = 10 men So, 18 – 10 = 8 men must be associated. A garrison of 2200 men is provisioned for 16 weeks at the rate of 45 dag per day per man. How many men must leave the garrison so that the same provisions may last 24 weeks at 33 dag per day per man? We use the following steps in reasoning:

Ratio and Proportion

163

(1) For more weeks, less men are needed. (2) For less dag, more men are needed. So, by the Rule of Three 24 weeks : 16 weeks : : 2200 men: the reqd. no.of days 33 dag : 45 dag

2200 16 45 2000 24 33 Hence, 2200 – 2000 = 200 men must leave the garrison. Ex.10: Two cogged wheels, of which one has 16 cogs and the other has 27, work into each other. If the latter turns 80 times in three quarters of a minute, how often does the other turn in 8 seconds? Soln: Reasoning to be used : (1) Less cogs, more turns. (2) Less time, less turns. x

16 cogs : 27 cogs : : 80 turns : x turns 45 sec : 8 sec By the Rule of Three

80 27 8 24 16 45 Ex. 11: If 30 men do a piece of work in 27 days, in what time can 18 men do another piece of work 3 times as great? Soln: x

18 : 30 : : 27 : the reqd. no.of days Work 1 : 3

Men

Less men, more days More work, more days

27 30 3 the reqd. no. of days = = 135 days 18 1 Ex. 12: If a family of 7 persons can live on 840 for 36 days, how long can a family of 9 persons live on 810? Soln: persons 9 : 7 : :36 : the reqd. no.of days 840 : 810

the reqd. no. of days =

More persons, less days Less money, less days

36 7 810 27 days 9 840

Ex. 13: If 1000 copies of a book of 13 sheets require 26 reams of paper, how much paper is required for 5000 copies of a book of 17 sheets? Soln:

More books, more paper Books 1000 : 5000 : :26 : x More sheets, more paper Sheets 13 : 17

26 5000 17 1000 13 = 170 reams Ex. 14: If 6 men can do a piece of work in 30 days of 9 hours each, how many men will it take to do 10 times the amount of work if they work for 25 days of 8 hours? Soln: We need three lines of reasoning in this question: (1) Less days, more men (i.e., if a work is to be finished in less days, there should be more men at the work). (2) Less working hours, more men (i.e., if the working hour is less, the number of persons at work should be more to complete the work in a stipulated time). (3) More work, more men (i.e., if the work is more, the number of persons should be more so that all the work can be finished within the given time). Following the Rule of Three: Step I:

the quantity of paper =

Days : .... : .... Hrs : ..... : ... : : 6 : the reqd. no. of men Work : ...... : ... Step II: The Rule of Three states that : (1) To do the work in less days (25 days) we need more men (Reasoning : Less days, more men), hence greater value will go at the second place and smaller value will go at the first place. Like this : Days : 25 : 30 (2) The working hours (8 hrs) is less now, so we need more men. Thus, the greater value will go at the 2nd place and the smaller value will go at the 1st place. Like: Hrs : 8 : 9 (3) If there is more work (10 times) we need more men. Thus, greater value will go at the 2nd place and the smaller value will go at

Quicker Maths

164 the 1st place. Like: Work : 1 : 10 Thus, we reach the stage where all the blanks in Step I can be filled up.

Days 25 : 30 Hrs 8 : 9 : : 6: x Work 1 : 10 Now, by the Rule of Three, we have

x

Third term Multiplication of means Multiplication of first terms

6 30 9 10 81 men 25 8 1 Now, we go for the Rule of Fractions, which is very much similar to the Rule of Three in theory. or, x

Some Basics of Fractions (1) When a fraction has its numerator greater than the denominator, its value is greater than one. Let us call it greater fraction. Whenever a number (say x) is multiplied by a greater fraction, it gives a value greater than itself. For example: When 15 is multiplied by

4 (greater fraction), we get 3

20, which is greater than 15. (2) When a fraction has its numerator less than the denominator, its value is less than one. Let us call it less fraction. Whenever a number (say x) is multiplied by a less fraction, it gives a value less than itself. For example:

3 15 9 , which is less than 15. 5 Note: We will use the above two basics as well as the reasoning used in Rule of Three while solving Ex. 14 by the Rule of Fractions. Soln: Step I: We look for our required unit. It is the number of men. So, we write down the number of men given in the question. It is 6. Step II: The number of days gets reduced from 30 to 25, so it will need more men (Reasoning : Less days, more men). It simply means that 6 should be multiplied by a greater fraction because we need a value greater than 6. So, we 30 25 Step III: Following in the same way, we see have: 6

that the above figure should be multiplied by a 9 ‘greater fraction’, i.e., by . 8

30 9 25 8 Step IV: Following in the same way, we see that the above figure should be multiplied by a ‘greater So, we have: 6

fraction’ i.e. by

10 . 1

30 9 10 81 men 25 8 1 Ex. 15: In a given period, 9 persons can make 108 toys. How many persons are needed to make 48 toys in the same period? Soln: We see that we require less number of toys. So, less number of persons is needed. It means the given number of persons should be multiplied by a less fraction. Thus, our answer should be: So, we have: 6

9

48 = 4 persons. 108

Ex. 16: If 8 men can reap 80 hectares in 24 days, how many hectares can 36 men reap in 30 days? Soln: Step I: The required unit is hectare, so we write down the given number of hectares, i.e., 80. Step II: The number of men increases, so now they will reap more hectares. So, 80 should be multiplied by a greater fraction, i.e., by Thus, we have 80

36 . 8

36 . 8

Step III: The number of days also increases and hence they will reap more hectares. Thus, we have: 80

36 30 450 hectares. 8 24

PROPORTIONAL DIVISION Proportion may be applied to divide a given quantity into parts which are proportional to the given numbers. Ex. 1: Divide 1350 into three shares proportional to the numbers 2, 3 and 4. Soln. 1st share = 1350

2 2 1350 = 300 2 3 4 9

2nd share = 1350 ×

3 = 450 9

Ratio and Proportion

Ex. 2:

4 3rd share = 1350 × = 600 9 Divide 391 into three parts proportional to the

165 Soln:

1 1 1 1st part = 4 , 2nd part = 5 , 3rd part = 7 1st part : 2nd part : 3rd part

1 2 3 , , 2 3 4 Multiplying these ratios by the LCM of the denominators 2, 3, 4 namely 12, we get fractions

Soln:

1 2 3 : : 6:8:9 2 3 4 Now 6 + 8 + 9 = 23 6 1st part = 23 391 = 102 2nd part =

Note:

Ex. 6:

Soln:

3rd part = Note: Ex. 3:

Soln:

8 As of the whole sum = 8 41 8 41 = 41 8 Divide 1540 among A, B, C so that A shall receive

the whole sum =

Ex. 4:

Ex. 7:

Soln:

2 3 as much as B and C together, and B of what 9 11 A and C together do. Soln:

A’s share : (B + C)’s share = 2 : 9 ---------- (1) B’s share : (A + C)’s share = 3 : 11 ---------- (2) Now, dividing 1540 in the ratio of 2 : 9 and 3 : 11, A’s share =

A’s share =

3 of 2400 + 5 = 605 3 4 5

B’s share =

4 of 2400 + 10 = 810 3 45

C’s share =

5 of 2400 + 15 = 1015 3 45

Divide 1320 among 7 men, 11 women and 5 boys such that each woman may have 3 times as much as a boy, and a man as much as a woman and a boy together. Find how much each person receives. 1 man = 1 woman + 1 boy 1 woman = 3 boys 1 man = 4 boys 7 man : 11 women : 5 boys = 28 boys : 33 boys : 5 boys = 28 : 33 : 5 Dividing 1320 in the ratio of 28, 33 and 5, we have 7 men’s share =

2 of 1540 = 280 11

3 of 1540 = 330 14 C’s share = 1540 – ( 280 + 330) = 930 Ex. 5: Divide 581 into three parts such that 4 times the first may be equal to 5 times the second and 7 times the third.

1 1 1 : : 35 : 28 : 20 4 5 7 Now, divide 581 in the proportion of these numbers. Remember that for such questions, the three parts are in the proportion of the reciprocals of the numbers 4, 5 and 7. Divide 2430 among three persons A, B and C such that if their shares be diminished by 5, 10, 15 respectively, the remainders shall be in the ratio 3 : 4 : 5. 2430 – ( 5 + 10 + 15) = 2400 Dividing 2400 in the ratio 3 : 4 : 5, we get =

8 391 = 136 23

9 391 = 153 23 The third part may also be found by subtracting the sum of 102 and 136 from 391. A certain sum of money is divided among A, B and C such that for each rupee A has, B has 65 paise and C has 40 paise. If C’s share is 8, find the sum of money. Here A : B : C = 100 : 65 : 40 = 20 : 13 : 8 Now, 20 + 13 + 8 = 41

4 times the 1st part = 5 times the 2nd = 7 times the 3rd = 1 (say)

28 1320 = 560 66

1 man’s share =

B’s share =

Ex. 8:

560 = 80 7

4 boys’ share = 80 (As 1 man = 4 boys) 1 boy’s share = 20 and 1 woman’s share = 3 × 20 = 60. How many one-rupee coins, fifty-paise coins and twenty-five-paise coins of which the numbers

Quicker Maths

166 1 are proportional to 2 , 3 and 4, are together 2 worth 210? Soln:

Here 2

1 :3:4=5:6:8 2

Their proportional value = 5 × 1 : 6

1 1 : 8 2 4

Ex. 1: In 40 litres mixture of milk and water the ratio of milk and water is 3 : 1. How much water should be added in the mixture so that the ratio of milk to water becomes 2 : 1? Soln: Solving the above question by the direct formula given in the above theorem: The quantity of water to be added to get the required ratio:

=5:3:2 Now, 5 + 3 + 2 = 10

the value of rupees =

5 of 210 = 105 10

3 The value of fifty-paise coins = of 210 = 63 10 2 of 210 = 42 10 Therefore, there are 105 rupees, 126 fifty-paise coins and 168 twenty-five paise coins. The value of 25-paise coins =

Miscellaneous Examples Theorem: If in x litres mixture of milk and water, the ratio of milk and water is a : b, the quantity of water to be added in order to make this ratio c : d is

x(ad – bc) c(a b) Proof: Quantity of milk in the mixture =

x . a ab

x . b ab Suppose we added y litres of water to get our required ratio (c : d). Then we have

Quantity of water in the mixture =

ax a b

bx : y c : d a b

ax bx y(a b) : c : d or, a b ab ax c or, bx y(a b) d x(ad – bc) or, y (a b)c Note:

The above result is very systematic. They should be remembered. It saves a lot of time.

Note:

40(3 1 – 1 2) 40 5 litres. (3 1)2 8

The above solution can be verified as follows: In 40 litres of mixture, milk

40 3 30 litres 31 and water = 40 – 30 = 10 litres. 5 litres water is added; so in the new mixture, milk is 30 litres and water is 10 + 5 = 15 litres. Thus, the new ratio is 30 : 15 = 2 : 1. This ratio is the same as given in the question. Another Quicker Approach: Initial ratio: Milk = 30 litres, Water = 10 litres Final ratio = 2 : 1 You have to add water. It means quantity of milk remains the same (30). The value of milk in final ratio is 2. This implies that 2 30 or, 1 15. This means, in the final mixture water should be 15. Therefore 5 litres of water should be added. Ex. 2: In 30 litres mixture of milk and water, the ratio of milk and water is 7 : 3. Find the quantity of water to be added in the mixture in order to make this ratio 3 : 7. Soln: Following the same theorem, we have, =

The reqd. answer = Note:

30(7 7 – 3 3) = 40 litres 3(7 3)

The above question is the special case of the above mentioned theorem. Here, we see that the first ratio is reversed in the second case. That is, a : b becomes b : a in the new mixture. Moreover, the total quality of initial mixture equals the denominator [c (a +

b)]. In this case, the water to be added a 2 b2 Another Quicker Approach: Initial ratio: Milk = 21 litres, Water = 9 litres. Final ratio: Milk : Water = 3 : 7 You have to add water. It means the quantity of milk remains the same (21 litres). As the value of milk in the final ratio is 3, we have 3 21 7 49. This means in the final mixture water should be 49 litres, ie 49 – 9 = 40 litres of water should be added.

Ratio and Proportion

167

Theorem: A mixture contains milk and water in the ratio a : b. If x litres of water is added to the mixture, milk and water become in the ratio a : c. Then, the ax quantity of milk in the mixture is given by and c–b bx that of water is given by cb Proof: Let the quantity of mixture be M litres. Then, the quantity of milk =

aM litres. ab

bM litres. ab When x litres of water are added to the mixture, we have aM bM : x a:c ab ab aM bM x(a b) : a:c or, ab ab and the quantity of water

Another Quicker Approach: Initial ratio = 3 : 2 Second ratio = 3 : 3 We see that in the first ratio water is 2 and it becomes 3 in the second ratio when 4 litres of water is added. This means, 1 in ratio = 4 litres of water. Therefore, milk = 3 in ratio 12 litres Water = 2 in ratio 8 litres. Ex. 4: A mixture contains milk and water in the ratio of 8 : 3. On adding 3 litres of water, the ratio of milk to water becomes 2 : 1. Find the quantity of milk and water in the mixture. Soln: To follow the above theorem, we change the ratios in the form a : b and a : c. Then the ratios can be written as 8 : 3 and 8 : 4. Thus, the quantity of milk in the mixture

83 24 litres 43 and the quantity of water in the mixture =

3 3 9 litres. 43 Another Quicker Approach: Initial: Milk : Water = 8 : 3 Final: Milk : Water = 2 : 1 or 8 : 4 Since milk remains the same, we change the second (final) ratio such that milk becomes same in the both ratios. Now, see the water part. There is an increase of 1. This means 1 3 litres. Therefore, milk = 3 × 8 = 24 litres Water = 3 × 3 = 9 litres =

aM a or, bM x(a b) c or, CM = bM + x(a + b)

x(a b) M (c b) Thus, the quantity of milk in the mixture

aM ax(a b) ax a b (a b)(c d) c b

Similarly, the quantity of water in the mixture

Ex. 3:

Soln:

bM bx(a b) bx a b (a b) (c b) c b

A mixture contains milk and water in the ratio of 3 : 2. If 4 litres of water is added to the mixture, milk and water in the mixture become equal. Find the quantities of milk and water in the mixture. If we want to solve the above question by the theorem stated above, we will have to change the form of ratios to a : b and a : c. In the above question, the initial ratio is 3 : 2. Thus, to equate the antecedents of the ratio, we write the second ratio as 3 : 3. Then by the above direct formula: The quantity of milk =

3 4 = 12 litres. 3– 2

and the quantity of water =

24 = 8 litres. 3 2

Theorem: If two quantities X and Y are in the ratio x : y. Then X + Y : X – Y :: x + y : x – y Proof: The above theorem can be proved by the rule of componendo-dividendo.

X x We are given that Y y Then, by the rule of componendo-dividendo:

XY xy XY xy or, X + Y : X – Y : : x + y : x – y Ex. 5: A sum of money is divided between two persons in the ratio of 3 : 5. If the share of one person is 20 less than that of the other, find the sum. Soln:

By the above theorem:

Sum 3 5 20 53

Quicker Maths

168

Note:

Ex. 6:

Soln:

8 Sum 20 = 80 2 The above question can also be solved as follows (this method is similar to the above theorem): 5 – 3 20

Soln:

Following the rule, we have,

Note:

ratio of their areas = 32 : 4 2 9 : 16. The above mentioned theorem is true for any twodimensional figure and for any measuring length related to that figure (see Examples 8 and 9).

20 5 + 3 5 3 (5 3) = 80 The prices of a scooter and a moped are in the ratio of 9 : 5. If a scooter costs 4200 more than a moped, find the price of the moped. Following the method mentioned in the above note, we have, 9 – 5 4200

Theorem: In any two 3-dimensional figure, if the corresponding sides or other measuring lengths are in the ratio a : b, then their volumes are in the ratio a3 : b3. Ex. 10: (a) The sides of two cubes are in the ratio 2 : 1. Find the ratio of their volumes. (b) Each side of a parallelopipe is doubled find the ratio of volume of old to new parallelopipe.

5

4200 5 = 5250 95

Theorem: In any two two-dimensional figure, if the corresponding sides are in the ratio a : b, then their areas are in the ratio a2 : b2. Proof: To prove the above theorem, we take the case of a rectangle. This will be true for every other twodimensional figure also. Suppose we have a rectangle whose sides are x and y. We are given that the sides of another rectangle are in the ratio a : b, therefore, the sides of

b b x and y. a a Therefore, the ratio of the two areas second rectangle are

2

b xy a 2 : b 2 a2 The sides of a hexagon are enlarged by three times. Find the ratio of the areas of the new and old hexagons. Following the above theorem, we see that the ratio of the corresponding sides of the two hexagons is a : b = 1 : 3. Therefore, the ratio of their areas is given by = xy :

Ex. 7:

Soln:

a 2 : b2 12 : 32 1 : 9. Ex. 8: The ratio of the diagonals of two squares is 2 : 1. Find the ratio of their areas. Soln: We should follow the same rule when the ratio of diagonals is given instead of the ratio of sides. Thus, the ratio of their areas = 22 : 12 = 4 : 1. Ex. 9: The ratio of the radius (or diameter or circumference) of two circles is 3 : 4. Find the ratio of their areas.

Soln:

3 3 (a) The required ratio = (2) : (1) = 8 : 1 3 3 (b) The required ratio = (1) : ( 2) = 1 : 8

Theorem: The ratio between two numbers is a : b. If each number be increased by x, the ratio becomes c : d. Then Sum of the two numbers

x(a b) (c d) ad bc

Difference of the two numbers =

x(a b)(c d) ad bc

And the two numbers are given as

xa(a d) xb (c d) and ad bc ad bc Proof: Let the sum of the two numbers be X.

aX bX and ab ab Now, when each number is increased by x then Then the numbers are

aX bX x : x c:d ab ab or,

aX x(a b) bX x(a b) : c:d ab ab

aX x(a b) c or, bX x(a b) d

X

x(a b)(c d) ad bc

And the numbers are

bX xb(c d) ab ad bc

aX xa(c d) and ab ad bc

Ratio and Proportion the difference of the two numbers x(a b)(c d) ad bc Ex. 11: The ratio between two numbers is 3 : 4. If each number be in creased by 2, the ratio becomes 7 : 9. Find the numbers. Soln: Following the above theorem, the numbers are 2 3(7 9) 2 4(7 9) and 3 9 4 7 3 9 4 7 or, 12 and 16. Another Quicker Approach: Initial ratio = 3 : 4 = 6 : 8 Final ratio = 7 : 9 Since each number is increased by the same value we have to change the first or the second ratio in such a way that the respective differences of antecedents and consequents become the same. So, if we change the first (initial) ratio to 6 : 8, we see that 7 - 6 = 1 = 9 - 8. This implies that 1 in the changed initial ratio is equivalent to 2. Therefore the numbers are 6 × 2 and 8 × 2, ie 12 and 16. Ex. 12: The ratio between two numbers is 3 : 4. If each number be increased by 6, the ratio becomes 4 : 5. Find the two numbers. Soln: The above question may be considered as a special case of the above theorem where c – a = d – b It is easy to distinguish this type of question. In such a question, there should be a uniform increase in ratio, i.e., the antecedent and consequent is increased by the same value. In the above question, we see that both the antecedent and the consequent are increased by 1 each, and the numbers are increased by 6. Therefore, we may say that 1 6 or, 3 3 × 6 = 18 and 4 4 × 6 = 24 Thus, the numbers are 18 and 24. Note: (1) The above general formula also works for the above example (Ex. 12). It is suggested that you apply that method because that method is universal and you should be familiar with it. (2) The above question may be rewritten as: The ratio of two numbers is 4 : 5. If each of them is decreased by 6, the ratio becomes 3 : 4. Find the two numbers. Apply the same rule in this case also. Ex. 13: The students in three classes are in the ratio 2 : 3 : 5. If 20 students are increased in each class, the ratio changes to 4 : 5 : 7. What was the total number of students in the three classes before the increase?

169 Soln:

In the above question also, we see that each term increases by the same value. That is, 4 – 2 = 5 – 3 = 7 – 5 = 2. Thus, we have 2 20 20 (2 3 5) 10 = 100 students. 2

Theorem: The incomes of two persons are in the ratio a : b and their expenditures are in the ratio c : d. If each of them saves X, then their incomes are given

Xa(d c) Xb(d c) and ad bc ad bc Proof: Try this yourself, because the proof for this theorem is similar to the above theorems. Ex. 14: The incomes of A and B are in the ratio 3 : 2 and their expenditures are in the ratio 5 : 3. If each saves 2000, what is their income? Soln: According to the above theorem, a : b = 3 : 2 (Income) c : d = 5 : 3 (Expenditure) X = 2000 (Savings) Xa (d c) Therefore, A’s income = ad bc 2000 3 (3 5) = 12,000 3 3 2 5 Xb(d c) and B’s income = ad bc 2000 2 (3 5) = 8,000 33 25 Note: I: If we are asked to find the expenditure, we have two options: (1) Expenditure = Income – Saving Thus, A’s expenditure = (12000 – 2000) = 10,000 and B’s expenditure = (8000 – 2000) = 6,000. (2) The direct formula is given by : Xc(b a) A’s expenditure = ad bc Xd(b a) B’s expenditure = ad bc II: If you note carefully, you will see the similarity between the direct formula for income and expenditure. III: Quicker Approach: A : B Income 3 : 2 = 6 : 4 Expenditure 5 : 3 A and B both save the same amount ( 2000). by

Quicker Maths

170 As Saving = Inc – Exp, we have to change the terms in the ratio in such way that Inc – Exp comes the same for both A and B. In the above case, if the ratio of Income, 3 : 2, is changed to 6 : 4 we see that 6 – 5 = 1 = 4 - 3. Thus, 1 in the changed ratio of income is equivalent to 2000. Therefore A's income = 6 × 2000 = 12000 B's income = 4 × 2000 = 8000 A's expenditure = 5 × 2000 = 10000 B's expenditure = 3 × 2000 = 6000 Ex. 15: The incomes of Ram and Shyam are in the ratio 8 : 11 and their expenditures are in the ratio 7 : 10. If each of them saves 500, what are their incomes and expenditures? (Only by the direct formula) Soln: Using the theorem: a : b = 8 : 11 (Income) c : d = 7 : 10 (Expenditure) X = 500 (Savings) Ram’s income Xa(d c) 500 8(10 7) = = 4000 ad bc 80 77 Shyam’s income Xb(d c) 500 11(10 7) = = 5500 ad bc 80 77 Ram’s expenditure Xc(b a) 500 7(11 8) = = 3500 ad bc 80 77 Shyam’s expenditure

Xd(b a) 500 10(11 8) = 5000 ad bc 80 77 Quicker Approach: A : B Income = 8 : 11 Exp = 7 : 10 Saving = Inc - Exp In the above case 8 – 7 = 1 = 11 – 10. Thus, 1 in the ratio is equivalent to 500. A's income = 8 × 500 = 4000 B's income = 11 × 500 = 5500 A's expenditure = 7 × 500 = 3500 B's expenditure = 10 × 500 = 5000 It is very easy to remember the above formulae. To be more familier with these, you need only good practice. Use them whenever you find such a question. You need not write the formula each time, but do only the digital values for calculations. You will get the answer within seconds. =

Note:

Note:

Theorem: If the ratio of any quantities be a : b : c : d, then the ratio of other quantities which are inversely proportional to that is given by

1 1 1 1 : : : a b c d

Ex. 16: The speed of three cars are in the ratio 2 : 3 : 4. What is the ratio among the times taken by these cars to travel the same distance? Soln: We know that speed and time taken are inversely proportional to each other. That is, if speed is more the time taken is less and vice versa. So, we can apply the above theorem in this case. Hence, ratio of time taken by the three cars

1 1 1 : : 2 3 4 Now, multiply each fraction by the LCM of denominators i.e., the LCM of 2, 3, 4, i.e., 12. So, the required ratio is given by =

12 12 12 : : 6:4:3 2 3 4 Ex. 17: The same type of work is assigned to three groups of men. The ratio of persons in the groups is 3 : 4 : 5. Find the ratio of days in which they will complete the works. Soln: We see that in this case also, man and days are inversely proportional to each other. So, the above rule can be applied in this case also. Therefore, the 1 1 1 : : 3 4 5 Multiplying the above fractions by the LCM of 3, 4 and 5, i.e., 60, required ratio is

we have,

60 60 60 : : 20: 15 : 12 3 4 5

Theorem: If the sum of two numbers is A and their difference is a, then the ratio of numbers is given by A+ a :A– a Proof: Let the two numbers be x and y. Then we have, x + y = A ---------- (1) and x – y = a --------- (2) (1) + (2) gives 2x = A – a

Aa 2 (1) – (2) gives 2y = A – a x

y

Aa 2

Ratio and Proportion Now, the ratio of two numbers

Aa Aa : =A+ a :A– a 2 2 Ex. 18: The sum of two numbers is 40 and their difference is 4. What is the ratio of the two numbers? Soln: Following the above theorem, the required ratio of numbers = 40 + 4 : 40 – 4 = 44 : 36 = 11 : 9 =x:y=

Theorem: A number which, when added to the terms of

ad - bc c-d ax c Proof: Let the required number be x, then bx d the ratio a : b makes it equal to the ratio c : d is

or, ad + dx = bc + cx or, x(c – d) = ad – bc

x=

ad bc cd

Ex. 19: Find the number which when added to the terms of the ratio 11 : 23 makes it equal to the ratio 4 : 7. Soln: Following the above rule : a : b = 11 : 23 c:d=4:7 the required number ad bc 11 7 23 4 (–)15 5 = cd 47 ( )3 Another Quicker Approach: Initial ratio = 11 : 23 Final ratio = 4 : 7 = 16 : 28 In this case, we have to change the second (final) ratio in such a way that Antecedent of final ratio – Antecedent of initial ratio = Consequent of final ratio – Consequent of initial ratio = Required number to be added. Here, 4 : 7 is changed to 16 : 28 such that 16 - 11 = 5 = 28 - 23. Thus the required number is 5. Theorem: A number which, when subtracted from the terms of the ratio a : b makes it equal to the ratio c : d is

bc - ad c-d

Proof: Try it yourself. Ex. 20: Find the number which when subtracted from the terms of the ratio 11 : 23 makes it equal to the ratio 3 : 7.

171 Soln:

Here, a : b = 11 : 23 c:d = 3:7 the required number

bc ad 23 3 11 7 8 2 cd 3 7 4 Another Quicker Approach: Initial ratio = 11 : 23 Final ratio = 3 : 7 = 9 : 21 Applying the same concept as in Ex. 19, we have changed the final ratio to 9 : 21 such that 11 - 9 = 23 - 21 = 2. Therefore, the required number is 2. Ex. 21: The contents of two vessels containing water and milk are in the ratio 1 : 2 and 2 : 5 are mixed in the ratio 1 : 4. The resulting mixture will have water and milk in the ratio ______. Soln: Change the ratios into fractions. Water : Milk 1 2 Vessel I 3 3 2 5 Vessel II 7 7 =

From Vessel I,

1 4 is taken and from Vessel II, 5 5

is taken. Therefore, the ratio of water to milk in the new

1 1 2 4 vessel = 3 5 7 5

2 1 5 4 : 3 5 7 5

31 74 1 8 2 20 : : 31 : 74 15 35 15 35 105 105 Ex. 22: The ratio of A’s and B’s income last year was 3 : 4. The ratio of their own incomes of last year and this year is 4 : 5 and 2 : 3 respectively. If the total sum of their present incomes is 4160, then find the present income of A. Soln: The ratio of present incomes = 3

5 3 15 12 : 30 : 48 5 : 8 : 4 = 4 2 4 2

4160 5 = 1600 58 Ex. 23: Three glasses A, B and C with their capacities in the ratio 2 : 3 : 4 are filled with a mixture of spirit and water. The ratio of spirit to water in A, B and C is 1 : 5, 3 : 5 and 5 : 7 respectively. If the contents of these glasses are mixed together, find the ratio of spirit to water in the mixture.

A’s present income =

Quicker Maths

172 Soln:

A : B : C 2 : 3 : 4 Sp : W = 1 : 5 3 : 5 5 : 7 When they are mixed, the ratio of spirit to water

1 3 5 3 4 = 2 1 5 35 5 7 5 5 7 3 4 : 2 1 5 35 5 7 1 9 5 5 15 7 : 3 8 3 3 8 3

25 47 : 25 : 47 8 8 Ex. 24: 465 coins consist of rupee, 50 paise and 25 paise coins. Their values are in the ratio 5 : 3 : 1. Find the number of each coin. Soln: The ratio of number of coins

100 100 100 :3 :1 5: 6: 4 100 50 25 the number of one-rupee coins 5

465 5 155 56 4

The number of 50P coins

The value of 50P coins = or 10 coins The value of 5P coins =

465 6 186 564

100 50 5 :5 :7 100 100 100 = 300 : 250 : 35 = 60 : 50 : 7 value of 1-rupee coins

the

11.70 60 6 or 6 coins 60 50 7

9 and 4 : 5 2

25 5 5 = 4 : 5 5 : 4 4 4

The number of 25P coins

= 3

11.70 7 0.7 60 50 7

or 0.7 × 20 = 14 coins Ex. 26: One year ago, the ratio between Laxman’s and Gopal’s salaries was 3 : 5. The ratio of their individual salaries of last year and present year are 2 : 3 and 4 : 5 respectively. If their total salaries for the present year are 4300, find the present salary of Laxman. Soln: The ratio of Laxman’s salary for the two years =2:3 The ratio of Gopal’s salary for the two years =4:5 We are also given that the ratio of their salary during the last year =3:5 Now, we change the antecedents (2 and 4) of the first two ratios so that the antecedent in the first becomes 3 (antecedent of the third ratio) and the antecedent in the second becomes 5 (consequent of the third ratio). Thus, 2 : 3 = 3 :

465 4 124 56 4 Ex. 25: A sum of 11.70 consists of rupee, 50 paise and 5 paise coins in the ratio 3 : 5 : 7. Find the number of each kind of coins. Soln : This question is different from Ex. 24. In Ex. 24, the ratio of values were given but in this case, the ratio of numbers is given. Now, the given ratio of numbers is to be changed in the ratio of values. (Whereas in the Ex. 24, the ratio of values was changed into ratio of numbers.) the ratio of values

11.70 50 = 5 60 50 7

Now, it is clear that the ratio of their salaries for the present year is

Note:

9 25 : 18 : 25 2 4

4300 the present salary of Laxman = 18 25 18 = 1800 You should understand the above method clearly. Once you do that, you need very few calculations to reach the answer. What happens when you are given: “The present ratio between their salaries is 18 : 25, and the ratios of their individual salaries for the two years is 2 : 3 and 4 : 5. Their total salary for last year was 3200. And you are asked to find the salary of Laxman for last year.” Now, we do the same for the consequents: 2 : 3 = 2 × 6 : 3 × 6 = 12 : 18 4 : 5 = 4 × 5 : 5 × 5 = 20 : 25

Ratio and Proportion Thus, the ratio of their salaries last year was 12 : 20 = 3 : 5

3200 3 = 1200 Laxman’s salary last year = 35 Ex. 27: A bucket contains a mixture of two liquids A and B in the proportion 7 : 5. If 9 litres of the mixture is replaced by 9 litres of liquid B, then the ratio of the two liquids becomes 7 : 9. How much of the liquid A was there in the bucket? Soln: Detailed Method: Suppose the two liquids A and B are 7x litres and 5x litres respectively. Now, when 9 litres of mixture are taken out, A remains

9 ×7 21 7 7x - 9 = 7x – = 7x - litres 7 + 5 12 4 and B remains

95 15 5 5x 9 5x 5x litres. 7 5 12 4 Now, when 9 litres of liquid B are added,

21 7x 4

15 : 5x 9 7 : 9 4

173 Common factor of first ratio

Quantity Replaced = + Sum of terms in 1st ratio Quantity replaced × term A in 2 nd ratio D 9 9 7 9 9 36 3 7 5 28 12 4 12 Quantity of A = 7 × 3 = 21 litres. Similarly, quantity of B = 5 × 3 = 15 litres. Ex. 28: The employer decreases the number of his employees in the ratio 10 : 9 and increases their wages in the ratio 11 : 12. What is the ratio of his two expenditures? Soln: The required ratio = 10 × 11 : 9 × 12 = 55 : 54 Ex. 29. A vessel contains liquids A and B in ratio 5 : 3. If 16 litres of the mixture are removed and the same quantity of liquid B is added, the ratio becomes 3 : 5. What quantity does the vessel hold? Soln: Detailed Method: Suppose the vessel contains 5x litres and 3x litres of liquids A and B respectively.

16 5 10 litres 53 of A and 16 – 10 = 6 litres of B. Now, (5x – 10) : (3x – 6 + 16) = 3 : 5 The removed quantity contains

21 4 7 or, 15 5x 9 9 4 7x

189 105 35x 63 or, 63x 4 4 or, 28x or, x

189 105 63 21 63 84 4 4

84 3 28

7 x 7 3 21 litres Quicker Method: If we ignore the intermediate steps, we find a formula which is fast-working as well as easier to remember. 1st ratio = 7 : 5, 2nd ratio = 7 : 9 D = Difference of cross-products of ratios = 7 9 7 5 63 35 28 Now, the formula is:

5x 10 3 3x 10 5 or, 25x – 50 = 9x + 30 or, 16x = 80 x=5 The vessel contains 8x = 8 × 5 = 40 litres. Quicker Method: When the ratio is reversed (i.e., 5 : 3 becomes 3 : 5), we can use the formula: or,

Total quantity

(5 3)2 × Quantity of A in the 52 32

64 10 40 litres. 16 Note: When the liquid B is used as a filler, the quantity of A is used in the formula. Another Quicker Approach: In the removed quantity, A is 10 litres and B is 6 litres. removed mixture

Quicker Maths

174 Again 16 litres of B is added. This implies that in the new mixture A is 10 litres less and B is 16 - 6 = 10 litres more. Now look at the ratios. Initial ratio = 5 : 3 Final ratio = 3 : 5 Note that in the final ratio the antecedent is less by (5 - 3 =) 2 and the consequent is more by (5 - 3 =) 2. This implies that 2 in the ratio is equivalent to 10 litres.

10 5 3 40 litres. 2 Ex. 30: If (a + b) : (b + c) : (c + a) = 6 : 7 : 8 and a + b + c = 14, then find a : b : c and the value of a, b and c. Soln: We should know that Therefore, total quantity =

a+b= =

6 [(a + b) + (b + c) + (a + c)] 678 6 6 [2(a b c)] 28 8 21 21

Similarly, b + c =

7 [2(a + b + c)] 678

7 28 28 21 3

and a + c =

Similarly, b 14 Thus, a

28 14 3 3

32 10 and c = 14 – 8 = 6 3 3

14 10 , b and c 6 3 3

14 10 : : 6 14 : 10 : 18 7 : 5 : 9 3 3 Quicker Method: (a + b) : (b + c) : (c + a) = 6 : 7 : 8 Now, [(a + b) + (b + c) + (c + a)] : (a + b) : (b + c) : (c + a) = (6 + 7 + 8) : 6 : 7 : 8 or, 2 (a + b + c) : (a + b) : (b + c) : (c + a) = 21 : 6 : 7 : 8 or (a + b + c) : (a + b) : (b + c) : (c + a) = 10.5 : 6 : 7 : 8

a:b:c=

14 14 a 759 7 3 b=

14 10 5 7 5 9 3

14 96 7 59 Ex 31: Two candles of the same height are lighted at the same time. The first is consumed in 7 hours and the second is consumed in 4 hours. Assuming that each candle burns at a constant rate, in how many hours, after being lighted, was the first candle four times the height of the second? Soln: Detail Method: Let the height of the candles be 'h' and after x hrs the height of the first candle be four times the height of the second. Now, the height of first candle after x hrs xh 7h xh = h 7 7 And the height of second candle after x hrs xh 4h xh = h 4 4 Now, as per question, c

7h xh 4h xh : 4 :1 7 4

8 32 28 21 3

Now, a = [(a + b + c) – (b + c)] = 14

Now, a : b : c = (10.5 – 7) : (10.5 – 8) : (10.5 – 6) = 3.5 : 2.5 : 4.5 = 7 : 5 : 9

(7h xh)4 4 or, 7(4h xh) 1 or, 28 – 4x = 112 – 28x or, 24x = 84

84 3.5 hrs = 3 hrs 30 min. 24 Direct Formula: If out of two candles of the same height, the first

x=

burns in T1 hrs and the second burns in T2 hrs, then after

T1T2 a b hrs the ratio of the height aT1 bT2

of remaining parts will be a : b. So in the above case, T1 7 hrs, T2 4 hrs a = 4 and b = 1

7 4(4 1) 84 required time = 4 7 1 4 24 3.5 hrs

Ratio and Proportion

175

Ex 32: A container contained 80 kg of milk. From this container 8 kg of milk was taken out and replaced by water. This process was further repeated twice. How much milk is now contained by the container? Soln: The best way to solve this type of question is to present the withdrawal of liquid as percentage or ratio of the original volume. For example, see the following explanation: As 8 kg is 10% of 80 kg, so in each withdrawal 10% of milk is withdrawn, which implies that after each operation 90% of the previous volume of milk remains in the mixture. Milk Water Originally: 80 kg 0 kg After first operation: 80 (90%) 80 (10%) = 72 kg = 8 kg After second operation: 72 (90%) 8 kg + 10% of 72 = 64.8 kg = 15.2 kg After third operation: 64.8 (90%) 15.2 kg + 10% of 64.8 = 58.32 kg = 21.68 kg If we simplify the above chart we may say that: The quantity of milk in the mixture after third 3

90 operation is 80 . 100 Similarly, the quantity of milk in the mixture after n

90 the nth operation is 80 100 The same thing can be presented in ratio as given below: The quantity of milk after the nth operation n

Note:

80 8 = 80 80 (Because after each operation, milk changes in the ratio 72 : 80 or 9 : 10, ie from 10 to 9.) The most generalised format of the above question: From a container containing 'X' litres of milk, x litres is withdrawn and replaced by water. This process was repeated 'n' times. Then quantity of milk left in the container after nth operation

Soln:

To solve the above example, we should understand the formula discussed in Ex 32. We have, quantity of milk left after nth operation =

X x Original quantity of milk ie X X or,

Quantity of milk left after nth operation Original quantity of milk

X x = X

n

16 X 9 In this case, 16 9 X

2

..... (*)

X9 4 X 5 X = 45 litres. Note: You may be confused in the equation (*) about the or,

16 . 16 9 Actually, the ratio of milk to water in the final mixture is 16 : 9. This implies that ratio of milk in the final mixture to total of milk and water (which is the same as the original quantity of milk) must be 16 : (16 + 9). Ex 34: A container is full of spirit. 40 litres of spirit is taken out and the container is filled with water. This process is repeated twice further. Now, the ratio of spirit to water in the container is 27 : 98. Find the capacity of the container. Soln: Apply the same formula as in Ex 33. LHS, ie,

Quantity of spirit after 3rd operation Capacity of container (or Original quantity of milk)

X 40 = X

3

or,

27 X 40 27 98 X

or,

3 X 40 5 X

n

X x = X litres . X Ex 33: A container is full of milk. Nine litres of milk is drawn and the container is filled with water. Again, nine litres of mixture are drawn and the container is again filled with water. The quantity of milk now left in the container to that of water in it is 16 : 9. Find the original quantity of milk in the container.

n

3

X 40 3 X 5 X = 100 litres.

or,

3

3

Quicker Maths

176

EXERCISE 1. Form the compound ratio of the ratios 45 : 75, 3 : 4, 51 : 68 and 256 : 81. 2. If A : B = 6 : 7 and B : C = 8 : 9, find A : B : C. 3. The sum of two numbers is 20, and their difference 1 is 2 . Find the ratio of the numbers. 2 4. If 0.7 of one number be equal to 0.075 of another, what is the ratio of the two numbers? 5. Find a fraction which shall bear the same ratio to

3 5 does to . 11 9 Two sums of money are proportional to 8 : 9. If the first is 20, what is the other? 5 Find two numbers in the ratio of 5 to 5 such that 7 1 when each is diminished by 12 , they shall become in 2 2 the ratio of 3 to 3. 3 Divide 37 into two parts such that 5 times one part and 11 times the other are together 227. 4 Find a ratio equal to whose antecedent is 9. 5 Find the value of x in the folowing proportions : (i) 5 : 15 = 2 : x. (ii) 75 : 3 = x : 9. Calculate a fourth proportional to the numbers : (i) 1, 2, 3. (ii) 490, 70, 69. (iii) 2.5, 1.5, 1.5. If 30 men do a piece of work in 27 days, in what time can 18 men do another piece of work 3 times as great? When wheat is 1.30 per kg, 60 men can be fed for 15 days at a certain cost. How many men can be fed for 45 days at the same cost, when wheat is Re 1 per kg? If a family of 7 persons can live on 840 for 36 days, how long can a family of 9 persons live on 810? If 5 horses eat 18 quintals of oats in 9 days, how long at the same rate will 66 quintals last for 15 horses? If 1000 copies of a book of 13 sheets require 26 reams of paper, how much paper is required for 5000 copies of a book of 17 sheets? If the carriage of 810 kg for 70 km costs 45, what will be the cost of the carriage of 840 kg for a distance of 63 km at half the former rate? that

6. 7.

8. 9. 10. 11.

12.

13.

14. 15. 16.

17.

1 27

18. If 300 men can do a piece of work in 16 days, how

1 of the same work in 15 days? 4 19. Divide 324.36 into three parts in the proportion of 5 : 6 : 7. 20. Divide 53.95 between A, B and C such that A gets thrice as much as B, and C one-third as much as B. 21. Divide 91.30 between A, B and C such that A gets 1 1 1 times as much as C and B 2 times as much as C. 2 2 2 22. Divide 625 among A, B and C such that A gets of 9 many men would do

3 of A’s share. 4 23. Divide 99 among A, B, C such that A may get 5 times B’s share and C gets

as much as B, and C gets

1 of what A and B together 2

get. 24. Divide 355 into three parts such that three times the first part may be equal to five times the second and seven times the third. 25. A body of 7300 troops is formed of 4 battalions, so that

1 2 3 of the first, of the second, of the third and 2 3 4

4 of the fourth are all composed of the same number 5 of men. How many men are there in each battalion? 26. The estate of a bankrupt person worth of 21000 is to be divided among four creditors whose debts are – A’s to B’s, as 2 : 3, B’s to C’s as 4 : 5, C’s to D’s as 6 : 7. What amount must each receive? 27. How many one-rupee coins, 50 P coins and 25 P coins, of which the numbers are proportional to 4, 5 and 6 are together worth 32? 28. A sum of 3115 is divided among A, B and C such that if 25, 28 and 52 be diminished from their shares respectively, the remainders shall be in the ratio of 8 : 15 : 20. Find the share of each. 29. What must be added to two numbers that are in the ratio of 3 : 4, so that they become in the ratio 4 : 5? 30. Find the number which, when subtracted from the terms of the ratio 19 : 23 makes it equal to the ratio of 3 : 4. 31. An employer reduces the number of his employees in the ratio 9 : 8 and increases their wages in the ratio

Ratio and Proportion

32.

33.

34.

35.

36.

177

14 : 15. State whether his bill of total wages increases or decreases, and in what ratio. 50 is divided among 6 men, 12 women and 17 boys so that 2 men get as much as 5 boys and 2 women as much as 3 boys. Find the share of a boy. Which of the following represents ab = 64? 1) 8 : a = 8 : b 2) a : 16 = b : 4 3) a : 8 = b : 8 4) 32 : a = b : 2 5) None of these Mr Pandit owned 950 gold coins, all of which he distributed amongst his three daughters Lalita, Amita and Neeta. Lalita gave 25 gold coins to her husband, Amita donated 15 gold coins and Neeta made jewellery out of 30 gold coins. The new ratio of the coins left with them was 20 : 73 : 83. How many gold coins did Amita receive from Mr Pandit? Mr X invested a certain amount in Debt and Equity Funds in the ratio of 4 : 5. At the end of one year, he earned a total dividend of 30% on his investment. After one year, he reinvested the amount including the dividend in the ratio of 6 : 7 in Debt and Equity Funds. If the amount reinvested in Equity Funds was 94,500, what was the original amount invested in Equity Funds? There was a science exhibition in an auditorium. On the first day 14 persons visited the exhibition, on the second day 12 persons and on the third day only 10 persons visited. The ratio of admission fees collected from each of them on these days was 2 : 3 : 5 respectively. If the total amount collected on these three days was 4560. What amount was collected on the first day?

37. A gave 25% of an amount to B. From the money B got, he spent 30% on a dinner. Out of the remaining amount, the respective ratio between the amount B kept as savings and the amount he spent on buying a book is 5 : 2. If B bought the book for 460, how much money did A have in the beginning ? 38. The monthly salary of Dex is

2 of their 5 father's monthly salary. Dex's sister pays 12,800 as salary. Dex's sister's monthly salary is

1 of her monthly salary. The 4 savings and expenses made out of the monthly salary by Dex are in the ratio of 3 : 5. How much does Dex save each month? 39. In an examination, the number of students who passed and the number of those who failed were in the ratio of 25 : 4. If five more students had appeared and the number of failed students was 2 less than earlier, the ratio of passed to failed students would have been 22 : 3. What is the number of students who appeared for the examination? 40. A certain sum is divided among A, B and C in such a study loan, which is

way that A gets 40 more than

45 3 51 256 16 16 : 15 75 4 68 81 15 2. A : B = 6:7 B:C = 8:9 A : B : C = 6 × 8 : 7 × 8 : 7 × 9 = 48 : 56 : 63 5 20 2 22.5 225 9 9 : 7 3. Ratio = 5 17.5 175 7 20 2 4. We have, 0.7x = 0.075y

x 0.075 75 3 y 0.7 700 28 3: 28

1 of the sum. B gets 2

3 of the sum and C gets 200. What 8 is the total sum? 120 less than

ANSWERS 1.

1 of his father's monthly 4

5.

x:

1 3 5 : 27 11 9

or, 27x

x

3 9 11 5

1 55

6.

20 9 45 = 22.5 8 2

7.

40 25 25 11 x– : 5x – :3 7 2 2 3

Quicker Maths

178 70 69 60 x 490 7

40 25 x 7 2 11 or, 25 9 5x 2 or,

40 9x 225 275 – 55x 7 2 2

385 – 360 50 x 7 2 or, x = 7 or,

Answer

Therefore, the numbers are

40 7 and 5 × 7 or 40 and 7

35. Quicker Method: This question can be simplified if we change the form of ratio as follow:

5 40 5 :5 : 5 40 : 35 8 : 7 7 7 2 11 and 3 : 3 : 3 11 : 9 3 3 12.5(11 9) 5 Common factor = 7 11 8 9 Thus, the numbers are 8 × 5 and 7 × 5 or 40 and 35. 8. x + y = 37 5x + 11y = 227 Solving these two equations, x = 30 and y = 7 9.

4 9 5 x

10. (i) x

15 2 6 5

75 9 225 3 11. (i) 1 : 2 :: 3 : x (ii) x

23 x 1 6 (ii) 490 : 70 : : 69 : x

30 3 27 = 135 days. 18

30 3 By rule of Fraction: 27 135 days 18 1 15 1.3 13. 60 26 men. 45 1 Note: I: Put the number of men (given) as you are asked to find the number of men. II: When the number of days increases, less persons could be fed, so multiply by a lessthan-one fraction, i.e.,

15 45

III: Since price decreases, more persons could be fed. Hence, multiply by a

1.3 1

more-than-one fraction i.e.,

7 810 14. 36 27days 6 840 Follow the same reasoning as in Q. 13.

45 or, x 4 Therefore, ratio 9 :

1.5 1.5 0.9 2.5 12. By the Rule of Proportion: 18 men : 30 men :: 27 : the reqd. no. of days 1 work : 3 work (iii) x

45 4:5 4

66 5 15. 9 11days 18 15 5000 17 16. 26 170 reams 1000 13 840 63 1 17. 45 = 21 810 70 2

16 1 18. 300 80 men 15 4 19. Three parts are

324.36 324.36 324.36 5, 6, 7 18 18 18 = 18.02 × 5, 18.02 × 6, 18.02 × 7 = 90.10, 108.12, 126.14

Ratio and Proportion 20. A : B = 3 : 1 B:C=3:1 A: B:C=3 × 3:1 ×3:1 ×1 =9 : 3 :1 Now, the process is the same as in Q. 19. 21. A : C = 3 : 2 C:B=2:5 A : C : B = 3 × 2 : 2 × 2 : 2 × 5 = 6 : 4 : 10 = 3 : 2 : 5 A:B:C= 3:5:2 22. A : B = 2 : 9 B : A = 9 : 2 A:C=4:3 B : A : C = 9 × 4 : 2 × 4 : 2 × 3 = 36 : 8 : 6 = 18 : 4 : 3 or, A : B : C = 4 : 18 : 3 Note: We have written the ratios in such a way that the consequent of the first ratio and the antecedent of the second ratio are the same. Like: A : B & B : C --------- (1) or, B : A & A : C --------- (2) or, B : C & C : A --------- (3) Then we apply the rule: For (1) A: B : C =A× B : B × B : B × C For (2) B :A: C = B ×A:A×A:A× C 23. A = 5B A : B = 5 : 1 C=

1 1 (A + B) = (6B) = 3B 2 2

B:C=1:3 A:B:C= 5×1:1 ×1:1 ×3=5:1:3 24. Try it yourself (follow the method used in Q. 23). 25.

x 2y 3z 4w k(say) 2 3 4 5 x 2k; y

3k 4k 5k ;z ; w 2 3 4

3 4 5 x : y : z : w 2 : : : = 24 : 18 : 16 : 15 2 3 4 Now, 24 + 18 + 16 + 15 = 73 the four battalions have 2400, 1800, 1600 and 1500. 26. A : B = 2 : 3 B:C=4:5 C:D=6:7 A : B : C = 2 × 4 : 3 × 4 : 3 × 5 = 8 : 12 : 15 C : D = 6 : 7 = 15 : 17.5 A : B : C : D = 8 : 12 : 15 : 17.5 = 16 : 24 : 30 : 35 Since 16 + 24 + 30 + 35 = 105 A’s share =

21000 16 = 3200 105

179 B’s share = 200 24 = 4800 C’s share = 200 × 30 = 6000 D’s share = 200 × 35 = 7000. 27. The ratio of values of a rupee, 50P and 25 P coins = 4 × 100 : 5 × 50 : 6 × 25 = 8 : 5 : 3 Since 8 + 5 + 3 = 16 The value of Re 1 coins

32 8 16 16

The value of 50 P coins

32 5 10 16

32 3 6 16 Therefore, the number of Re 1 coins = 16 × 1 = 16 The number of 50 P coins = 10 × 2 = 20 The number of 25 P coins = 6 × 4 = 24 28. The total sum after deduction = 3115 – (25 + 28 + 52) = 3010 Their diminished share is in the ratio 8 : 15 : 20 The value of 25 P coins

3010 8 = 560 43 B’s diminished share = 70 × 15 = 1050 C’s diminished share = 70 × 20 = 1400 A’s share = 560 + 25 = 585 B’s share = 1050 + 28 = 1078 C’s share = 1400 + 52 = 1452 29. 3 : 4 4:5

A’s diminished share

The number =

4 4 35 1 1 5 4 1

30. 19 : 23 3:4

The number =

19 4 23 3 7 7 43 1

31. 9 : 8 14 : 15 We know that the total bill = wage per person × no. of total employees. Therefore, the ratio of change in bill = 9 × 14 : 8 × 15 = 126 : 120 = 21 : 20 The ratio shows that there is a decrease in the bill. Note: For a detailed method let the no. of employees in two cases = 9x & 8x. Wages in two cases be 14y & 15y Initial wage = 9x × 14y = 126xy Changed wage = 8x × 15y = 120 xy This shows the decrease in bill and ratio is 126xy : 120xy = 21 : 20.

Quicker Maths

180 32. 2m = 5b 2w = 3b Combining the two relations: (Follow the rule) 2m = 5b 3b = 2w 2 × 3m = 5 × 3b = 5 × 2w 6m = 15b = 10w Now, to find the ratio of wages of a man, a woman and a boy, let 6m = 15b = 10 w = k (say)

k k k m ; b ; w 6 15 10 1 1 1 : : 5: 3: 2 6 10 15 The ratio of wages of 6 men, 12 women and 17 boys = 6 × 5 : 12 × 3 : 17 × 2 = 30 : 36 : 34 50 34 = 17 17 boys get 30 36 34 1 boy gets 1. Note: For a quicker approach, if 6m = 10w = 15b then m : w : b = 10 × 15 : 6 × 15 : 6 × 10 = 5 : 3 : 2 6m : 12w : 17b = 6 × 5 : 12 × 3 : 17 × 2 = 30 : 36 : 34 m : w : b

50 34 = 17 30 36 34 b= 1 17b

32 b a 2 ab = 64 34. Let Lalita, Amita and Neeta have 20x, 73x and 83x gold coins respectively at present. Initially, Lalita had (20x + 25) gold coins. Amita had (73x + 15) gold coins. And Neeta had (83x + 30) gold coins. Then, 20x + 25 + 73x + 15 + 83x + 30 = 950 176x + 70 = 950 176x = 880 33. 4;

880 5 176 Amita had 73x + 15 = 73 × 5 + 15 = 365 + 15 = 380 gold coins. Alternative Method: Total no. of coins left with them = 950 – (25 + 15 + 30) = 880 73 So, after donation Amita had 880 20 73 83 x=

=

880 73 = 5 × 73 = 365 coins. 176

So, initially she had (365 + 15 =)380 coins. 35. Amount reinvested in Equity Funds = 94500 Amount reinvested in Debt + Equity Funds

13 = 175500 7 Amount invested earlier in Debt + Equity Funds 175500 = = 135000 1.3 Original amount invested in equity funds = 94500 ×

5 × 135000 = 75000 9 36. Ratio of amount collected = (14 × 2) : (12 × 3) : (10 × 5) = 28 : 36 : 50 = 14 : 18 : 25 Sum of ratios = 14 + 18 + 25 = 57 =

14 × 4560 = 1120 57

Amount collected on day one = 37. Let the amount got by B be x. Expense on dinner =

3x 10

Remaining amount = x –

3x 10 x 3x 7x = 10 10 10

Expense on book = 460

7x 2 = 460 10 7

x 460 5 x = 5 × 460 = 2300 Initial amount with A = (2300 × 4) = 9200 Quicker Approach: Instead of 'x' we suppose B got 100. After spending 30 on dinner he is left with 70. Out of 70 he spent 50 on savings and 20 on books. Now, since B got 100, A had 400 in the beginning. Now compare these assumed values with actual values. We are given that B spent 460 on books. So, 20 460

460 400 20 400 = 9200 Note: Solving by this approach will save your writing work, because most of the calculations are done mentally. 38. Let the monthly salary of Dex’s father be x. Then Dex’s monthly salary =

x 4

Ratio and Proportion Dex’s sister’s monthly salary = Now,

181 2x 5

2x 1 = 12800 5 4

x = 12800 10 x = 128000 or,

128000 = 32000 4 Again, the ratio of savings to expenses of Dex is 3 : 5. Dex’s monthly salary =

32000 3 = 12000 8 Quicker Approach: Suppose, Father's monthly salary = 100 Dex's monthly salary = 25 Sister's monthly salary = 40 Sister's study loan payment = 10 Now given, 10 12800 Dex's salary = 25 1280 × 25 Savings of Dex =

3 Dex's savings = 1280 × 25 8 = 12000 39. Let the number of students who appear be (25x + 4x) = 29x Now, Appeared Passed Failed 29x 25x 4x Now, 29x + 5 (29x + 5) 4x – 2 – (4x – 2)

Then,

( 29x 5) – ( 4x 2) 22 4x 2 3

25x 7 22 4x 2 3 or, 88x – 44 = 75x + 21 or, 13x = 21 + 44 = 65 x=5 Hence total number of appeared students = 29x = 29 × 5 = 145 Quicker Approach: In such questions, we do not need to solve in first attempt. First check the given choices and select the choice where no. of students is perfectly divisible by (25 + 4 =) 29. In this case, only option (1) 145 is such a number. So it is our answer. 40. Let the total sum be x. or,

Then, A gets

x + 40 2

3x – 120 8 C gets 200. B gets

Now,

x 40 3x 120 200 x 2 8

4x 320 3x 960 1600 x 8 or, 8x – 7x = 1920 – 960 or, x = 960 or,

182

Quicker Maths

Chapter 19

Partnership A partnership is an association of two or more persons who put their money together in order to carry on a certain business. It is of two kinds: (i) Simple (ii) Compound Simple partnership: If the capital of the partners are invested for the same period, the partnership is called simple. Compound Partnership: If the capitals of the partners are invested for different lengths of time, the partnership is called compound. In a group of n persons invested different amount for different period then their profit ratio is: At1 : Bt2 : Ct3 : Dt4 .......... : Xtn [Here first person invested amount A for t1 period, second persons invested amount B for t2 period, and so on.] Ex. 1: Three partners A, B and C invest 1600, 1800 and 2300 respectively in business. How should they divide a profit of 1938? Soln: The profit should be divided in the ratios of the capitals, i.e. in the ratio 16 : 18 : 23. Now, 16 + 18 + 23 = 57 16 A’s share = of 1938 = 544 57

18 of 1938 = 612 57 23 C’s share = of 1938 = 782 57 A, B and C enter into partnership. A advances 1200 for 4 months, B 1400 for 8 months, and C 1000 for 10 months. They gain 585 altogether. Find the share of each. 1200 in 4 months earns as much profit as 1200 × 4 or 4800 in 1 month. 1400 in 8 months earns as much profit as 1400 × 8 or 11200 in 1 month. 1000 in 10 months earns as much profit as 1000 × 10 or 10,000 in 1 month. Therefore, the profit should be divided in the ratios of 4800, 11,200 and 10,000 i.e. in the ratios of 12, 28 and 25.

Now, 12 + 28 + 25 = 65

Soln:

12 585 = 108 65

B’s share =

28 585 = 252 65

25 585 = 225 65 In compound partnership, the ratio of profits is directly proportional to both money and time, so they are multiplied together to get the corresponding shares in the ratio of profits. A starts a business with 2,000. B joins him after 3 months with 4,000. C puts a sum of 10,000 in the business for 2 months only. At the end of the year, the business gave a profit of 5600. How should the profit be divided among them? Ratio of their profits (A’s : B’s : C’s) = 2 × 12 : 4 × 9 : 10 × 2 = 6 : 9 : 5 Now, 6 + 9 + 5 = 20 C’s share =

Note:

Ex. 3:

Soln:

Then A’s share = B’s share =

B’s share =

Ex. 2:

A’s share =

5600 6 = 1680 20

5600 9 = 2520 20

5600 5 = 1400 20 A and B enter into a partnership for a year. A contributes 1500 and B 2000. After 4 months they admit C, who contributes 2250. If B withdraws his contribution after 9 months, how would they share a profit of 900 at the end of the year? A’s share : B’s share : C’s share = 1500 × 12 : 2000 × 9 : 2250 × 8 = 15 × 12 : 20 × 9 : 22.5 × 8 = 180 : 180 : 180 = 1 : 1 : 1 C’s share =

Ex. 4:

Soln:

Therefore, each of them gets

900 = 300. 3

Quicker Maths

184 Ex. 5:

Soln:

A, B and C enter into partnership. A advances onefourth of the capital for one-fourth of the time. B contributes one-fifth of the capital for half of the time. C contributes the remaining capital for the whole time. How should they divide a profit of 1140? A’s share : B’s share : C’s share 1 1 1 1 1 1 1 1 11 : : 1 – 1 : : 4 4 5 2 4 5 16 10 20 Multiplying each fraction by LCM of 16, 10 and 20, i.e., 80. We have 5 : 8 : 44

Soln:

450 12 1 300 x 2 or, 300 × 2x = 450 × 12

=

1140 A’s share = 57 5 = 100 B’s share =

Ex. 8:

Soln:

Then,

C’s share = Ex. 6:

Soln:

50 45 8 : 45 6 6 : 70 6 2 2 = 400 : 405 : 420 = 80 : 81 : 84 Therefore, the profit will be divided in the ratio of 80 : 81 : 84. Now, you must have understood both simple partnership and compound partnership. The formula for compound partnership can also be written as = 50 4

A 's Capital A 's Time in partnership B's Capital B's Time in partnership

Ex. 7:

A 's Pr ofit B's Pr ofit

The above relationship should be remembered because it is used very often in some types of question. A began a business with 450 and was joined afterwards by B with 300. When did B join if the profits at the end of the year were divided in the ratio 2 : 1?

450 12 x 2 300 = 9 months Therefore, B joined after (12 – 9 =) 3 months. A and B rent a pasture for 10 months. A puts in 100 cows for 8 months. How many cows can B put in for the remaining 2 months, if he pays half as much as A? Suppose B puts in x cows. The ratio of A’s and B’s 1 rents 1 : 2 : 1 2

1140 8 = 160 57

1140 44 = 880 57 A and B enter into a speculation. A puts in 50 and B puts in 45. At the end of 4 months, A withdraws half his capital and at the end of 6 months B withdraws half of his capital. C then enters with a capital of 70. At the end of 12 months, in what ratio will the profit be divided? A’s share : B’s share : C’s share

Suppose B joined the business for x months. Then using the above formula, we have

Ex 9:

Soln:

100 8 2 x2 1

100 8 1 or, x = 400 cows 22 A and B enter into a partnership with their capitals in the ratio 7 : 9. At the end of 8 months, A withdraws his capital. If they receive the profits in the ratio 8 : 9, find how long B’s capital was used. Suppose B’s capital was used for x months. Following the same rule, we have,

78 8 9x 9

789 7 89 Therefore, B’s capital was used for 7 months. Ex 10: A, B and C invested capitals in the ratio 2 : 3 : 5; the timing of their investments being in the ratio 4 : 5 : 6. In what ratio would their profit be distributed? Soln: We should know that if the three investments be in the ratio a : b : c and the duration for their investments be in the ratio x : y : z, then the profit would be distributed in the ratio ax : by : cz. Thus, following the same rule, the required ratio = 2 × 4 : 3 × 5 : 5 × 6 = 8 : 15 : 30 Ex 11: A, B and C invested capitals in the ratio 5 : 6 : 8. At the end of the business term, they received the profits in the ratio 5 : 3 : 12. Find the ratio of time for which they contributed their capitals? Soln: Following the same rule: If investment is in the ratio a : b : c and profit in the ratio p : q : r or, x

Partnership

185

p q r : : then the ratio of time = a b c Therefore, the required ratio 5 3 12 1 3 : : 1: : 2 : 1 : 3 5 6 8 2 2 Ex 12: A and B enter into a partnership with capitals in the ratio 5 : 6. At the end of 8 months, A withdraws his capital. If they receive profits in the ratio of 5 : 9, find how long B’s capital was used. Soln: This question is similar to the one in Ex. 9. You may solve it by the method used in Ex. 9. But, following the rule defined in Ex. 11, we see that the ratio of time of investment 5 9 3 : 1: 2 : 3 5 6 2 Now, we are given that A invested for 8 months.

Soln:

Detail Method: The difference counts only due to the 40% of the profit which was distributed according to their investments. Let the total profit be x. Then 40% of x is distributed in the ratio 125,000 : 85,000 = 25 : 17 Therefore, the share of the first partner

25 = 40% of x 25 17 25 40x 25 5x 40% of x 42 100 42 21 and the share of the second partner

17 17x = 40% of x 42 105

=

8 3 = 12 months 2 The validity of the above rules can be checked thus: Suppose A and B invested 5 and 6 respectively. A invested for 8 months and B invested for 12 months. Then, the ratio of their profit = 5 × 8 : 6 × 12 = 10 : 18 = 5 : 9 Which is the same as given in the question.

B invested for

Note:

2 of the profit 5 and B and C share the remaining profit equally. A’s income is increased by 220 when the profit rises from 8% to 10%. Find the capitals invested by A, B and C. For A’s share: (10% – 8%) 220

Ex. 13: A, B and C are partners. A receives

Soln:

220 100 = 11000 2 A’s capital = 11000 100%

For B’s & C’s share:

2 11000 5

3 11000 3 = 16500 5 2 B’s and C’s capitals are 8250 each. Ex. 14: Two partners invest 125,000 and 85,000 respectively in a business and agree that 60% of the profit should be divided equally between them and the remaining profit is to be treated as interest on capital. If one partner gets 300 more than the other, find the total profit made in the business.

Now, from the question, the difference in shares = or,

5x 17x 300 21 105

x(25 17) 300 105

300 105 = 3937.50 8 Direct Method: The ratio of profit = 125,000 : 85,000 = 25 : 17 x

100 25 17 total profit = 300 = 3937.50 40 25 17 Ex. 15: A and B entered into a partnership, investing 16,000 and 12,000 respectively. After 3 months, ‘A’ withdrew 5000, while B invested 5000 more. After 3 more months, C joins the business with a capital of 21,000. After a year, they obtained a profit of 26,400. By what value does the share of B exceed the share of C? Soln: The above question may be restated as: A invested 16,000 for 3 months and (16,000 – 5000) for 9 months. B invested 12,000 for 3 months and (12,000 + 5000) for 9 months. C invested 21,000 for 6 months. (These steps should be calculated mentally by you and not in writing.) Now, A’s share : B’s share : C’s share = (16 × 3 + 11 × 9) : (12 × 3 + 17 × 9) : (21 × 6) = 147 : 189 : 126 = 7 : 9 : 6

Quicker Maths

186 Therefore, B’s share exceeds that of C by

26400 26400 3 (9 6) = 3600 796 22 Note: During the calculation, we did not carry the zeroes of thousand because they are of no use in calculating the ratio. Ex. 16: A, B and C are partners in a business. A, whose 1 of 8 the profit. B, whose money has been used for 6 money has been used for 4 months, claims

1 rd of the profit. C had invested 3 1560 for 8 months. How much money did A and B contribute? Ratio of their shares in profit months, claims

Soln:

=

1 1 1 1 1 1 13 : : 1 : : 3 : 8 : 13 8 3 8 3 8 3 24

Now, for A and C A × 4 : 1560 × 8 = 3 : 13 3 1560 8 = 720 13 4 Now, for B and C B × 6 : 1560 × 8 = 8 : 13

A=

8 1560 8 = 1280 B 13 6 Ex. 17: Two partners invested 50,000 and 70,000 respectively in a business and agreed that 70% of the profits should be divided equally between them and the remaining profit in the ratio of investment. If one partner gets 90 more than the other, find the total profit made in the business. Soln: The difference comes only due to the 30% of the profit which was distributed in the ratio of their investments. Suppose the total profit is x. Then, 30% of x is distributed in the ratio 50,000 : 70,000 = 5 : 7 Therefore, the share of the first partner

5x x 5 x 30% of = 30% of 5 7 12 8 and the share of the second partner

7 7x 7x x = 30% of = 30% of 57 12 40

Now, the difference in shares = or,

7x x = 90 40 8

7x 5x 90 40

90 40 = 1800 2 Quicker Method (Direct Formula): Ratio of profits = 50,000 : 70,000 = 5 : 7

x

100 5 7 the total profit = 90 = 1800 30 7 5 Ex. 18: A, B and C invested capitals in the ratio 2 : 3 : 4. At the end of the business term, they received the profits in the ratio 3 : 6 : 10. Find the ratio of the periods for which they contributed their capitals. Soln: If the investments are in the ratio x : y : z and the profits in the ratio P : Q : R, then the ratio of periods

P Q R = x: y : z

3 6 10 : : 2 3 4 Multiply each term by the LCM of 2, 3 & 4, i.e., 12. Therefore, the required ratio =

3 6 10 12 : 12 : 12 = 18 : 24 : 30 = 3 : 4 : 5 2 3 4 Ex. 19: A and B invested in the ratio 3 : 2 in a business. If 5% of the total profit goes to charity and A’s share is 855, find the total profit. Soln: Suppose the total profit is 100. Then 5 goes to charity. Now, 95 is divided in the ratio 3 : 2. 95 3 = 57 3 2 But, we see that A’s actual share is 855.

A’s share =

100 Actual total profit = 855 = 1500 57 Direct Formula: In the above case:

100 3 2 Total profit = 855 100 – 5 3 100 5 855 = 1500 95 3

Partnership

Ex. 20: In a partnership, A invested for

187 1 th of the capital 6

Soln:

1 1 1 1 C invested 1 1 part of the 6 3 2 2 capital Now, ratio of profit = A : B : C

1 1 th of the time, B invested rd of the capital 6 3

1 rd of the time, and C invested the rest of 3 the capital for the whole period. At the end of the period, they earned a profit of 4600. Find the share of B.

=

for

1 1 1 1 1 1 1 1 : : 1 : : 1: 4 :18 6 6 3 3 2 36 9 2

4 4 B's share 4600 4600 1 4 18 23 = 800

EXERCISE 1. How should a profit of 450 be divided between two partners, one of whom has contributed 1200 for 5 months and the other 750 for 4 months? 2. There are three partners A, B and C in a business. A puts in 2000 for 5 months, B 1200 for 6 months and C 2500 for 3 months; and the profit is 508.82. How ought it to be divided? 3. A and B enter into a partnership for a year. A contributes 1500 and B 2000. After 4 months, they admit C, who contributes 2250. If B withdraws his contribution after 9 months, how would they share a profit of 900 at the end of the year? 4. A, B and C enter into a partnership. A advances onefourth of the capital for one-fourth of the time; B advances one-fifth of the capital for half of the time; and C, the remainder of the capital for the whole time. How should they divide a profit of 3420? 5. Three partners altogether invested 114,000 in a business. At the end of the year, one got 337.50, the second 1125.00 and the third 675 as profit. How much amount did each invest? What is the percentage of profit? 6. A and B enter into a speculation; A puts in 50 and B puts in 45. At the end of 4 months, A withdraws half 1 his capital and at the end of 5 months, B withdraws 2 of his; C, then, enters with a capital of 70; at the end of 12 months, the profits of the concern are 254; how ought it to be divided? 7. A and B enter into a partnership with capitals as 5 : 6; and at the end of 8 months, A withdraws. If they receive profits in the ratio of 5 : 9, find how long B’s capital was used. 8. A and B rent a pasture for 10 months; and A puts in 90 oxen for 7 months. How many oxen can B put in for the remaining 3 months, if he pays half as much as A?

9. Three men A, B and C start a business together. They invest 30000, 24000 and 42000 respectively in the beginning. After 4 months, B took out 6000 and C took out 10000. They get a profit of 11960 at the end of the year. What is the B’s share in the profit? 10. A starts a business with an initial investment of 18000. After 4 months, B enters into the partnership with an investment of 24000. Again after two months C enters with an investment of 30000. If C receives 1845 in the profit at the end of the year, what is the total annual profit? 11. A started a business with an investment of 14,000.

6 of the amount that A 7 in vested and A withdraws 4000. After 2 more months, C joins with 8000 and A again withdraws 2000. After an year, if C received 2,656 as his share then what was the total profit? 12. ‘A’ began a small business by investing a certain amount of money. After four months from the start of the business, ‘B’ joins the business with an amount which is 6,000 less than A’s initial investment. ‘C’ joins the business after seven months from the start of the business with an amount which is 2,000 less than A’s initial investment. At the end of the year total investment reported was 1,42,000. What will be A’s share in the profit if B received 8,000 as profit share? 13. A starts a business by investing 28,000. After 2 months, B joins with 20,000 and after another two months C joins with 18,000. At the end of 10 months from the start of the business, if B withdraws 2,000 and C withdraws 2,000, in what ratio should the profit be distributed among A, B and C at the end of the year? After 2 months B joins in with

Quicker Maths

188 14. A and B started a business by investing 18,000 and 24,000 respectively. At the end of the 4th month from the start of the business, C joins with 15,000. At the end of the 8th month B quits at which time C invests 3000 more. At the end of the 10th month B rejoins with the same investment. If the profit at the end of the year is 12,005, what is B’s share of profit? 15. A, B and C started a business by investing 20000, 28000 and 36000 respectively. After 6 months, A and B withdrew an amount of 8000 each and C invested an additional amount of 8000. All of them invested for equal periods of time. If at the end of the year, C got 12550 as his share of profit, what was the total profit earned? 16. A started a business. After 4 months from the start of the business, B and C joined him. The ratio of the investments of A, B and C was 4 : 6 : 5. If A’s share in annual profit was 250 more than that of C, what was the total annual profit earned? 17. A starts a business with a capital of 1500. B joins the business 6 months after the start of the business and C joins the business 8 months after the start of the business. At the end of the year their respective shares in the profit was in ratio of 5 : 3 : 3. What is the sum of amount put in the business by B and C together?

19.

20.

21.

22.

23.

as much as A and C together. Total profit at the end of the year was 5,200. What was C’s share in the profit? A and B started a business with an investment of 2,800 and 5,400 respectively. After 4 months, C joined with 4,800. If the difference between C’s share and A’s share in the annual profit was 400, what was the total annual profit? A and B are partners in a business. They invest in the ratio of 5 : 6; at the end of 8 months A withdraws. If they receive profits in the ratio of 5 : 9, then find how long B’s investment was used. A, B and C started a business with investments of 1600, 2100 and 1500 respectively. After 8 months from the start of the business, B and C invested additional amounts in the ratio of 3 : 5 respectively. If the ratio of total annual profit to C’s share in the annual profit was 3 : 1 then what was the additional amount invested by B after 8 months? A, B and C started a business with their investment in the ratio of 1 : 3 : 5. After 4 months, A invested the same amount as before but B as well as C withdrew half of their investments. Find the ratio of their profits at the end of the year. A, B and C started a business and invested in the ratio

1 of the amount 12 of what B and C had invested. If the annual income was 9200 then what was the share of B? of 3 : 4 : 5. After 4 months A withdrew

1 18. A, B and C start a small business. A contributes of 5 the total capital invested in the business. B contributes

Answers 1. The ratio of profit = 12 × 5 : 7.5 × 4 = 60 : 30 = 2 : 1

450 2 = 300 3

1st partner gets

450 1 = 150 3 2. The ratio of profits = 20 × 5 : 12 × 6 : 25 × 3 = 100 : 72 : 75 Find their shares. 3. A’s share : B’s share : C’s share = 15 × 12 : 20 × 9 : 22.5 × 8 = 180 : 180 : 180 = 1 : 1 : 1 Find their shares. 4. A’s share : B’s share : C’s share 2nd partner gets

1 1 1 1 : 4 4 5 2

1 1 1 1 11 : 1 1 : : 4 5 16 10 20

Multiply each by the LCM of the denominators i.e. 80. = 5 : 8 : 44 Find the shares. 5. The ratio of investments = Ratio of profits = 337.5 : 1125 : 675 = 3375 : 11250 : 6750 Dividing each by 1125, we have the ratio = 3 : 10 : 6 Find the shares. The reqd. percentage of profit =

337.5 1125 675 100% 11400

2137.5 % 1.875% 1140 6. A’s share : B’s share : C’s share = 50 × 4 + 25 × 8 : 45 × 5 + 22.5 × 7 : 70 × 7 = 400 : 382.5 : 490 Find the shares.

Partnership

189

7. The ratio of capitals = 5 : 6 Let the ratio of time = 8 : x Then 5 × 8 : 6x = 5 : 9

or, 12x + 8x – 48000 + 5x – 10000 = 142000 or, 25x = 142000 + 58000

40 9 12 months. 65 By Direct Formula: Capitals: 5 : 6; Profit: 5 : 9 x

5 9 8 5 9 B's time A 's time 6 5 65 12 months 8.

A 's share 90 7 2 B's share x 3 1

90 7 105 oxen 3 2 9. Ratio of capital = 30000 × 12 : (24000 × 4 + 18000 × 8) : (42000 × 4 + 32000 × 8) = 36000 : (96000 + 144000) : (168000 + 256000) = 360000 : 240000 : 424000 = 360 : 240 : 424 = 45 : 30 : 53 Sum of terms of ratio = 45 + 30 + 53 = 128 x

30 11960 = 2803.125 2803 128 10. Ratio of equivalent capitals for 1 month = (18000 × 12) : (24000 × 8) : (30000 × 6) = (18 × 12) : (24 × 8) : (30 × 6) = 18 : 16 : 15 Sum of terms of ratio = 18 + 16 + 15 = 49 Now, B’share =

1845 49 = 6027 15 11. Ratio of equivalent capitals of A, B and C for 1 month Total profit =

= (14000 × 2 + 10000 × 2 + 8000 × 8) : (

6 × 14000 × 7

10) : (8000 × 8) = (28000 + 20000 + 64000) : 120000 : 64000 = 112000 : 120000 : 64000 = 112 : 120 : 64 = 14 : 15 : 8 Sum of terms of ratio = 14 + 15 + 8 = 37

2656 37 = 12284 8 12. Suppose A starts business by investing an amount of x. Then B’s investment = (x – 6000) C’s investment = (x – 2000) Now, 12x + (x – 6000) × 8 + 5(x – 2000) = 142000 Total profit =

200000 = 8000 25 Ratio of profits = Ratio of investments = 12 × 8000 : 16000 : 30000 = 96 : 16 : 30 = 48 : 8 : 15 Now, 8 8000 48 48000 13. Ratio of profit of A to B to C = 28000 × 12 : 20000 × 8 + 18000 × 2 : 18000 × 6 + 16000 ×2 = 28 × 12 : 160 + 36 : 108 + 32 = 336 : 196 : 140 = 12 : 7 : 5 14. Ratio of profit of A to B to C = 18000 × 12 : 24000 × 8 + 24000 × 2 : 4 × 15000 + 4 × 18000 = 216000 : 192000 + 48000 : 60000 + 72000 = 216000 : 240000 : 132000 = 216 : 240 : 132 = 18 : 20 : 11 Sum of ratio terms = 18 + 20 + 11 = 49 x=

40 5 6x 9

20 12005 = 4900 49 15. Ratio of profits of A : B : C = 20000 × 6 + (20000 – 8000) × 6: 28000 × 6 + (28000 – 8000) × 6 : 36000 × 6 + (36000 + 8000) × 6 = (120 + 72) : (168 + 120) : (216 + 264) = 192 : 288 : 480 = 2 : 3 : 5 Sum of ratio terms = 2 + 3 + 5 = 10 Share of profit of B =

12550 10 = 25100 5 16. Ratio of profit of share A : B : C = 4 × 12 : 6 × 8 : 5 × 8 = 48 : 48 : 40 = 6 : 6 : 5 Total profit =

250 Total profit = (6 5) (6 6 5) = 250 × 17 = 4250 17. Ratio of profits = 1500 × 12 : 6B : 4C : : 5 : 3 : 3

Capital of B =

1500 12 3 = 1800 6 5

1500 12 3 = 2700 4 5 Amount of (B + C) = 2700 + 1500 = 4500 18. Let the total investment in business be x. Capital of C =

1 Then A invested x = 5 Remaining amount = x –

x 5

x = 5

4x 5

Quicker Maths

190 Now, suppose B invested y.

20. Ratio of profit = 5 : 9

Share of A' s investment

Pr ofit of A

4x C invested 5 y

Now, Share of B' s investment Pr ofit of B

x 4x Then y = 5 5 y

58 5 or, 6 month( x ) 9

or, 2y = y=

5x 5

5 8 9 72 = 12 5 6 6 x = 12 months Thus, B invests the amount for 12 months 21. Ratio of profit A : B : C = 1600 × 12 : 2100 × 8 + (2100 + 3x) × 4 : 1500 × 8 + (1500 + 5x) × 4 = 19200 : 16800 + 8400 + 12x : 12000 + 6000 + 20x

or, x =

x 2

4x x 8x 5x 3x 5 2 10 10 Then, ratio of amounts = A : B : C C invested

=

x : x : 3x =2:5:3 5 2 10

Hence, C’s share in profit = 5200×

3 = 1560 10

Quicker approach: B's contribution is equal to the sum of contribution by A and C together and also the total contribution is 100%. B's contribution is 50% and A's + C's contribution is 50%. Also given that A's contribution is 1 / 5 or 20%. C's contribution is 30%. Therefore, the ratio of contribution of A, B and C is 2 : 5:3

3 Hence C's share in profit = 5200 10 = 1560 19. Ratio of profit of A : B : C = 2800 × 12 : 5400 × 12 : 4800 × 8 = 14 : 27 : 16 Now, let the profit of A, B and C be 14x, 27x and 16x respectively. Then, 16x – 14x = 400 or, 2x = 400 x = 200 Total profit = 14x + 27x + 16x = 57x= 57 × 200 = 11400

62400 32x 3 = 18000 20x 1 or, 60x + 54000 = 32x + 62400 or, 28x = 8400 x = 300 B’s investment after 8 months = 3 × 300 = 900 22. Ratio of profit Now,

= x × 4 + 2x × 8 : 3x × 4 +

3x 5x × 8 : 5x 4 8 2 2

= 20x : 24x : 40x = 5 : 6 : 10 23. Initial investment A : B : C 3x : 4x : 5x Ratio of profit = 3x × 4 + 3x – 9 x 8 : 4x×12 : 5x × 12 12

9x 8 : 48x : 60x 4 12x + 18x : 48x : 60x 30x : 48x : 60x 5x : 8x : 10x 23x 9200 12x +

8x =

9200 8 = 3200 23

Chapter 20

Percentage The term per cent means ‘for every hundred’. It can best be defined as: “A fraction whose denominator is 100 is called a percentage, and the numerator of the fraction is called the rate per cent.” The following examples illustrate the percents and their fractional values: 1) A student gets 60 per cent marks in Arithmetic means that he obtained 60 marks out of every hundred of full marks. That is, if the full marks be 500, he gets 60 + 60 + 60 + 60 + 60 = 300 marks in mathematics. The above five 60s are one 60 for every hundred. The total marks obtained by the student can be calculated in other ways, like, 0 500 300 100 The above calculations can be made easier by reducing the fractional value to its prime. As, in the above case;

60% of 500 =

60% =

60 3 100 5

3) A tradesman makes a profit of 15 per cent means that he makes a profit of 15 when he invests 100. But what does he gain when he invests 900? Which of the applications, mentioned above, is easier to deal with? Don’t you think that the fraction containing hundred is more helpful in this case! (Why?) Because two complete hundreds cancel out easily, giving quick 15 900 15 9 135 (I think you need 100 not write anything to calculate such problems.) We see that our key operator in this chapter is the prime fraction of per cent value. So, we should collect some of the important (most-used) prime fractions:

result; like:

1 1 3 8 32 5%

1 20

1 1 8 % 3 12

3 , our calculation 5 becomes easier. In that case, the total marks obtained

12%

3 by the student = 500 300 5 2) A man invests 5% of his income into shares. It means: i) he invests 5 out of every 100 of his income into shares.

15%

If we remember that 60% =

or, ii) he invests

5 of his income into shares. 100

1 or, iii) he invests th of his income into shares. 20 Now, if his income is 1050, how does he invest in shares? 1050 Your quick answer should be = 52.5 20 We suggest you not to move with the fraction contain 100, if possible.

3 25

1 2 13 % 3 15 3 20

2 1 16 % 3 6 25%

1 4

1 3 37 % 2 8 60%

3 5

1 1 6 % 4 16 8%

2 25

10%

1 10

1 1 12 % 2 8 2 1 14 % 7 7 16%

4 25

20%

1 5

1 1 33 % 3 3 40%

2 5

1 5 62 % 2 8

Quicker Maths

192 2 2 66 % 3 3

75%

3 4

Soln:

1 7 87 % 2 8 Ex. 1: Find 8 per cent of 625.

Soln:

8% of 625 =

2 625 = 50 25

Ex. 2: What fraction is 12

Soln:

58 2 6 % 6 3 Second Method: % of sugar in the original solution = 5% of 8 litres = 0.4 litres After evaporation of 2 lt of water, the quantity of the remaining solution = 8 – 2 = 6 litres the required percentage of sugar x

1 per cent? 2

1 12 1 25 1 12 % 2 2 100 200 8

3 Ex. 3: What percentage is equivalent to ? 8

Soln:

3 75 1 100 37 % 8 2 2

Ex. 4: What per cent is equivalent to

7 ? or express 11

0.4 2 100% 6 % 6 3 Ex. 9: One type of liquid contains 25% of milk, the other contains 30% of milk. A cane is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of milk in the new mixture. Soln: The reqd. percentage of milk in the new mixture

=

7 as rate per cent. 11 7 700 7 100 63 % 11 11 11 Ex. 5: The population of a town has increased from 60,000 to 65,000. Find the increase per cent. Soln: Increase in population = 65, 000 – 60,000 = 5000 Percentage increase

Soln:

5000 25 1 = 60, 000 100 3 8 3 % Ex. 6: Ram’s salary is increased from 630 to 700. Find the increase per cent. Soln: Increase in salary = 700 – 630 = 70 70 1 100 11 % 630 9 Ex. 7: In an election of two candidates, the candidate who gets 41% is rejected by a majority of 2412 votes. Find the total no. of votes polled. Soln: (59% – 41% =) 18% 2412

Percentage increase =

2412 100 13400 18 Ex. 8: If 2 litres of water is evaporated on boiling from 8 litres of sugar solution containing 5% sugar, 100%

find the percentage of sugar in the remaining solution. As sugar has not been evaporated from the solution, the quantity of sugar in the original 8 litres of solution = the quantity of sugar in the remaining 8 – 2 = 6 litres of solution i.e., 5% of 8 = x% of 6

Quantity of milk in the new mixture 100 Quantity of the new mixture

6 parts of 25% milk 4 parts of 30% milk 100 (6 parts 4 parts) of the liquid

25 30 4 100 100 100 (15 12) 27 10 This equation can be solved by the method of Alligation. 6

Note:

25 6

30

x

4

30 x 6 3 x 25 4 2 or, 60 – 2x = 3x – 75 or, 5x = 60 + 75 x = 27% Ex. 10: Due to fall in manpower, the production in a factory decreases by 25%. By what per cent should the working hour be increased to restore the original production? Soln: Decrease in production is only due to decrease in manpower. Hence, manpower is decreased by 25%.

Percentage

193

Now, suppose that to restore the same production, working hours are increased by x%. Production = Manpower × Working hours = M × W (say) Now, M × W = (M – 25% of M) × (W + x% of W) 75 100 x M W or, M × W = 100 100 or, 100 × 100 = 75 (100 + x) 400 100 x or, 3 100 1 x 33 % 3 3 Method II: To make the calculations easier, suppose Manpower = 100 units and Working hours = 100 units Suppose working hours increase by x%. Then, (100 – 25) (100 + x) = 100 × 100

or, 100 + x =

400 3

100 1 33 % 3 3 Direct Formula: Required % increase in working x

25 100 1 100 33 % 100 – 25 3 3 To find how much per cent one quantity is of another Ex.11: Express the fraction which 1.25 is of 10 as a percentage.

hours

Soln:

The fraction =

1.25 125 1 10 1000 8

1 1 100 12 1 1 8 Now, 2 12 % 8 100 100 2 Note: The above question is often put as “what rate per cent is 1.25 of 10?” Ex. 12: What rate per cent is 6P of 1?

3 100 6P 6 3 50 The fraction = 6% 1 100 50 100 Ex. 13: 12% of a certain sum of money is 43.5. Find the sum. Soln:

12 1 of a sum = 43 100 2

the sum

87 100 = 362.50 2 12

Theorem: If two values are respectively x% and y% more than a third value, then the first is the

100 x 100% of the second. 100 y Proof: Let the third value be 100. Then, the first is 100 + x% of 100 = 100 + x and the second is 100 + y% of 100 = 100 + y the first is

100 x 100% of the second. 100 y

Ex. 14: Two numbers are respectively 20% and 50% more than the third. What percentage is the first of the second? Soln: Following the above theorem, we have the required value 120 100 80% 150 Theorem: If A is x% of C and B is y% of C, then A is

x 100% of B. y Proof: Try this and for yourself. Ex. 15: Two numbers are respectively 20% and 25% of a third number. What percentage is the first of the second? Soln: Following the above theorem, we have the 20 100 80% 25 Note: The above relationships are very simple. When “What is the first of second” is asked, put the first as the numerator and the second as the denominator and vice versa. Ex. 16: Two numbers are respectively 30% and 40% less than a third number. What per cent is the second of the first? Soln: At first, you should find the formula yourself. If you can’t find it, go through the following remarks. (1) Since the two numbers are less than the third; and (2) we have to find the per cent of the second with respect to the first, our formula should be:

required value

100 – 40 60 5 100 100 85 % 100 – 30 70 7

Quicker Maths

194 Ex. 17: A positive number is divided by 5 instead of being multiplied by 5. What % is the result of the required correct value? Soln: Let the no. be 1, then the correct answer = 5 1 The incorrect answer that was obtained = 5 1 100% 4% The reqd. % 55 Ex. 18: A positive no. is by mistake multiplied by 5 instead of being divided by 5. By what per cent more or less than the correct answer is the result obtained? Soln:

Let the no. be 1, then the correct answer =

1 5

The incorrect answer that was obtained = 5 The result is more than the correct answer by 1 24 5 5 5

24 5 100% 2400% The reqd. 1 5

Percentage Expenditures and Saving 1 Ex. 19: A man loses 12 % of his money and, after 2 spending 70% of the remainder, he is left with 210. How much had he at first? Soln: The above question can be solved in many ways. We will discuss a few of them. I: Let the man be supposed to have x at first. 1 1 After losing 12 % or , he is left with 2 8 x 7x = 8 8 After spending 70% of the money, he is left with 30% of the remainder, i.e., 7x 3 210 8 10 210 10 8 x = 800 3 7 II: Suppose he had 100 at first. After losing 1 1 12 he would have 87 left. He spent 2 2 1 70% of 87 . 2 x

1 105 he would have 30% of 87 2 or 4 left. But he has 210 left. Thus, we have the following proportions: 105 : 210 : : 100 4 : the required money (By the rule of three)

the required money =

4 210 100 = 800 105

III: Quicker Method: It is a very short and fastcalculating method. The only thing is to remember the formula well.

210 100 100 His initial money (100 – 12.5)(100 – 70) 210 100 100 = 800 87.5 30 Note: As his “initial money” is definitely more than the “left money”, there should not be any confusion in putting the larger value (100) in the numerator and the smaller value (100 – 12.5) in the denominator. Ex. 20: 3.5% of income is taken as tax and 12.5% of the remaining is saved. This leaves 4,053 to spend. What is the income? Soln: Quicker Maths gives the solution as:

4053 100 100 Income (100 – 3.5)(100 – 12.5) = 4,800 Thus, we derive a general formula in the form of the following theorem: Theorem: x% of a quantity is taken by the first, y% of the remaining is taken by the second and z% of the remaining is taken by third person. Now, if A is left in the fund, then there was

A 100 100 100 (100 – x)(100 – y)(100 – z) in the beginning Proof: The initial amount must be more than ‘A’. So, by the rule of fraction, ‘A’ should be multiplied by the fractions which are more than one. Now, the problem is to find these fractions. Since in each step the amount is lessened, (100 – x), (100 – y) and (100 – z) should be in our dealing fractions apart from 100.

Percentage

195

Thus, the fractions (more than one) by which A

100 20 100 25 and 100 100

100 100 100 is multiplied are 100 – x , 100 y and 100 z 100 100 100 A 100 – x 100 y 100 z Theorem: x% of a quantity is added. Again, y% of the increased quantity is added. Again, z% of the increased quantity is added. Now, it becomes A, then the initial amount is given by

A 100 100 100 (100 x)(100 y)(100 z)

Note:

Population Formula Ex 23: If the original population of a town is P, and the annual increase is r%, what will be the population in n years? Soln: Population after one year becomes

Pr r P 1 100 100 That is, the population P at the beginning of the

Proof: The initial amount must be less than the final amount ‘A’. So, by the rule of fraction, A should be multiplied by less-than-one fractions. Since in each step the amount is increased, (100 + x), (100 + y) and (100 + z) should be our dealing values apart from 100. Thus, the fractions (less than one) by which A is multiplied are

P

r year is multiplied by 1 in the course of 100 the year. Now, the population at the beginning of the

100 100 100 , and 100 x 100 y 100 z

r second year is P 1 . 100

Therefore, the required initial amount

r the population in 2 years = P 1 100

100 100 100 A 100 x 100 y 100 z Some more examples based on the above theorems: Ex. 21: After deducting 10% from a certain sum, and then 20% from the remainder, there is 3600 left. Find the original sum. Soln: The original sum is naturally more than 3600. Therefore, it should be multiplied by

100 100 and (100 10) (100 20) 3600 100 100 = 5000 90 80 Ex. 22: A man had 4800 in his locker two years ago. In the first year, he deposited 20% of the amount in his locker. In the second year, he deposited 25% of the increased amount in his locker. Find the amount at present in his locker. Soln: The amount is certainly more than 4800. And each year, the new amount is added. So, the sum should be multiplied by

the required sum =

4800 120 125 100 100 = 7200 The above example is different from others. Mark it.

the required amount =

Therefore, the required initial amount

2

n

Note:

r the population in n years P 1 100 If the annual decrease be r%, then the population in n years n

r = P 1 100 Ex. 24: If the annual increase in the population of a town is 4% and the present number of people is 15,625, what will the population be in 3 years? Soln:

4 The required population = 15625 1 100

3

26 26 26 17576 25 25 25 Ex. 25: If the annual increase in the population of a town be 4% and the present population be 17576, what was it three years ago? 15625

Quicker Maths

196 3

Soln:

26 The population 3 years ago = Present 25 population the population 3 years ago 17576 25 25 25 15625 26 26 26 Ex.24 and Ex. 25 are the best examples to look at the game of fractions. In Ex.24, the required population was definitely more, so we had put the higher value (26) in the numerator and the lower value (25) in the denominator. In Ex. 25, the opposite of this can be seen.

Note:

When the Rate of Growth is Different for Different Years Theorem: The population of a town is P. It increases by x% during the first year, increases by y% during the second year and again increases by z% during the third year. The population after 3 years will be

P (100 x)(100 y)(100 z) 100 100 100 Proof: By the rule of fraction, P should be multiplied by more-than-one fractions. Thus, the population after 3 years is

Note:

100 x 100 y 100 z P 100 100 100 Mark that the multiplying fraction is 100 x 100 and not Why ? 100 100 – x Because the population is increased by x%, so we should deal with (100 + x) and 100. And since after one year, the population is more than

100 x . 100 We proceed similarly for the succeeding years. Ex. 26: The population of a town is 8000. It increases by 10% during the first year and by 20% during the second year. What is the population after two years? Soln: The required population

P, so P should be multiplied by

8000 110 120 10,560 = 100 100

When Population Increases for One Year and then Decreases for the Next Year. Theorem: In the above theorem, when the population decreases by y% during the second year, while for the first and third years, it follows the same, the population after 3 years will be

P(100 x)(100 y)(100 z) 100 100 100 Proof: Try to prove it by yourself. Ex. 27: The population of a town is 10,000. It increases by 10% during the first year. During the second year, it decreases by 20% and increased by 30% during the third year. What is the population after 3 years? Soln: The required population 10000 110 80 130 11440 100 100 100 Ex. 28: During one year, the population of a locality increases by 5% but during the next year, it decreases by 5%. If the population at the end of the second year was 7980, find the population at the beginning of the first year. Soln: The required population

100 100 7980 100 5 100 5 7980 100 100 8000 95 105 Note: In the above example, the population after two years is given and the population in the beginning of the first year is asked. That is why, the fractional values are inversed. Mark that point. The same thing happens to the next example. Ex. 29: The population of a town increases at the rate of 10% during one year and it decreases at the rate of 10% during the second year. If it has 29,700 inhabitants at present, find the number of inhabitants two year ago. Soln: The required population

29700 100 100 (100 10) (100 10)

29700 100 100 30, 000 90 110 Ex. 30: The population of a town is 8000. If the males increase by 6% and the females by 10%, the population will be 8600. Find the number of females in the town.

Percentage Soln:

197

Let the population of females be x. Then 110% of x + 106% of (8000 – x) = 8600

or, 120x = 120 × 100 – 100 × 100

110x 106(8000 x) 8600 or, 100 100 or, x(110 – 106) = 8600 × 100 – 8000 × 106

or,

8600 100 8000 106 12,000 3,000 110 106 4 If we ignore the intermediate steps, we can get the population of females and males directly thus: The population of females x

Note:

8600 100 8000(100 6) 3,000 (10 6) The population of males

120 of the former price 100 The householder must now consume

II: The raised price =

120 i.e. the reciprocal of 100 original amount 100 120

of

the

100 the reduction in consumption 1 of 120 1 th of the original 6

2 consumption = 16 % 3 III: Quicker Method: Theorem: If the price of a commodity increases by r%, then the reduction in consumption so as not to

20, 000 5, 000 4 By Method of Alligation

=

600 15 100 7.5% 8000 2

Now,

r 100 % increase the expenditure, is 100 r Proof: The formula can be written in the form 100 r . If you watch carefully, you can 100 r

Male: Female = 2.5 : 1.5 = 5 : 3 the population of females =

100(120 – 100) 2 16 % 120 3

the original consumption =

8600 100 8000(100 10) (6 10)

Average % increase

x

8000 3 3000 53

Reduction in Consumption Ex. 31: If the price of a commodity be raised by 20%, find by how much per cent must a householder reduce his consumption of that commodity so as not to increase his expenditure. Soln: I: Let the price and consumption each be 100 units. Then, his earlier expenditure was = (100 × 100) Now, the new price = 120 units To maintain the expenditure, suppose he reduces his consumption by x%, then his total expenditure = [120 × (100 – x)] From the question, we have, 100 × 100 = 120 (100 – x)

100 see the fractional value is less than 1, 100 r i.e., the numerator is less than the denominator. Why? Because our required value, in this case, is less than the supplied value (20%). Thus, in this case also we applied the rule of fraction. 20 50 2 100 16 % 100 20 3 3 Ex. 32: If the price of sugar falls down by 10%, by how much per cent must a householder increase its consumption, so as not to decrease expenditure in this item? Soln: This question is similar to the previous example. It can also be solved by all the three methods given above. But we will discuss only method III. Try to solve it by the other two methods also on your own.

Thus, answer =

Quicker Maths

198 Quicker Method Theorem: If the price of a commodity decreases by r%, then increase in consumption, so as not to decrease expenditure on this item, is

r (100 r ) 100 % Proof: If we write the formula in the form r

100 , 100 – r

100 we see that the fractional value is more 100 r than 1. Why? (Try to define it yourself.) So, in this case,

Ex. 33: If A’s salary is 25% more than that of B, then how much per cent is B’s salary less than that of A? Soln: I: Suppose B’s salary is 100 per month. Then A’s salary is 125 per month. We see that B’s salary is 25 less than that of A, when A’s salary is 125. Thus, when A’s salary is 100, B’s salary is 25 100 = 20 less than that of A i.e., B’ss 125 salary is 20% less than that of A. II: Quicker Method: If A’s income is r% more than B’s income, then B’s income is less

r 100 % . than A’s income by 100 r Thus, in this case, answer

10 1 Answer (100 – 10) 100 11 9 %

Percentage Relationship If first value is r% more than the second value, then

r

the second is 100 % less than the first 100 r value. Proof: By the rule of fraction, r should be multiplied by a fraction which is less than one. And that 100 . 100 r By general mathematical calculation Let the second value be 100. Then first value is (100 + r). Now, we see that when the first is (100 + r), the second is more by r. Therefore, when the first is 100 the second is

fraction should be

r 100 more than the first. 100 r

r The second is 100 + r ×100 % than the first. Theorem: If the first value is r% less than the second r 100 % 100 – r

value then, the second value is more than the first value. Proof: Try it yourself.

25 100% 20% 100 25 Note: Do you get the similarity between this formula and the formula given in Ex. 26? Ex. 34: If A’s salary is 30% less than that of B, then how much per cent is B’s salary more than that of A? Soln: Quicker Method: If A’s salary is r% less than B’s, then B’s salary

=

r 100 % . is more than A’s salary by 100 r Thus, in this case, answer 30 6 100 42 % 100 – 30 7 Note: Do you get the similarity between Ex. 32 and Ex. 34? Ex. 35: A number is 50% more than the other. Then how much per cent is the second number less than the first? Soln. We can apply the above discussed formula in this case also. Then the second number is

=

1 50 100 % 33 % less than the first. 3 100 50 Ex. 36: A number is 20% less than the other; then by how much per cent is the second more than the first? Soln: Apply the formula given in Ex. 32 or Ex. 34. One value as a percentage of another Ex. 37: Express 20 as a percentage of 80. Soln: If one is 80, the other is 20.

Percentage if one is 100, the other is

199 20 100 25 80

20 is 25% of 80. Thus, without going for details, we may say that x “ x as a percentage of y y 100% 250 100 500% 50 Ex. 39: Express 160 as a percentage of 120.

=

A(100 x)(100 x) [A(100)2 x 2 ] 100 100 (100) 2

And % change

Soln: Answer =

Ax 2 100 x 2 % (100) 2 A 100

Thus, we see that there is always a decrease

160 400 1 100 133 % 120 3 3 Ex. 40: Express 20 as a per thousand fraction of 200.

Soln: Answer =

20 1000 100 per thousand. Answer = 200

First increase and then decrease Ex. 41: The salary of a worker is first increased by 10% and thereafter it was reduced by 10%. What was the change in his salary? Soln: Let the salary of the worker be 100. After increase, it becomes 100 + 10% of 100 = 110 After decrease, it becomes 110 – 10% of 110 = 99 the % reduction = 100 – 99 = 1 % By Quicker Method: Theorem: If the value of a number is first increased by x% and later decreased by x%, the net change is always a decrease which is equal to x% of x or

[A(100 x)][100 x] 100 100

A[(100) 2 x 2 ] Ax 2 A Now, net change = (100) 2 (100) 2

Ex. 38: Express 250 as a percentage of 50.

Soln:

=

x2 . 100

(because sign is -ve) and is given by

x2 %. 100

Thus, in the above example,

(10)2 1% decrease % = 100 Ex. 42: A shopkeeper marks the price of his goods 12% higher than its original price. After that, he allows a discount of 12%. What is his percentage profit or loss? Soln: In this case, there is always a loss. And the % (12)2 1.44% 100 Ex. 43: If the population of a town is increased by 15% in the first year and is decreased by 15% in the next year, what effect can be seen in the population of that town? value of loss

Soln: There is a decrease of

(15)2 % i.e., 2.25% 100

When both values are different

Proof: Let the number be A. When it is increased by x%, it becomes A + x%

Theorem: If the value is first increased by x% and then decreased by y% then there is

Ax A(100 x) 100 100 Now, when the increased value is decreased by x%, then it becomes

xy x y 100 % increase or decrease,

of A = A

A(100 x) A(100 x) x% of . 100 100 A(100 x) Ax(100 x) 100 100 100 100A(100 x) Ax(100 x) = 100 100

=

according to the +ve or -ve sign respectively. Proof: Let the value be A. When it is increased by x%, it becomes Ax A(x 100) 100 100 Now, when the increased value is decreased by y%, it becomes

A + x% of A = A

Quicker Maths

200 A(x 100) A(x 100) y% of 100 100

=

A(x 100) Ay(x 100) 100 (100) 2

A(x 100)(100 y) (100)2

A(x 100)(100 y) A Now, net change = (100) 2

=

A (100) 2 100x 100y xy (100)2 A (100) 2

A 100x 100y xy

Thus, in this case, 10 5

increase as the sign is +ve. Ex. 45: A shopkeeper marks the prices of his goods at 20% higher than the original price. After that, he allows a discount of 10%. What profit or loss did he get? 20 10 8% 100 he gets 8% profit as the sign obtained is +ve. Ex. 46: If the salary of a worker is first decreased by 20% and then increased by 10%. What is the percentage effect on his salary? Soln: By Quicker Maths: % effect = % increase – % decrease

Soln:

By the theorem: 20 10

(100) 2

% change

A 100x 100y xy 100 (100) 2 A

xy x y % 100 Theorem: If the order of increase and decrease is changed, the result remains unaffected. Proof: Try this yourself. So, combining the above two theorem, we may write as: Effect = % increase – % decrease % increase % decrease 100 The use of the above formula will clear your doubts. Ex. 44: The salary of a worker was first increased by 10% and thereafter, decreased by 5%. What was the change in his salary? Soln: Let the salary of the worker be 100. After increase, it becomes 100 + 10% of 100 = 110 After decrease, it becomes 110 – 5% of 110 = 104.5 The % increase = 104.5 – 100 = 4.5%

Quicker Method: By the above theorem: If the value firstly increases by x% and then xy decreased by y% then there is x y % 100 increase or decrease, according to the sign +ve or -ve respectively.

10 5 4.5% 100

–

% increase % decrease 100

10 20 12% 100 His salary is decreased by 12% (because the sign is -ve). Note: Change of order of increase and decrease means that in the above example, firstly an increase of 10% is performed and then the decrease of 20% is performed. In both the cases, the result remains the same. Ex. 47: The population of a town was reduced by 12% in the year 1988. In 1989, it was increased by 15%. What is the percentage effect on the population in the beginning of 1990? Soln: % effect = % increase – % decrease

= 10 – 20 –

– = 15 – 12 –

% increase % decrease 100

15 12 3 1.8 1.2 100

Thus, the population was increased by 1.2%.

Successive increase or decrease Theorem: If the value is increased successively by x% and y% then the final increase is given by

xy x y 100 %. Proof: If we put -y in place of y in the previous theorem, we get the required result. This is done because we may say that Increase = –Decrease.

Percentage

201

Ex. 48: A shopkeeper marks the prices at 15% higher than the original price. Due to increase in demand, he further increases the price by 10%. How much % profit will he get? Soln: By above theorem: 15 10 26.5% 100 Ex. 49: The population of a town is decreased by 10% and 20% in two successive years. What per cent population is decreased after two years? Soln: Put x = –10 and y = –20

% profit = 15 + 10 +

(–10) (–20) 28% then, –10 –20 + 100 Therefore, the population decreases by 28%.

Effect on revenue Theorem: (i) If the price of a commodity is diminished by x% and its consumption is increased by y%, (ii) or, if the price of a commodity is increased by x% and its consumption is decreased by y% then the effect on revenue = Inc. % value – Dec. % value – Inc.% value Dec.% value and the value is 100 increased or decreased according to the +ve or -ve sign obtained. Note: The above written formula is the general form for both the cases. For case (i), it becomes:

y–x–

yx 100

xy 100 Thus, we see that it is more easy to remember the general formula which works in both the cases equally. Proof: We are discussing the proof for case (i). Let the price of the commodity be A/unit and the consumption be B units. Then, total revenue expenses = AB Now, the new price = A – x% of A

Whereas, for case (ii), it becomes: x – y –

Ax A(100 x) 100 100 And new consumption = B + y% of B

=A–

=B+

By B(100 y) 100 100

the new revenue expenses =

A(100 x) B(100 y) 100 100

AB(100 x)(100 y) 100 100 Change in revenue expenses

=

AB(100 x)(100 y) AB 100 100

AB1002 100y –100x – xy – 1002 AB 1002 AB100y 100x xy

1002 the % effect on revenue AB100y – 100x xy 100 xy =yx 2 100 AB 100 Similarly, we can prove for case (ii) also. Ex. 50: The tax on a commodity is diminished by 20% and its consumption increases by 15%. Find the effect on revenue. Soln: New Revenue = Consumption × Tax = (115% × 80%) of the original

115 80 = of the original 100 100 115 80 % of original = 92% of original = 100 Thus, the revenue is decreased by (100 – 92) = 8% By Theorem: Effect on revenue = Inc. % value – Dec. % value

–

Inc.% value Dec.% value 100

15 20 8% 100 Therefore, there is a decrease of 8%. Ex. 51: If the price is increased by 10% and the sale is decreased by 5%, then what will be the effect on income? Soln: Let the price be 100 per good and the sale is also of 100 goods. So, the money obtained after selling all the 100 goods = 100 × 100 = 10,000

= 15 – 20 –

Quicker Maths

202 Now, the increased price is 110 per good and the decreased sale is 95 goods. So, the money obtained after selling all the 95 goods = 110 × 95 = 10,450. profit = 10,450 – 10,000 = 450

Soln:

% effect = 25 – (–10) –

450 100 % profit = 10000 4.5% By Theorem: % effect = Inc. % value – Dec. % value

–

10 5 4.5% 100 his income increases by 4.5%. Ex. 52: If the price is decreased by 12% and sale in increased by 10% then what will be the effect on income? Soln: By Theorem: 12 10 3.2% 100 his income is decreased by 3.2%. Ex. 53: The landholding of a person is decreased by 10%. Due to late monsoon, the production decreases by 8%. Then what is the effect on revenue? Soln: Change any of the decreased value in the form of an increase. Suppose we change it as 10% decrease = –10% increase Now, putting the values in the above formula, we have

% effect = 10 – 12 –

(–10) 8 100 = –18 + 0.8 = –17.2% Therefore, his revenue is reduced by 17.2%. Now, suppose we change this as follow 8% decrease = –8% increase Now, putting the values in the above formula, we have % effect

% effect = –10 – 8 –

(–8) 10 = –8 – 10 – = –18 + 0.8 = –17.2% 100 Thus, we get the same result in both the cases. So, it hardly matters which of the values changes its form. Ex. 54: The number of seats in a cinema hall is increased by 25%. The price on a ticket is also increased by 10%. What is the effect on the revenue collected?

25 (–10) 100

35 2.5 37.5 Thus, there is an increase of 37.5% in the revenue. This example can also be solved by changing the form of any increased value to decreased value. Try it yourself by changing the other value.

Inc.% value Dec.% value 100

= 10 – 5 –

Since there is an increase in the seats as well as in the price, we use: Decrease = – (Increase) Thus, the formula becomes:

Note:

xy is our key formula. All 100 of the above forms can be obtained by only changing the signs. For example: i) Increase of x% and increase of y%

We see that, x y

xy = x y % increase or decrease 100 according to its sign. ii) Increase of x% and decrease of y% (put y = –y)

= x – y

xy % 100

increase or decrease

according to its sign. iii) Decrease of x% and increase of y% (Put x = –x)

= x y

xy % increase or decrease 100

according to its sign. iv) Decrease of x% and decrease of y% (Put x = –x and y = –y) xy = x y % increase or decrease 100 according to its sign. Theorem: The passing marks in an examination is x%. If a candidate who scores y marks fails by z

marks, then the maximum marks, M

100(y z) . x

Proof: It is very much easy to prove the above theorem. Let the maximum marks be M, then there exists a relation: x% of M = y + z or, M

y z 100(y z) x% x

Percentage

203

Note:

If you have understood the relationship, you don’t need to remember the formula. But some of the students get confused in finding the relationship in the examination hall, so we have given the direct formula. Ex. 55: A student has to score 40% marks to get through. If he gets 40 marks and fails by 40 marks, find the maximum marks set for the examination. Soln: By the above theorem, Maximum marks

100(40 40) 200 40

Ex. 56: In an examination, a candidate must get 80% marks to pass. If a candidate who gets 210 marks fails by 50 marks, find the maximum marks. Soln: By the above theorem: 100(210 50) 325 80 Theorem: A candidate scoring x% in an examination fails by ‘a’ marks, while another candidate who scores y% marks gets ‘b’ marks more than the minimum required pass marks. Then, the maximum marks for

Maximum marks =

100(a b) . yx

that examination are M

Proof: Let the maximum marks for the examination be M. Thus, marks scored by the first candidate = x% of M and marks scored by the second candidate = y% of M. Now, passing marks for both the candidates are equal; so, x% of M + a = y% of M – b or,

Mx My a b 100 100

or,

M(y x) ab 100

M

100(a b) (y x)

Ex. 57: A candidate scores 25% and fails by 30 marks, while another candidate who scores 50% marks, gets 20 marks more than the minimum required marks to pass the examination. Find the maximum marks for the examination. Soln: By the theorem: Maximum marks =

100(30 20) 200 50 25

Note:

(i) The above formula can be written as Maximum marks =

100(Diff. of their scores) Diff . of their % marks

(ii) Difference of their scores = 30 + 20. Because the first candidate gets 30 less than the required pass marks, while the second candidate gets 20 more than the required passing marks. Ex. 58: A candidate who gets 30% of the marks in a test fails by 50 marks. Another candidate who get 320 marks fails by 30 marks. Find the maximum marks. Soln: We can’t use the above said direct formula in this case. (Why ?) So, we use the fact that pass marks for both the candidates are the same. If x is the maximum marks, then the passing marks for the first candidate = 30% of x + 50 and the pass marks for the second candidate = 320 + 30 Therefore, 30% of x + 50 = 320 + 30 or,

3x 300 10

300 10 1, 000 3 Note: The above method should be understood well, because it works when the form of question is changed. Theorem: In measuring the sides of a rectangle, one side is taken x% in excess and the other y% in deficit. The error per cent in area calculated from the x

measurement is x – y –

xy in excess or deficit, 100

according to the +ve or -ve sign. Proof: You must be familiar with the above formula. The proof for this is being given below. Let the sides of the rectangle be a and b. area = ab New sides are: a + x% of a and b – y% of b or,

a(100 x) b(100 y) and 100 100

new area Error

ab(100 x)(100 y) (100) 2

ab(100 x)(100 y) ab (100) 2

ab[100x 100y xy] (100) 2

Quicker Maths

204 % error

ab[100x – 100y xy] 100 (100)2 ab

xy 100 or, % error = % Excess – % Deficit xy

% Excess % Deficit 100 Ex. 59: In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error per cent in area calculated from the measurement. Soln: By the above theorem:

–

% error = 5 – 4 –

5 4 1 4 1 % excess 100 5 5

because sign is +ve Ex. 60: If one of the sides of a rectangle is increased by 20% and the other is increased by 5%, find the per cent value by which the area changes. Soln: The above theorem will also work in this case. The only change required is to change the form of one of increased % value to the decreased % value, ie., 5% increase = – 5% decrease Now, % effect = 20 – (–5) –

20 (5) 100

20 5 1 26% Since the sign is +ve, there is 26% increase in area. Note: We can also use 20% increase = –20 decrease. Try to solve by using the above change. Ex. 61: If the sides of a square are increased by 30%, find the per cent increase in its area. Soln: We can use the same method that has been used in the previous example.

% increase in area = 30 – (–30) –

30 (–30) 100

60 9 69% Theorem: If the sides of a triangle, rectangle, square, circle, rhombus (or any 2-dimentional figure) are increased by x%, its area is increased by

x(x 200) x2 % or 2x % . 100 100 Soln:

For a triangle:

Suppose a triangle has three sides a, b and c. Then its area A s(s a)(s b)(s c) abc 2 Now, when all the three sides are increased by x%, the sides become:

Where, s

a(100 x) b(100 x) c(100 x) , , 100 100 100

Now, s1

(100 x) a b c (100 x) 100 s 100 2

New area, A1 =

a(100 x) b(100 x) c(100 x) s1 s1 s1 s1 100 100 100 4

100 x s(s a)(s a)(s b)(s c) 100 2

100 x A1 A 100 Now, % increase in area 100 x 2 – 1 A A1 A 100 100 100 A A 100 x 2 1 100 100

2 x 1 – 1 100 100

x2 2x 1 – 1 100 2 (100) 100

x(x 200) x2 2x 100 100 Therefore, we may say that if all the sides of a triangle are increased by x%, then its area increases by

2

x(x 200) x % or 2x . 100 100 For a Rectangle: Let the sides of a rectangle be a and b. Then its area; A = ab Now, suppose, its sides become: a(100 x) b (100 x) and 100 100

Percentage

205

And its new area; 2

2

100 x 100 x A1 ab A 100 100

% increase in area

A1 A 100 A

100 x 2 1 A 100 100 A 100 x 2 1 100 100

x(x 200) x2 2x 100 100 Therefore, we may say that if sides of a rectangle are increased by x%, then its area is increased by

x(x 200) x2 % . And this is the same % or 2x 100 100 as for the triangle. Note: Now, we don’t need to calculate for square and rhombus. That will give the same result. For a Rectangle: Let us have a circle with radius r metres. 2 So, its area = A r When its radius is increased by x%, it becomes:

r(100 x) 100

r(100 x) So, its new area = A1 100 2

2

2

(100 x) 100 x r A 100 100 2

% increase in area

A1 A 100 A

100 x 2 x(x 200) x2 2x 1 100 100 100 100

Therefore, we see that it gives the same result for a circle also. Final conclusion: Now we conclude that this general formula is applicable for any 2-dimensional figure. Take an example: If the sides of a rectangle are increased

by 10%, what is the percentage increase in its area? Soln: The required answer (10)2 20 1 21% = 2 10 100 Note: (1) What will be the effect on the area when the sides are decreased by x% ? (Try this yourself.) (2) What is the effect on the volume of a threedimensional object when its sides are increased by x%? (Try it yourself.) Theorem: In an examination, x% failed in English and y% failed in Maths. If z% of students failed in both the subjects, the percentage of students who passed in both the subjects is 100 – (x + y – z) or, (100 – x) + (100 – y) + z Proof: Percentage of students who failed in English only = (x – z)% Percentage of students who failed in Maths only = (y – z) % Percentage of students who failed in both the subjects = z% (given) the percentage of students who passed in both the subjects = 100 – [(x – z) + (y – z) – z] = 100 – (x + y - z) Ex. 62: In an examination, 40% of the students failed in Maths, 30% failed in English and 10% failed in both. Find the percentage of students who passed in both the subjects. Soln: Following the above theorem: The required % = 100 – (40 + 30 – 10) = 40% Note: We should know that the following sets complete the system, i.e., 100% = % of students who failed in English + % of students who failed in Maths – % of students who failed in both the subjects + % of students who passed in both the subjects Ex. 63: A man spends 75% of his income. His income increases by 20% and his expenditure also increases by 10%. Find the percentage increase in his savings. Soln: Detailed Method: Suppose his monthly income = 100 Thus, he spends 75 and saves 25. His increased income = 100 + 20% of 100 = 120 His increased expenditure = 75 + 10% of 75 = 82.50 his new savings = 120 – 82.5 = 37.50 % increase in his savings

37.50 25 100 50% 25

Quicker Maths

206 Quicker Method (Direct Formula): Percentage increase in savings 20 100 10 75 1250 50% 100 75 25 Ex. 64: A solution of salt and water contains 15% salt by weight. Of it, 30kg water evaporates and the solution now contains 20% of salt. Find the original quantity of salt. Soln: Suppose there was x kg of solution initially.

15x 3x kg 100 20 Now, after evaporation, only (x – 30) kg of

The quantity of salt = 15% of x =

mixture contains

3x kg of salt. 20

or, 20% of (x – 30) =

3x 20

x 30 3x 5 20 or, 15x = 20x – 600

or,

600 x 5 120kg Quicker Method (Direct Formula): Original quantity of solution = Quantity of

Final % of salt evaporated water % Diff. of salt 20 = 30 120 kg 20 – 15 Ex. 65: In a library, 20% of the books are in Hindi, 50% of the remaining are in English and 30% of the remaining are in French. The remaining 6300 books are in regional languages. What is the total number of books in the library? Soln: Suppose there are x books in the library. x 5 50% of the remaining, i.e., 50% of

Then, the no. of books in Hindi = 20% of x

x 4x 2x are in English. x 50% of 5 5 5 Now, 30% of the remaining, i.e., 30% of

x 2x x 5 5

= 30% of

2x 3x books are in French. 5 25

x 2x 3x Now, x – 6300 5 5 25

or,

7x 6300 25

6300 25 22500 7 Quicker Method (Direct Formula): No. of total books x

100 100 100 6300 100 20 100 50 100 30 6300 100 100 100 22,500 80 50 70 Ex. 66: 40 litres of a mixture of milk and water contain 10% of water. How much water must be added to make the water 20% in the new mixture? Soln: Quantity of water = 10% of 40 = 4 litres. Now, suppose x litres of water are added, then 4 + x = 20% of (40 + x)

40 x 5 or, 20 + 5x = 40 + x

or, 4 x

20 5 litres. 4 Quicker Method (Direct Formula): The quantity of water to be added

x

40(20 10) 5 litres. (100 20)

Ex. 67: The manufacturer of an article makes a profit of 25%, the wholesale dealer makes a profit of 20%, and the retailer makes a profit of 28%. Find the manufacturing price of the article if the retailer sold it for 48. Soln: By the rule of fraction: Cost of manufacturing

100 100 100 48 100 28 100 20 100 25 100 100 100 48 = 25 128 120 125 Ex. 68: What quantity of water should be added to reduce 9 litres of 50% acidic liquid to 30% acidic liquid?

Percentage Soln:

207

Detailed Method: Acid in 9 litres = 50% of 9 = 4.5 litres. Suppose x litres of water are added. Then, there are 4.5 litres of acid in (9 + x) litres of diluted liquid. Now, according to the question, 30% of (9 + x) = 4.5 3 or, (9 + x) = 4.5 10 or, 27 + 3x = 45 or, 3x = 18

18 6 litres. 3 Quicker Method: The quantity of water to be added x

9(50 30) 6 litres. 30 Ex. 69: In an examination the percentage of students qualified to the number of students appeared from school ‘A’ is 70%. In school ‘B’ the number of students appeared is 20% more than the students appeared from school ‘A’ and the number of students qualified from school ‘B’ is 50% more than the students qualified from school ‘A’. What is the percentage of students qualified to the number of students appeared from school ‘B’? 1) 30% 2) 70% 3) 87.5% 4) 78.5% 5) None of these Soln: Detailed method: Suppose 100 students appeared from school A. Then we have Appeared Passed A 100 70 B 120 70 + 50% of 70 = 105

Required % =

105 100 87.5% 120

Direct Formula (Quicker Method): Required %

70 (100 50)% 100 100 (100 20)%

70 150 100 87.5% 100 120 Ex. 70: In 1 kg mixture of sand and iron, 20% is iron. How much sand should be added so that the proportion of iron becomes 10%? 1) 1 kg 2) 200 gm 3) 800 gm 4) 1.8 kg 5) None of these

Soln:

In 1 kg mixture, iron = 20% of 1000 gm = 200 gm and sand = 800 gm Suppose x gm sand is added to the mixture Then, total mixture = (1000 + x) gm

200 Now, % of iron = (1000 x) 100 10 (given) or, 1000 + x = 2000 x = 1000 gm = 1 kg Quicker Method: When a certain quantity of goods B is added to change the percentage of goods A in a mixture of A and B then the quantity of B to be added is

Pr evious % value of A Mixture quantity Changed % value of A – Mixture Quantity In above question, required quantity of sand 20 1 1 2 1 1 kg 10 Ex. 71: Weights of two friends Ram and Shyam are in the ratio of 4 : 5. Ram’s weight increases by 10% and the total weight of Ram and Shyam together becomes 82.8 kg, with an increase of 15%. By what per cent did the weight of Shyam increase? 1) 12.5% 2) 17.5% 3) 19% 4) 21% 5) None of these Soln: 3;Let the weights of Ram and Shyam be 4x and 5x. Now, according to question,

to be added

4x 110 + Shyam’s new wt = 82.8 100

... (i)

(4x 5x ) 9x 115 82.8 100

... (ii)

and

From (ii), x = 8 Putting in (i), we get Shyam’s new wt = (82.8 – 35.2 =) 47.6 % increase in Shyam’s wt

47.6 40 100 19% 40 Quicker Method: If Shyam’s weight increase by x% then there exist a relationship 4(100 + 10) + 5(100 + x) = (4 + 5) [100 + 15] .... (*) or, 440 + 5(100 + x) = 1035 or, 100 + x = 119 x = 19

Quicker Maths

208 OR By method of Alligation:

Ram 10%

By the rule of alligation

x 15 4 15 10 5

Shyam x%

or, x – 15 = 4 x = 19

15% 4

5 (given)

EXERCISES 1. Mr Shamin’s salary increases every year by 10% in June. If there is no other increase or reduction in the salary and his salary in June 2011 was 22,385, what was his salary in June 2009? 1) 18,650 2) 18,000 3) 19,250 4) 18,500 5) None of these 2. Ramola’s monthly income is three times Ravina’s monthly income. Ravina’s monthly income is fifteen per cent more than Ruchira’s monthly income. Ruchira’s monthly income is 32,000. What is Ramola’s annual income? 1) 1,10.400 2) 13,24,800 3) 36,800 4) 52,200 5) None of these 3. An HR Company employs 4800 persons, out of which 45 per cent are males and 60 per cent of the males are either 25 years or older. How many males are employed in that HR Company who are younger than 25 years? 1) 2640 2) 2160 3) 1296 4) 864 5) None of these 4. In a test, a candidate secured 468 marks out of maximum marks ‘A’. Had the maximum marks ‘A’ converted to 700, he would have secured 336 marks. What was the maximum marks of the test? 1) 775 2) 875 3) 975 4) 1075 5) None of these 5. Six-elevenths of a number is equal to 22 per cent of the second number. The second number is equal to one-fourth of the third number. The value of the third number is 2400. What is 45% of the first number? 1) 109.8 2) 111.7 3) 117.6 4) 123.4 5) None of these 6. In an entrance examination, Ritu scored 56 per cent marks, Smita scored 92 per cent marks and Rina scored 634 marks. The maximum marks of the examination is 875. What is the average marks scored by all the three girls together?

7.

8.

9.

10.

11.

1) 1929 2) 815 3) 690 4) 643 5) None of these Akash scored 73 marks in subject A. He scored 56% marks in subject B and X marks in subject C. Maximum marks in each subject were 150. The overall percentage marks obtained by Akash in all the three subjects together was 54%. How many marks did he score in subject C? 1) 84 2) 86 3) 79 4) 73 5) None of these If twentyfive per cent of three-sevenths of twenty six per cent of a number is 136.5, what is the number? 1) 6300 2) 5600 3)4800 4) 4900 5) None of these Two-thirds of Ranjit’s monthly salary is equal to Raman’s monthly salary. Raman’s monthly salary is thirty per cent more than Pawan’s monthly salary. Pawan’s monthly salary is 32000. What is Ranjit’s monthly salary? 1) 64200 2) 62500 3) 64500 4) 62400 5) None of these In a class there are 60 students, out of whom 15 per cent are girls. Each girl’s monthly fee is 250 and each boy’s monthly fee is 34 per cent more than a girl. What is the total monthly fees of girls and boys together? 1) 19335 2) 18435 3) 19345 4) 19435 5) None of these In an examination, 30% of the total students failed in Hindi, 45% failed in English and 20% failed in both the subjects. Find the percentage of those who passed in both the subjects. 1) 35.7% 2) 35% 3) 40% 4) 45% 5) 44%

Percentage 12. Ganesh’s monthly income is

209

4 th of Suresh’s monthly 5

income. The respective ratio between Suresh’s and Sunil’s monthly income is 13 : 17. Sunil spends 12% of his monthly income on food expenses which is 8,160. What is the annual income of Ganesh? 1) 4,99,100 2) 4,92,900 3) 4,99,200 4) 4,99,300 5) 4,94,900 13. One-fourth of two-fifths of 30 per cent of a number is 15. What is 20 per cent of that number? 1) 100 2) 50 3) Data provided are not adequate to answer the question 4) 200 5) 75 14. In a company ‘XYZ’, the ratio of the total number of undergraduate employees to the total number of graduate employees is 13 : 23. The company has only two branches – one in Mumbai and another in Delhi. If the total number of undergraduate employees in Mumbai branch is 351, which is 30% of the total undergraduate employees in the company, what is the total number of graduate employees in the company? 1) 2185 2) 1955 3) 2070 4) 1970 5) 2170 15. The monthly salaries of Pia and Som are in the ratio of 5 : 4. Pia, from her monthly salary, gives

3 5 to her

mother, 15% towards her sister’s tuition fees, 18% towards a loan and she shops with the remaining amount, which is 2100. What is the monthly salary of Som? 1) 25000 2) 30000 3) 15000 4) 20000 5) 24000 16. In the year 2013, the population of village A was 20% more than the population of village B. The population of village A in 2014 increased by 10% as compared to the previous year. If the population of village A in 2014 was 5610, what was the population of village B in 2013? 1) 4650 2) 5550 3) 4250 4) 5800 5) 4500 17. Mohan gave 25% of a certain amount of money to Ram. From the money Ram received, he spent 20%

on buying books and 35% on buying a watch. After the mentioned expenses, Ram has 2700 remaining. How much did Mohan have initially? 1) 16000 2) 15000 3) 24000 4) 27000 5) 20000 18.Meena Kumari goes to a shop and buys a saree costing 5225, including a sales tax of 12%. The shopkeeper gives her a discount so that the price is decreased by an amount equivalent to sales tax. The price is decreased by (nearest value) 1) 615 2) 650 3) 560 4) 580 5) 680 19. Raja gives 30% of his salary to his mother, 40% of the remaining salary he invests in an insurance scheme and PPF in the ratio of 4 : 3 and the remaining he keeps in his bank account. If the difference between the amount he gives to his mother and that he invests in insurance scheme is 8400, how much is Raja’s salary? 1) 60,000 2) 62,000 3) 64,000 4) 65,000 5) 54,000 20. The ratio of the monthly salary of Om to that of Pihu is 7 : 9. Om and Pihu both save 20% and 40% of their respective monthly salary respectively. Om 1 7 of his savings in PPF and Pihu invests 2 9 of his savings in PPF. If Om and Pihu together saved 17500 in PPF, what is Pihu’s monthly salary? 1) 72000 2) 36000 3) 45000 4) 35000 5) 54000 21. Out of her monthly salary, Ridhi spends 34% on various expenses. From the remaining, she gives

invests

1 2 th to her brother, to her sister and the remaining 3 6 she keeps as savings. If the difference between the amounts she gave to her sister and brother was 10,560, what was Ridhi’s savings? 1) 3,740 2) 3,420 3) 4,230 4) 3,230 5) Other than those given as options 22. Ramesh has 20% savings with him from his monthly salary. If the expenditure on clothing is 25% of overall expenditure and his total expenditure except clothing is 3600, then find his savings (in ). 1) 1000 2) 1500 3) 1600 4) 1200 5) 900

Quicker Maths

210

Solutions 1. 4; Salary in June 2011 = 22385 or, First number =

100 100 Salary in June 2009 = 22385 110 110

2. 2;

100 = 185 ×100 =2035 11 = 18500 Ravina’s monthly income 115 = 36800 100 Ramola’s monthly income = 3 × 36800 = 110400 Ramola’s annual income = 12 × 110400 = 1324800 Total number of persons = 4800 Number of males = 45% of 4800

= 32000 ×

3. 4;

45 4800 = 2160 100 Now, according to the question, Number of males who are younger than 25 years

=

40 2160 = 864 100 Converted maximum marks = 700 Converted marks = 336

= (100 – 60 =) 40% of 2160 = 4. 3;

=

336 100 = 48% 700 468 is 48% of maximum marks ‘A’

% marks =

468 100 975 48 Quicker Approach: 336 marks 700

A=

468 marks

700 700 78 468 = 336 56

700 39 = 25 × 39 = 975 28 5. 5; According to the question,

=

6 × First number = 22% of second number 11 1 Second number = × Third number 4 1 = × 2400 = 600 4

22 Second number 11 100 6 22 600 11 242 100 6

required answer = 45% of 242 = 6. 4;

45 242 = 108.9 100

56 Ritu’s marks = 875 × = 490 100

Smita’s marks = 875 ×

92 = 805 100

Rina’s marks = 634 Total marks = 490 + 805 + 634 =1929 1929 643 3 7. 2; Akash scored in subject A = 73 marks

Average =

56 150 56 3 = 100 2 = 28 ×3 = 84 marks Total marks Akash got in all the three subjects

In subject B =

54 9 450 = 54 × 100 2 = 27 × 9 = 243 marks Akash’s marks in subject C = 243 – (84 + 73) = 243 – 157 = 86 marks 8. 4; Let the number be x.

together =

Then, x

25 3 26 136.5 100 7 100

136.5 100 100 7 = 4900 25 3 26 9. 4; Pawan’s monthly salary = 32,000

x=

Raman's monthly salary = 32000 × = Ranjit’s monthly salary = 10. 1; Number of girls = 60×

130 100

41600

3 41600 = 2

15 =9 100

62400

Percentage Total monthly fee of girls = 250 × 9 = 2250 Number of boys = 60 – 9 = 51 134 = 335 100 Total monthly fees of boys = 51 × 335 = 17085 Sum = 17085 + 2250 = 19,235 Let the number of students be 100. Number of students who failed in Hindi is 30%. n(H) = 30 Number of students who failed in English is 45% n(E) = 45 Number of students who failed in both the subjects is 20% n(HE) = 20 Applying the rule, n(HE) = n(H) + n(E) – n(HE) = 30 + 45 – 20 = 55 Percentage of students who failed in Hindi or English or both the subjects = 55% Number of students who passed in both the subjects = 100 – 55 = 45% Quicker Approach (Venn Diagram Method):

Monthly fee of one boy = 250 ×

11. 4;

211 20 500 = 100 100 14. 3; Let the no. of undergraduate employees be 13x and that of graduate employees be 23x. Then, according to the question, 30% of 13x = 351

Now, 20% of 500 =

or,

30 13x = 351 100

351 100 = 90 390 Now, total no. of graduate employees in company XYZ = 23 × 90 = 2070 Alternative Method:

x=

100 1170 No. of UG students = 351 30 23

No. of graduates = 1170 13 90 23 15. 5;

= 2070 Let the monthly salary of Pia and Som be 5x and 4x respectively. Then, money given by Pia to her mother 3 = 3x 5 Remaining amount = 5x – 3x = 2x Money given by Pia as sister’s tuition fees = 15% of 5x = 0.75x Money given by Pia towards loan = 18% of 5x = 0.9x Total money given = 3x + 0.75x + 0.90x = 4.65x Remaining amount = 5x – 4.65x = 0.35x Now, 0.35x = 2100

= 5x

Pass in both = 100 – (10 + 20 + 25) = 100 – 55 = 45 Note: If the concept of Venn diagrams is clear this question can be solved mentally without writing anything. 8160 100 = 68000 12. 3; Sunil’s monthly income = 12 68000 13 = 52000 Suresh’s monthly income = 17 4 = 41600 5 Ganesh’s annual income = 41600 ×12 = 499200 13. 1; Let the number be x.

Ganesh’s monthly income = 52000 ×

30 2 1 =15 100 5 4 15 5 4 100 or, x = = 500 60

Then, x

2100 210000 x = = 6000 0.35 35 Monthly salary of Som = 4 × 6000 = 24000 Quicker Approach: Suppose Pia's monthly salarly is 500 and Som's monthly salary is 400.

x=

2 Now, Pia is left with 500 200 after 5 giving money to her mother. Left with (200 – 75 =) 125 after sister's tuition fees. Left with (125 – 90 =) 35 after loan

Quicker Maths

212 35 2100 2100 400 = 300 × 80 = 24000 400 = 35 16. 3; Population of village B in 2013 100 100 = 4250 110 120 17. 3; Ram spent 20% on books and 35% on buying a watch. Remaining percentage = (100 – 20 – 35)% = 45% Remaining amount of Ram = 2700. Now, 45% = 2700

= 5610×

2700 100 = 6000 100% = 45 Mohan gave 25% of a certain amount to Ram. 25% of Mohan’s money = 6000 6000 100 25 = 24000 18. 3; Before sale tax cost of saree

100% of Mohan’s money =

100 = 4665 = 5225 112 Sale tax = 5225 – 4665 = 560 And it is given that discount is equal to sales tax, so discount = 560 Alternatively think like: Let the cost of saree be 100. Sales tax = 12 = Discount Selling price = 112 Now, 112 5225

19. 1;

5225 12 560 12 112 Let Raja’s salary be x. Raja gives 30% of his salary to his mother. x 30 3x to his mother Raja gives 10 100

3x 7x = 10 10 Investments of Raja in insurance and PPF is 40% of the remaining salary.

Remaining salary of Raja = x –

Insurance + PPF =

7x 40 7x 10 100 25

Remaining salary of Raja =

7x 7x – 10 25

35x –14x 21x = 50 50 Raja’s investment in insurance scheme

=

7x 4 4x 25 7 25 Now, according to the question,

=

3x 4x – 8400 10 25 15x – 8x 8400 50 or, 7x = 8400 × 50

or,

8400 50 = 1200 × 50 = 60000 7 Quicker Approach: Suppose Raja's salary = 100; Given to mother = 30 Amount remaining after giving to mother = 70 He invested 28 in Insurance and PPF in 4 : 3 ratio.

x=

Invested in insurance =

28 4 = 16 7

Now, the given difference = 30 – 16 = 14 8400 100 60, 000 20. 3; Let Pihu’s monthly salary be 9x and Om’s 7x. Then, 7 1 of 20% of 7x + of 40% of 9x = 17500 9 2 or, 10% of 7x + 40% of 7x = 17500 or, 0.7x + 2.8x = 17500 or, 3.5x = 17500 x = 5000 Hence Pihu’s monthly salary = 9 × 5000 = 45000 Quicker Approach: Om Pihu Salary 70 : 90 Savings 14 : 36 Investment (PPF) 7 28 According to the question, 7 + 28 17500

90

17500 90 = 45000 35

Percentage

213

21. 5; Quicker Approach: Let the total income of Ridhi be 100. 34% is spent on various expenses Remaining = 100 – 34 = 66 Ridhi gave to her brother = 66 ×

1 = 111 6

2 = 111 3 Her savings = 66 – (11 + 44) = 11 As given in qustion (44 – 11=) 33 10560

Ridhi gave to her sister = 66

10560 11 3520 33 22. 4; Total expenditure except clothing is 3600. 75% of expenditure = 3600 11

100 = 4800 Total expenditure = 3600 75 As Ramesh saves 20% His expenditure is 80% of salary If 80% 4800 Savings = 20% 1200 Alternative Method: Suppose salary = 100 Savings = 20 Expenditure = 80 Expenditure on clothing = 20 Other expenditure = 60 Now, 60 3600 20 1200

214

Quicker Maths

Chapter 21

Average An average, or more accurately an arithmetic mean is, in crude terms, the sum of n different data divided by n. For example, if a batsman scores 35, 45 and 37 runs in first, second and third innings respectively, then his average runs in 3 innings is equal to

35 45 37 = 39 3

runs. Therefore, the two mostly used formula in this chapter are: Average =

Total sum of data No.of data

And, Total = Average × No. of data Ex. 1: The average age of 30 boys of a class is equal to 14 yrs. When the age of the class teacher is included the average becomes 15 yrs. Find the age of the class teacher. Soln: Total ages of 30 boys = 14 × 30 = 420 yrs. Total ages when class teacher is included = 15 × 31 = 465 yrs Age of class teacher = 465 – 420 = 45 yrs. Direct Formula: Age of new entrant = New average + No. of old members × Increase in average = 15 + 30(15 – 14) = 45 yrs Ex. 2: The average weight of 4 men is increased by 3 kg when one of them who weighs 120 kg is replaced by another man. What is the weight of the new man? Soln: Quicker Approach: If the average is increased by 3 kg, then the sum of weights increases by 3 × 4 = 12 kg. And this increase in weight is due to the extra weight included due to the inclusion of new person. Weight of new man = 120 + 12 = 132 kg Direct Formula: Weight of new person = weight of removed person + No. of persons × increase in average = 120 + 4 × 3 = 132 kg

Ex. 3: The average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of the failed candidates is 15, what is the number of candidates who passed the examination? Soln: Let the number of passed candidates be x. Then total marks = 120 × 35 = 39x + (120 – x) × 15 or, 4200 = 39x + 1800 - 15x or, 24x = 2400 x = 100 number of passed candidates = 100. Direct Formula: Number of passed candidates Total candidates (Total average – Failed average) Passed average – Failed average

and Number of failed candidates

Total candidates (Passed average Total average) Passed average Failed average No. of passed candidates 120(35 – 15) 100 39 15 By Method of Alligation:

Passed 39

Failed 15 35

20

4

No. of passed candidates: No. of failed candidates = 20 : 4 = 5 : 1 120 No. of passed candidates = 5 1 5 100 Ex. 4: The average of 11 results is 50. If the average of first six results is 49 and that of last six is 52, find the sixth result.

Quicker Maths

216 Soln:

The total of 11 results = 11 × 50 = 550 The total of first 6 results = 6 × 49 = 294 The total of last 6 results = 6 × 52 = 312 The 6th result is common to both; Sixth result = 294 + 312 – 550 = 56 Direct Formula: 6th result = 50 + 6{(52 – 50) + (49 – 50)} = 50 + 6 {2 – 1} = 56 Ex. 5: The average age of 8 persons in a committee is increased by 2 years when two men aged 35 yrs and 45 yrs are substituted by two women. Find the average age of these two women. Soln: By the direct formula used in Ex. 2, the total age of two women = 2 × 8 + (35 + 45) = 16 + 80 = 96 yrs.

Note:

96 = 48 yrs. 2 Ex. 6: The average age of a family of 6 members is 22 yrs. If the age of the youngest member be 7 yrs, then what was the average age of the family at the birth of the youngest member? Soln: Total ages of all members = 6 × 22 = 132 yrs. 7 yrs ago, total sum of ages = 132 – (6 × 7) = 90 yrs. But at that time there were 5 members in the family. Average at that time = 90 ÷ 5 = 18 yrs. Ex. 7: A man bought 13 shirts of 50 each, 15 pants of 60 each and 12 pairs of shoes at 65 a pair. Find the average value of each article. Soln: Direct Method:

2 10 44 2 14 yrs = 18 12 3 3 3 = 14 yrs 8 months. Ex. 11: A batsman in his 17th innings makes a score of 85, and thereby increases his average by 3. What is his average after 17 innings? Soln: Let the average after 16th innings be x, then 16x + 85 = 17 (x + 3) = Total score after 17th innings. x = 85 – 51 = 34 average after 17 innings = x + 3 = 34 + 3 = 37. Direct Formula: Average after 16 innings = 85 – 3 × 17 = 34 Average after 17 innings = 85 – 3(17 – 1) = 37 Ex. 12: A cricketer has completed 10 innings and his average is 21.5 runs. How many runs must he make in his next innings so as to raise his average to 24? Soln: Total of 10 innings = 21.5 × 10 = 215 Suppose he needs a score of x in 11th innings;

average age of two women =

13 50 15 60 12 65 13 15 12 = 58.25 Ex. 8: The average score of a cricketer in two matches is 27 and in three other matches is 32. Then find the average score in all the five matches. Soln: Direct Method: Average in 5 matches

Average =

2 27 3 32 54 96 30 23 5 Ex. 9: The average of 11 results is 30, that of the first five is 25 and that of the last five is 28. Find the value of the 6th number. Soln: Direct Formula: 6th number = Total of 11 results – (Total of first five + Total of last five results) = 11 × 30 – (5 × 25 + 5 × 28) = 330 – 265 = 65

=

Ex 4 and Ex 9 are different. In Ex 4 the 6th result is common to both the groups but in Ex 9 the 6th result is excluded in both the results. Ex. 10: In a class, there are 20 boys whose average age is decreased by 2 months, when one boy aged 18 years is replaced by a new boy. Find the age of the new boy. Soln: This example is similar to Ex. 2. The only difference is that in Ex 2 the average increases after replacement whereas in this case the average decreases. Thus, you can see the difference in direct formula. Direct Formula: Age of new person = Age of removed person – No. of persons × Decrease in average age = 18 – 20 ×

215 x 24 11 or, x = 264 – 215 = 49 Direct Formula: Required score = 11 × 24 – 21.5 × 10 = 49 Note: The above formula is based on the theory that the difference is counted due to the score in last innings.

then average in 11 innings

Average related to speed Theorem: If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/

Average

217

hr, then the average speed during the whole

2xy journey is given by x y km/hr.. or, If half of the journey is travelled at a speed of x km/hr and the next half at a speed of y km/hr, then average speed during the whole journey is

2xy x y km/hr.. or, If a man goes to a certain place at a speed of x km/hr and returns to the original place at a speed of y km/hr, then the average speed during up-

Ex. 13: A train travels from A to B at the rate of 20 km per hour and from B to A at the rate of 30 km/hr. What is the average rate for the whole journey? Soln: By the formula: Average speed 2 20 30 24 km / hr. 20 30 Ex. 14: A person divides his total route of journey into three equal arts and decides to travel the three parts with speeds of 40, 30 and 15 km/hr respectively. Find his average speed during the whole journey. Soln: By the theorem: Average speed

2xy and-down journey is x y km/hr.. Note:

In all the above three cases the two parts of the journey are equal; hence the last two may be considered as a special case of the first. That’s why all the three lead to the same result. Proof: Proof for this is given in “Time and Distance” Chapter. Theorem: If a person travels three equal distances at a speed of x km/hr, y km/hr and z km/hr respectively, then the average speed during the whole journey is

3xyz xy yz xz km/hr..

Proof: Let the three equal distances be A km. Time taken at the speed of x km/hr =

A hrs. x

A Time taken at the speed of y km/hr = hrs. y A Time taken at the speed of z km/hr = hrs. z

A B C Total distance travelled in time x y z = 3A km Average speed during the whole journey

3A 3xyzA A A A Ayz Axz Axy x y z

3xyz km / hr xy yz xz

3 40 30 15 40 30 30 15 40 15

3 40 30 15 24 km / hr 2250

Ex.15: One-third of a certain journey is covered at the rate of 25 km/hr, one-fourth at the rate of 30 km/hr and the rest at 50 km/hr. Find the average speed for the whole journey. Soln:

Let the total journey be x km. Then speed of 25 km/hr and

x km at the 3

x km at 30 km/hr and 4

x x 5 the rest distance x x at the speed 3 4 12 of 50 km/hr. Total time taken during the journey of x km =

x x 5x hrs hrs hrs 3 25 4 30 12 50

18x 3x hrs hrs 600 100

x 100 1 33 km / hr 18x 3 3 600 Note: In this example the three parts of the journey are not equal; so we didn’t apply the theorem. Ex. 16: The average salary of the entire staff in a office is 120 per month. The average salary of officers is 460 and that of non-officers is 110. If the number of officers is 15, then find the number of non-officers in the office. Soln: Let the required number of non-officers = x average speed

Quicker Maths

218 Then, 110x + 460 × 15 = 120 (15 + x) or, 120x – 110x = 460 × 15 – 120 × 15 = 15 (460 – 120) or, 10x = 15 × 340 x = 15 × 34 = 510 Quicker Method: (Method of alligation):

Officers 460

Soln:

Non-officers 110

35x 42 35x 42 35 7 42

35x 42 x 1 42 or, 35x + 42 = 42x – 42 or, 7x = 84 x = 12 Thus the original expenditure of the mess = 35 × 12 = 420

Now, we have

120 10

Suppose the average expenditure was x. Then total expenditure = 35x. When 7 more students join the mess, total expenditure = 35x + 42 Now, the average expenditure

340

Therefore, ratio of officers to non-officers = 10 : 340 = 1 : 34. 34 number of non-officers 15 1 510 OR, Direct Formula: No. of non-officers = No. of officers ×

Av. salary of officers – Mean average Mean average – Av. salary of non – officers

Direct formula: If decrease in average = x increase in expenditure = y increase in no. of students = z and number of students (originally) = N, then

x(N z) y z

the original expenditure = N

460 – 120 15 510 120 – 110 Ex. 17: There were 35 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by 42 per day while the average expenditure per head diminishes by Re 1. Find the original expenditure of the mess.

Note:

1(35 7) 42 In this case, 35 35 12 7 = 420 This formula may be used in different cases of such examples. Try it.

EXERCISES 1. In an examination, a batch of 60 students made an average score of 55 and another batch of 40 made it only 45. What is the overall average score? 1) 52 2) 40 3) 51 4) 56 5) None of these 2. The average marks of a student in four subjects is 75. If the student obtained 80 marks in the 5th subject then the new average is 1) 80 2) 76 3) 92 4) 95 5) None of these 3. The average of first 61 natural numbers is 1) 30 2) 30.5 3) 31 4) 32 5) None of these 4. In an exam, the average was found to be 50 marks. After deducting computational errors the marks of the 100 candidates had to be changed from 90 to 60 each and the average came down to 45 marks. The total number of candidates who took the exam were

1) 300 2) 600 3) 200 4) 150 5) None of these 5. The average age of a group of 16 persons is 28 yrs and 3 months. Two persons each 58 yrs old left the group. The average age of the remaining persons is 1) 26 2) 24 3) 22 4) 20 5) None of these 6. The average weight of 50 balls is 5 gm. If the weight of the bag be included the average weight increases by 0.05 gm. What is the weight of the bag? (in gm) 1) 5.05 2) 6.05 3) 7.05 4) 7.55 5) None of these 7. The average age of a group of 10 students is 15 yrs. When 5 more students joined the group the average age rose by 1 yr. The average age of the new students is 1) 18 yrs 2) 17 yrs 3) 16 yrs 4) 12 yrs 5) None of these

Average 8. The average weight of 8 persons is increased by 2.5 kg when one of them who weighs 56 kg is replaced by a new man. The weight of the new man is 1) 73 kg 2) 72 kg 3) 75 kg 4) 80 kg 5) None of these 9. The average weight of A, B and C is 84 kg. If D joins the group, the average weight of the group becomes 80 kg. If another man E who weighs 3 kg more than D replaces A, then the average of B, C, D and E becomes 79 kg. What is the weight of A? 1) 64 kg 2) 72 kg 3) 75 kg 4) 80 kg 5) None of these 10. The average of 11 results is 50. If the average of first 6 results is 49 and that of last 6 is 52, find the 6th result. 1) 50 2) 52 3) 56 4) 60 5) None of these 11. A man drives to his office at 60 km/hr and returns home along the same route at 30 km/hr. Find the average speed. 1) 50 km/hr 2) 45 km/hr 3) 40 km/hr 4) 55 km/hr 5) None of these 12. Find the average of five consecutive even numbers a, b, c, d and e. ac cd 1) 2) c 3) 2 2 2bc ce 4) 5) 5 2 1 13. A man covers of his journey by train at 60 km/hr,, 3 1 next by bus at 30 km/hr and the rest by cycle at 10 3 km/hr. Find his average speed during whole journey. 1 2) 33 km/hr 3) 20 km/hr 3 4) 50 km/hr 5) None of these 14. The average marks in English of a class of 24 students is 56. If the marks of three students were misread as 44, 45 and 61 in lieu of the actual marks 48, 59 and 67 respectively, then what would be the correct average? 1) 56.5 2) 59 3) 57.5 4) 58 5) None of these 15. A person travels from P to Q at a speed of 40 kmph and returns to Q by increasing his speed by 50%. What is his average speed for both the trips? 1) 36 kmph 2) 45 kmph 3) 48 kmph 4) 50 kmph 5) None of these 16. In a class, the average age of both male and female

219

17.

18.

19.

20.

1) 30 km/hr

21.

22.

students together is 18 years. The total age of the 15 female students is 240. How many male students are definitely there in the class? 1) 20 2) 40 3) 30 4) Data provided are inadequate to answer the question 5) 25 In a yoga class there were 12 members . Two members left the class and 4 new members joined. If the average age decreased by 4 years and the total age decreased by 2 years, what is the new average age of the class? (in years) 1) 22 2) 27 3) 23 4) 28 5) 18 A professional institute’s total expenditure on students for a particular course is partly fixed and partly varies linearly with the number of students. The average expense per student is 615 when there are 24 students and 465 when there are 40 students. What is the average expense when there are 60 students? 1) 370 2) 450 3) 350 4) 420 5) 390 The average weight of boys in a class is 45 kg while that of girls is 36 kg. The average weight of the whole class is 42.25 kg. What is the respective ratio between the number of boys and girls in the class ? 1) 11 : 25 2) 25 : 11 3) 25 : 12 4) 12 : 25 5) None of these 15 years ago the average age of a family of four members was 40 years. Two children were born in this span of 15 years. The present average age of the family remains unchanged. Among the two children who were born during the 15 years, if the older child at present is 8 years older than the younger one, what is the ratio of the present age of the older child to the present age of the younger child? 1) 9 : 4 2) 7 : 3 3) 7 : 6 4) 7 : 4 5) 9 : 5 In a primary school the average weight of male students is 65.9 kg and the average weight of female students is 57 kg. If the average weight of all the students (both male and female) is 60.3 kg and the number of male students in the school is 66, what is the number of female students in the school? 1) 154 2) 162 3) 168 4) 180 5) 112 There are three positive numbers. One-third of the average of all the three numbers is 8 less than the value of the highest number. The average of the lowest and the second lowest number is 8. What is the highest number?

Quicker Maths

220 1) 11 2) 14 3) 10 4) 9 5) 13 23. A batsman played three matches in a tournament. The ratio of the score of 1st to 2nd match was 8 : 9 and that of the score of 2nd to 3rd match was 3 : 2. The difference between the 1st and the 3rd match was 16 runs. What was the batsman’s average score in all the three matches? 1 1) 40 2) 58 3) 60 4 1 1 4) 45 5) 61 5 3 24. Three Science classes A, B and C take a Life Science test. The average score of students of class A is 83.

The average score of students of class B is 76. The average score of class C is 85. The average score of class A and B is 79 and the average score of class B and C is 81. Then the average score of classes A, B and C is 1) 80 2) 80.5 3) 81 4) 81.5 5) None of these 25. The average weight of boys in a class of students is 58 kg, while that of girls is 50 kg. The average weight of the entire class is 53 kg. The number of girls is approximately what per cent of the number of boys in the class? 1) 167 2) 178 3) 69 4) 100 5) 129

ANSWERS 1. 3; Average of combined group 60 55 40 45 51 60 40 4 75 80 76 2. 2; 5 3. 3; Sum of first 61 natural numbers

Average of new students

61(61 1) 2

61(62) Average 2 61 31 By Direct Formula:

The average of first ‘n’ natural numbers is

n1 . 2

61 1 31 2 4. 2; Let the total number of candidates = x 50x 100(90 – 60) 45 x

Hence in this case, average

x 600 1 16 28 2 58 4 5. 2; 24 14 6. 4; 51 × 5.05 – 50 × 5 = 7.55 gm. By Direct Formula: Wt of bag = Old average + Increase in average (Total no. of objects) = 5 + 0.05 (51) = 5 + 2.55 = 7.55 gm 7. 1; Total age of 10 students = 15 × 10 = 150 yrs Total age of 15 students = 15 × 16 = 240 yrs

240 150 18 yrs. 5

8. 5; 56 + 8 × 2.5 = 76 yrs 9. 3; A + B + C = 3 × 84 = 252 kg A + B + C + D = 4 × 80 = 320 kg D = 320 – 252 = 68 kg E = 68 + 3 = 71 kg 320 A 71 79 Now, 4 A = 75 kg 10. 3; 6 × 49 + 6 × 52 – 11 × 50 = 294 + 312 – 550 = 56 11. 3; By Direct Formula: 2 60 30 2 60 30 40 km / hr Average 60 30 90 12. 3; Average of five consecutive even numbers or odd numbers is the middle term. In this case, the average is c. 3 60 30 10 13. 3; Average 60 30 60 10 30 10 3 60 30 10 20 km / hr 2700 14. 5; Total marks = 24 × 56 = 1344 Total of actual marks = 1344 – (44 + 45 + 61) + (48 + 59 + 67) = 1368 1368 57 Actual Average = 24 Quicker approach: Instead of finding the total marks, we should calculate the deviation in calculation of marks.

Average

221

Here the deviation is (48 – 44) + (59 – 61) + (67 – 61) = 4 + 14 + 6 = 24 24 1 24 Correct average is 56 + 1 = 57 15. 3; Speed of man from P to Q = 40 kmph

Deviation in average is

Speed of man from Q to P =

40 150 = 60 kmph 100

2 40 60 = 48 kmph 40 60 Alternative Method: Suppose the distance PQ = QP = LCM of 40 and 60 = 120 km.

Average speed =

Time taken from P to Q =

120 = 3 hrs and 40

120 time taken from Q to P = = 2 hrs 60 240 Average speed = km = 48 km/hr 3 2 16. 4; Suppose there are 'M' male students and their average is x years. According to the question, Total age of 15 female students = 240 Now, total age of (male + female) students in the class = (M + 15) × 18 = Mx + 240 Hence we can’t determine the no. of male students in the class. 17. 3; Let the new average age of the class be x years. Now, according to the question, (x + 4) × 12 = x × 14 + 2 12x + 48 = 14x + 2 14x – 12x = 48 – 2 = 46 2x = 46 x = 23 years 18. 5; Let the partly fixed expenditure be x. And that partly varying be y. Then, x + 24y = 615 × 24 ... (i) Again, x + 40y = 465 × 40 ... (ii) Solving equations (i) and (ii), we get

Putting the value of y in equation (i), we get x = 24(615 – 240) = 24 × 375 = 9000 Now, when there are 60 students 9000 240 60 Average expenditure = 60 9000 14400 23400 = = = 390 60 60 19. 2; Let the number of boys and girls be x and y respectively. Now, according to the question,

45 x 36 y = 42.25 xy 45x + 36y = 42.25x + 42.25y 45x – 42.25x = 42.25y – 36y 2.75x = 6.25y

x 6.25 25 y 2.27 11 Quicker Method (By Alligation) : Apply alligation on average weight

= 625 : 275 = 25 : 11 20. 2; Fifteen years ago the average age of a fourmember family was 40 years. Then, total present age of the four members = 4 × 40 + 15 × 4 = 160 + 60 = 220 Now, two children are born in a span of 15 years. Then, the total present age of the family of six members = 6 × 40 = 240 years Now, sum of the present ages of the two children = 240 – 220 = 20 years Given that difference of their ages = 8 yrs Older child's age =

20 8 14 yrs 2

20 8 6 yrs 2 reqd ratio = 14 : 6 = 7 : 3 Direct Method:

Younger child's age =

y=

3840 = 240 16

Reqd ratio =

20 8 28 7 = = =7:3 20 8 12 3

Quicker Maths

222 21. 5; Let the number of female students be x. Then, 60.3(66 + x) = 66 × 65.9 + 57x or, 60.3x – 57x = 66 × 65.9 – 66 × 60.3 or, 3.3x = 66(65.9 – 60.3) or, 3.3x = 66 × 5.6 x=

3B ... (i) 4 Again, 76B + 85C = 81B + 81C or, 4C = 5B 5B C= ... (ii) 4

A=

66 5.6 = 2 × 56 = 112 3.3

Now, average =

Quicker Method (By Alligation):

33 : 56 If male = 66, female = 56 × 2 = 112 22. 1; Let the three positive numbers in increasing order

=

83

83A 76B 85C ABC

3B 5B 76B 85 4 4 3B 5B B 4 4

249B 304B 425B 978 = = 81.5 3B 4B 5B 12

Quicker Method (By Alligation):

xyz 3 z 8 be x, y and z then, 3 xyz z8 9 x + y + z = 9z – 72

... (i)

xy 8 x + y = 16 ... (ii) 2 Put x + y = 16 in (i), we have, 16 + z = 9 z – 72 8z = 88 z = 11 23. 5; Ratio of 1st Match to 2nd Match is 8 : 9 and 2nd Match to 3rd Match is 3 : 2.

Also,

A : B : C = 3 : 4 : 5 (This ratio represents the ratio of the number of students in classes A, B and C, Reqd average (83 3) (76 4) (85 5) 978 = = 81.5 3 45 12 25. 1; Let the number of boys in the class be x and that of girls be y.

=

Then,

x 58 50 y = 53 xy

or, 58x + 50y = 53x + 53y or, 5x = 3y

x 3 y 5

Since 1st – 3rd = 8 – 6 16 or 2 16

Reqd % =

16

Total runs 23 2 23 184 184

5 500 2 100 166 % 167% 3 3 3

Quicker Method (By Alligation):

1

reqd average = 3 61 3 24. 4; Let there be A students in Class A, B students in Class B and C students in Class C. According to the question, 83A + 76B = 79A + 79B or, 4A = 3B

reqd % =

5 500 100 = 167% 3 3

Chapter 22

Problem Based on Ages To solve the problems based on ages, students require the knowledge of linear equations. This method needs some basic concepts as well as some more time than it deserves. Sometimes it is easier to solve the problems by taking the given choices in account. But this hit-and-trial method proves costly sometimes, when we reach our solution much later. We have tried to evaluate some easier as well as quicker methods to solve this type of questions. Although, we are not able to cover each type of questions in this section, our attempt is to minimise your difficulties.

Have a look at the following questions Ex. 1: The age of the father 3 years ago was 7 times the age of his son. At present, the father’s age is five times that of his son. What are the present ages of the father and the son? Ex. 2: At present, the age of the father is five times the age of his son. Three years hence, the father’s age would be four times that of his son. Find the present ages of the father and the son. Ex. 3: Three years earlier, the father was 7 times as old as his son. Three years hence, the father’s age would be four times that of his son. What are the present ages of the father and the son?

By the conventional method: Soln: 1.Let the present age of son = x yrs Then, the present age of father = 5x yrs 3 years ago, 7 (x – 3) = 5x – 3 or, 7x – 21 = 5x – 3 or, 2x = 18 x = 9 yrs Therefore, son’s age = 9 yrs Father’s age = 45 yrs Soln:2. Let the present age of son = x yrs Then, the present age of father = 5x yrs 3 yrs hence, 4(x + 3) = 5x + 3 or, 4x + 12 = 5x + 3 x = 9 yrs. Therefore, son’s age = 9 yrs and father’s age = 45 yrs Soln. 3. Let the present age of son = x yrs and the present age of father = y yrs

3 yrs earlier, 7 (x – 3) = y – 3 or, 7x – y = 18 ...(1) 3 yrs hence, 4(x + 3) = y + 3 or, 4x + 12 = y + 3 or, 4x – y = –9 ...(2) Solving (1) & (2) we get, x = 9 yrs & y = 45 yrs Quicker Method: Soln. 1:Son’s age =

3 (7 1) 9 yrs 7 5

and father’s age = 9 × 5 = 45 yrs. Undoubtably you get confused with the above method, but it is very easy to understand and remember. See the following form of question: Q: t1 yrs earlier the father’s age was x times that of his son. At present, the father’s age is y times that of his son. What are the present ages of the son and the father?

Son' s age Soln. 2:Son’s age =

t1 (x 1 ) x y

(4 1) 3 = 9 yrs 54

and father’s age = 9 × 5 = 45 yrs To make more clear, see the following form: Q: The present age of the father is y times the age of his son. t2 yrs hence, the father’s age become z times the age of his son. What are the present ages of the father and his son?

Son' s age Soln. 3:Son’s age

(z 1 )t 2 yz

3(4 1) 3(7 1) 9 18 9 yrs 74 3

To make the above formula clear, see the following form of question: Q: t1 yrs earlier, the age of the father was x times the age of his son. t2 yrs hence, the age of the father becomes z times the age of his son. What are the present ages of the son and the father?

Quicker Maths

224

Son' s age

t 2 (z 1 ) t1(x 1 ) (x z)

In the tabular form the above three types can be arranged as: t1yrs earlier Father’s age that of Son’s

x times

Son' s age

P resent y times

t2 yrs hence z times

t1 (x 1) (z 1)t 2 Son' s age x–y (y – z)

Son' s age

t 2 (z 1) t 1 ( x 1) (x – z)

Ex. 4: The age of a man is 4 times that of his son. 5 yrs ago, the man was nine times as old as his son was at that time. What is the present age of the man? Soln: By the table, we see that formula (1) will be used. 5(9 1) Son’s age = (9 4) = 8 yrs Father’s age = 4 × 8 = 32 yrs Note: The relation between ‘earlier’ and ‘present’ ages are given; so we look for the formula derived from the two corresponding columns of the table. That gives the formula (1). Ex. 5: After 5 yrs, the age of a father will be thrice the age of his son, whereas five years ago, he was 7 times as old as his son was. What are their present ages? Soln: Formula (3) will be used in this case. So 5(7 1) 5(3 1) Son’s age = = 10 yrs 73 From the first relationship of ages, if F is the age of the father then F + 5 = 3(10 + 5) F = 40 yrs Ex. 6: 10 yrs ago, Sita’s mother was 4 times older than her daughter. After 10 yrs, the mother will be two times older than the daughter. What is the present age of Sita? Soln: In this case also, formula (3) will be used. Daughter’s age =

10(4 1) 10(2 1) 20 yrs 42

Ex. 7: One year ago the ratio between Samir’s and Ashok’s age was 4: 3. One year hence the ratio of their ages will be 5 : 4. What is the sum of their present ages in years?

Soln:

One year ago, Samir’s age was age. One year hence, Samir’s age will be

4 of Ashok’ss 3 5 of Ashok’ss 4

age. Ashok’s age (by formula (3)); 4 5 1 1 1 1 1 1 3 4 3 4 A 7 yrs 4 5 1 3 4 12 Now, by the first relation: (S 1) 4 7 1 3 S = 8 + 1 = 9 yrs. Total of ages = A + S = 9 + 7 = 16 yrs where A = Ashok’s present age and S = Samir’s present age Ex. 8: Ten yrs ago, A was half of B in age. If the ratio of their present ages is 3 : 4, what will be the total of their present ages? 1 Soln: 10 yrs ago, A was of B’s age. 2 At present, A is

3 of B’s age. 4

B’s age [use formula (1)] 1 10 1 2 = 1 3 = 20 yrs 2 4 3 A’s age = of 20 = 15 yrs 4 Ex. 9: The sum of the ages of a mother and her daughter is 50 yrs. Also 5 yrs ago, the mother’s age was 7 times the age of the daughter. What are the present ages of the mother and the daughter? Soln: Let the age of the daughter be x yrs. Then, the age of the mother is (50 – x) yrs. 5 yrs ago, 7 (x – 5) = 50 – x – 5 or, 8x = 50 – 5 + 35 = 80 x = 10 Therefore, daughter’s age = 10 yrs and mother’s age = 40 yrs

Quicker Method (Direct Formula): Daughter’s age =

Total ages No. of yrs ago(Times – 1) Times 1

Problem Based on Ages =

50 5(7 1) 10 yrs 7 1

Thus, daughter’s age = 10 yrs and mother’s age = 40 yrs. Ex. 10: The sum of the ages of a son and father is 56 yrs. After 4 yrs, the age of the father will be three times that of the son. What is the age of the son? Soln: Let the age of the son be x yrs. Then, the age of the father is (56 – x) yrs. After 4 yrs, 3(x + 4) = 56 – x + 4 or, 4x = 56 + 4 – 12 = 48 x = 12 yrs. Thus, son’s age = 12 yrs.

Quicker Method (Direct Formula): Son’s age Total ages – No. of yrs after (Time – 1) = Times 1 = Note:

56 – 4(3 1) 48 12 yrs 3 1 4

Do you get the similarities between the above two direct methods? They differ only in signs in the numerator. When the question deals with ‘ago’, a +ve sign exists and when it deals with ‘after’, a -ve sign exists in the numerator. Ex. 11: The sum of the present ages of the father and the son is 56 yrs. 4 yrs hence, the son’s age will 1 be that of the father. What are the present 3 ages of the father and the son? 1 Soln: Son’s age is that of father.. 3 Father’s age is 3 times that of son. Now we use the formula as in Ex.10. Try it. Important Note: We can solve the above question without changing the form of ‘times’. When the question remains in its original form, we find the age of father directly as: Father’s age

1 56 4 1 56 8 3 3 176 44 yrs 4 4 1 1 3 3 And hence we may get the age of son. Thus, this reveals an important fact that in each of the examples from 1 to 11, we may get the second age directly by inverting the ratio. For example,

225 for 4 times we may use

1 4 , for we may use 4 3

3 , and so on. Try to solve each of the above 4 examples by inverting the ratio. Ex. 12: The ratio of the father’s age to the son’s age is 4 : 1. The product of their ages is 196. What will be the ratio of their ages after 5 years? Soln: Let the ratio of proportionality be x, then 4x × x = 196 or, 4x2 = 196 or, x = 7 Thus, Father’s age = 28 yrs, Son’s age = 7 yrs After 5 yrs, Father’s age = 33 yrs. Son’s age = 12 yrs Ratio = 33 : 12 = 11 : 4 Ex. 13: The ratio of Rita’s age to the age of her mother is 3 : 11. The difference of their ages is 24 yrs. What will be the ratio of their ages after 3 yrs? Soln: Difference in ratios = 8 Then 8 24 1 3 ie, value of 1 in ratio is equivalent to 3 yrs Thus, Rita’s age = 3 × 3 = 9 yrs. Mother’s age = 11 × 3 = 33 yrs. After 3 years, the ratio = 12 : 36 = 1 : 3 Ex. 14: The ratio of the ages of the father and the son at present is 6 : 1. After 5 years, the ratio will become 7 : 2. What is the present age of the son? Soln: Father : Son Present age = 6 : 1 After 5 yrs = 7 : 2

Son’s age = 1

5(7 2) 5 yrs. 6 2 7 1

Father’s age = 6

5(7 2) = 30 yrs. 6 2 7 1

Then what direct formula comes? Father : Son Present age = x : y After T yrs = a : b Then, Son’s age T(a b) = y difference of cross product and, Father’s age T(a b) = x difference of cross product

Quicker Maths

226 Ex. 15:The ratio of the ages of the father and the son at present is 3 : 1. 4 years earlier, the ratio was 4 : 1. What are the present ages of the son and the father? Soln: Father : Son Present age = 3 : 1 4 yrs before = 4 : 1 Son’s age = 1

4(4 1) 12 yrs. 4 1 3 1

Father’s age = 3

4(4 1) 36 yrs. 4 1 31

Now, the direct formula comes as: Father : Son Present age = x : y T yrs before = a : b Then, Son’s age

T(a b) = y difference of cross product and, Father’s age

T(a b) = x difference of cross product Note:

1. While evaluating the difference of crossproduct, always take the +ve sign. 2. Both the above direct formulas look similar. The only difference you can find is in the denominators. But it has been simplified as “difference of cross-products” to make it easier to remember. So, with the help of one formula only you can solve both the questions. 3. The above two questions (Q. 14 and Q. 15) are similar to the questions discussed earlier (Q. 1 & Q. 2). Do you get the point? If you change ‘ratio’ in ‘times’ you will get the same thing. Now, you have two simple methods. Use the method which you feel easier to remember. Thus, you may solve Q. 14 as Q. 2 and vice versa. Q. 15 as Q. 1 and vice versa. 4. We suggest you to go through both the methods and choose the better of the two. Ex. 16: A man’s age is 125% of what it was 10 years 1 ago, but 83 % of what it will be after 10 years. 3 What is his present age?

Soln:

Detail Method: Let the present age be x yrs. Then 1 125% of (x – 10) = x; and 83 % of (x + 10) = x 3 1 125% of (x – 10) = 83 % of (x 10) 3 5 5 (x 10) (x 10) 4 6

or,

5 5 50 50 x x 4 6 6 4

or,

5x 250 12 12

x 50 yrs Direct Method: With the help of the above detail method, we can define a general formula as: Present age

1 125 No. of yrs ago 83 No. of yrs after 3 = 1 125 – 83 3 375 250 1 125 10 83 10 10 3 3 1 375 250 125 83 3 3

10 625 50 yrs 125

Try Yourself 1. The ratio of the ages of Geeta and her mother is 1 : 5. After 7 years, the ratio becomes 3 : 8. What are the present ages of Geeta and her mother? (Ans.: 5 yrs, 25 yrs) 2. The ratio of the ages of A and B is 3 : 11. After 3 years, the ratio becomes 1 : 3. What are the ages of A and B? (Ans.: 9 yrs, 33 yrs) 3. The ratio of the ages of Mohan and Meera is 3 : 4. Four years earlier, the ratio was 5 : 7. Find the present ages of Mohan and Meera. (Ans.: 24 yrs, 32 yrs)

Problem Based on Ages

227

EXERCISES 1. The age of the father is 30 years more than the son’s age. Ten years hence, the father’s age will become three times the son’s age that time. What is the son’s present age in years? 1) 8 2) 7 3) 5 4) Cannot be determined 5) None of these 2. The ratio of the present age of Manisha and Deepali is 5 : X. Manisha is 9 years younger than Parineeta. Parineeta’s age after 9 years will be 33 years. The difference between Deepali’s and Manisha’s age is the same as the present age of Parineeta. What should come in place of X? 1) 23 2) 39 3) 15 4) Cannot be determined 5) None of these 3. The sum of the ages of 4 members of a family 5 years ago was 94 years. Today, when the daughter has been married off and replaced by a daughter-inlaw, the sum of their ages is 92. Assuming that there has been no other change in the family structure and all the people are alive, what is the difference between the age of the daughter and that of the daughter-inlaw? 1) 22 years 2) 11 years 3) 25 years 4) 19 years 5) 15 years 4 Rita’s present age is four times her daughter’s present age and two-thirds of her mother’s present age. The total of the present ages of all of them is 154 years. What is the difference between Rita’s and her mother’s present age? 1) 28 years 2) 34 years 3) 32 years 4) Cannot be determined 5) None of these 5. Farah was married 8 years ago. Today her age is 1

2 7

times that at the time of marriage. At present her 1 of her age. What was her 6 daughter’s age 3 years ago? 1) 6 years 2) 7years 3) 3 years 4) Cannot be determined 5) None of these 6. The respective ratio between the present ages of Parag and Sapna is 21 : 19. Six years ago, the respective ratio between their ages was 9 : 8. How old is Lina if her present age is 12 years less than Sapna’s present age ?

daughter’s age is

1) 38 years 2) 28 years 3) 26 years 4) 30 years 5) 42 years 7. 8 years ago, Jyoti’s age was equal to the Swati’s present age. If the sum of Jyoti’s age 10 years from now and Swati’s age 6 years ago is 88 years. What was Kusum’s age 14 years ago if Kusum is 8 years younger to Swati? (in years) 1) 22 2) 14 3) 25 4) 24 5) 16 8. The ratio of the present ages of A and B is 7 : 9. Six 1 1 of A’s age at that time and 3 3 of B’s age at that time was 1 : 2. What will be the ratio of A's to B's age 6 years from now? 1) 4 : 5 2) 14 : 15 3) 6 : 7 4) 18 : 25 5) 22 : 25 9. The ratio of present ages of P and Q is 8 : 5. After 4 years their ages will be in the ratio 4 : 3 respectively. What will be the ratio of P’s age after 7 years from now and Q’s age now ? 1) 3 : 2 2) 1 : 2 3) 2 : 1 4) 3 : 1 5) None of these 1 10. 4 years ago, the ratio of 2 of A’s age at that time and

years ago the ratio of

four times of B’s age at that time was 5 : 12. Eight 1 of A’s age at that time will be less 2 than B’s age at that time by 2 years. What is B’s present age? 1) 10 years 2) 14 years 3) 12 years 4) 5 years 5) 8 years 11. The present age of Bob is equal to Abby’s age 8 years ago. Four years hence, the ratio of Abby’s age to Bob’s age will be 5 : 4. What is Bob’s present age? 1) 24 years 2) 32 years 3) 40 years 4) 20 years 5) 28 years 12. At present, the ratio of the ages of A to B is 3 : 8; and that of A to C is 1 : 4. Three years ago, the sum of the ages of A, B and C was 83 years. What is the present age (in years) of C? 1) 32 2) 12 3) 48 4) 54 5) 15 13. B is 3 years older than A and B is also 3 years younger than C. 3 years hence, the respective ratio between

years hence,

Quicker Maths

228 the ages of A and C will be 4 : 5. What is the sum of the present ages of A, B and C ? 1) 48 years 2) 56 years 3) 63 years 4) 84 years 5) 72 years 14. At present, Priya is 6 years older than Ray. The ratio of the present ages of Priya to Mini is 3 : 4. At present Ray is 14 years younger than Mini. What is Ray’s present age? 1) 16 years 2) 20 years 3) 14 years 4) 18 years 5) 24 years 2 of his father’s present age. 7 Joe’s brother is 3 years older then Joe. The ratio of the present age of Joe’s father to that of Joe’s brother is 14 : 5. What is Joe’s present age? 1) 6 years 2) 15 years 3) 12 years 4) 18 years 5) 20 years 16. Five years ago, the ratio of the age of Opi to that of Mini was 5 : 3. Nikki is 5 years younger than Opi. Nikki is five years older than Mini. What is Nikki’s present age? 1) 35 years 2) 25 years 3) 20 years 4) 10 years 5) 30 years 17. At present, Akki is seven years younger then Binny. Binny’s age sixteen years hence will be equal to twice that of Akki two years ago. What will be the sum of their present ages? 1) 72 years 2) 61 years 3) 54 years 4) 62 years 5) 46 years

1 of Amit’s age at 3 that time. The ratio of Amit’s age six years hence to Somi’s age twelve years hence will be 7 : 4. What was Somi’s age three years ago? (in years) 1) 13 2) 29 3) 17 4) 25 5) 27 Two years ago, the ratio of A’s age at that time to B’s age at that time was 5 : 9. A’s age three years ago was 13 years less than B’s age six years ago. What is B’s present age? 1) 38 years 2) 30 years 3) 34 years 4) 32 years 5) 36 years A’s age eight years ago was equal to twice B’s age two years ago. C is six years older than B. If the ratio of the present age of A to that of C is 8 : 5, then what is B’s present age? (in years) 1) 18 2) 12 3) 15 4) 20 5) 14 B is eighteen years younger than A. The ratio of B’s age six years hence to C’s present age is 3 : 2. If at present A’s age is twice the age of C, then what was B’s age four years ago? 1) 24 years 2) 28 years 3) 26 years 4) 20 years 5) 16 years The sum of the present ages of A, B, C and D is 76 years. After 7 years the ratio of their ages becomes 7 : 6 : 5 : 8. What is C’s present age? 1) 14 2) 12 3) 13 4) 8 5) 10

18. Five years ago, Somi’s age was

19.

15. Joe’s present age is

20.

21.

22.

Solutions 1. 3; Let the son’s present age be x years. Then the father’s present age is (x + 30) years. Father’s age after 10 years = (x + 40) years Son’s age after 10 years = (x + 10) years (x + 40) = 3(x + 10) or, x + 40 = 3x + 30 or, 2x = 10 x=5 2. 5; Parineeta’s present age = (33 – 9 =) 24 yrs. Manisha’s present age = (24 – 9 =) 15 yrs Deepali’s present age = 15 + 24 = 39 yrs. Ratio of the present age of Manisha and Deepali = 15 : 39 = 5 : 13 X = 13 3. 1; There are four members in a family. five years ago the sum of ages of the family members = 94 years

Now, sum of present ages of family members = 94 + 5 × 4 = 114 years Daughter is replaced by daughter-in-law. Thus, sum of family member’s ages becomes 92 years. Difference = 114 – 92 = 22 years 4. 1; Let Rita's present age be x years. Her daughter’s age =

x years 4

3 x years. 2 Now, total sum of ages of Rita, her daughter and her mother = 154 x 3 or, x x 154 4 2

Her mother’s age =

Problem Based on Ages

229

4x x 6x 154 4 or, 11x = 154 × 4 x = 56 years. or,

Rita’s daughter’s age =

56 14 years 4

3 56 = 84 years 2 Difference = 84 – 56 = 28 years Quicker Method: Suppose age of Rita = x. Then, the ratio of the ages of Rita, Rita’s mother’s age =

x 3 : x 4 2 =4:1:6 Now, we have 4 + 1 + 6 = 11 154 yrs

6. 3; Let the present age of Parag and Sapna be 21x and 19x respectively. Six years ago, their age was 21x – 6 and 19x – 6 years. Now, according to the question, 21x 6 9 19x 6 8 168x – 48 = 171x – 54 3x = 6 x = 2 Sapna’s present age =19 × 2 = 38 years Lina’s age = 38 – 12 = 26 years Quicker Method:

daughter and mother is R : D : M = x :

154 2 28 yrs. 11 5. 3; Let Farah’s age 8 years ago be x years. Farah’s present age = (x + 8) years Now, according to the question, 9x 7 x 56 9 x x+8= 7 2x = 56 x = 28 Farah’s present age = (28 + 8 =) 36 years Her daughter’s age 3 years ago 1 = 36 × –3 years 6 Hence 3 years age = 6 – 3 =) = 3 years Quicker Approach:

6–4=2

2 times of her age at the 7 time of marriage (ie 8 yrs ago).

Farah's present age is 1

Her present age is

2 times more than her age 7

8 years ago. 2 of her age at the time of marriage = 8 yrs 7 Her age at the time of marriage 7 = 8 × = 28 yrs 2 Her present age = 28 + 8 = 36 yrs 36 Present age of her daughter = = 6 yrs 6 Her daughter's age 3 yrs ago = 3 yrs

In ratio terms 3 6 yrs in ratio terms 19 38 yrs Lina's age = 38 – 12 = 26 yrs Note : 9 : 8 is multiplied by the difference of 21 and 19 (the other ratio terms) and 21 : 19 is multipled by the difference of 9 and 8. By doing this, we make difference in the respective terms of ratio equal (ie 3 in this case) 7. 5; Let the Swati’s present age be x years. Jyoti’s present age = (x + 8) years Now, according to question, x + 8 + 10 + x – 6 = 88 2x + 12 = 88 2x = 88 – 12 = 76 76 = 38 years 2 Kusum’s present age = (38 – 8 =) 30 years Kusum’s age 14 years ago = (30 – 14 =) 16 years 8. 3; Let the present age of A be 7x years and that of B be 9x years. 3(7 x – 6) 1 Now, 6 years ago, 3(9 x – 6) 2 14x – 12 = 9x – 6 or, 5x = 6

x=

6 years 5 Ratio after 6 years 76 6 42 30 72 5 6:7 6 9 6 54 30 84 5 Reqd ratio = 6 : 7

x=

Quicker Maths

230 9. 4; Let P’and Q’s present ages be 8x and 5x years respectively. 8x 4 4 5x 4 3 24x + 12 = 20x + 16 24x – 20x = 16 – 12 4x = 4 x = 1 P’s age 7 years hence = 8x + 7 = 8 + 7 = 15 years Required ratio = 15 : 5 = 3 : 1 Quicker Method:

After 4 years,

in ratio terms 4 = 4 yrs P’s present age = 8 yrs and Q’s present age = 5 yrs required ratio = 8 + 7 : 5 = 15 : 5 = 3 : 1 10. 1; Let the present age of A be x years and that of B be y years. Then, 4 years ago, A’s age = (x – 4) years B’s age = (y – 4) years Now, according to the question, x–4 5 2 4(y – 4) 12

92 12 = 48 yrs 23 13. 5; From the question, B =A+ 3 A= B – 3 and B = C – 3 C = B + 3 Again, after 3 years,

12

B 4 B6 5 5B = 4B + 24 5B – 4B = 24 B = 24 A + B + C = B – 3 + B + B + 3= 3B = 3 × 24 = 72 years 14. 4; Let the present age of Ray be x years. Then, Priya’s age = (x + 6) years Ray is 14 years younger than Mini. Mini’s age = (x + 14) years

x–4 5 or, 4y –16 6

or, 6x – 24 = 20y – 80 or, 6x – 20y = –56 or, 10y – 3x = 28 ... (i) After 8 years,

x 8 2 y8 2 x 4 2 y 8 2

x = –2 2 or, 2y – x = –4 or, x = 2y + 4

y4 5 x4 4 or, 4y + 16 = 5x + 20 or, 4y + 16 = 5(y – 8) + 20 or, 4y + 16 = 5y – 40 + 20 or, y = 36 years Hence Bob’s present age = 36 – 8 = 28 years 12. 3; According to the question, A:B=3:8 A : C = 1 : 4 = 3 : 12 A : B : = 3 : 8 : 12 Given, A + B + C = 23 (83 + 9 =) 92 yrs

B33 4 B33 5

x–4 5 or, 2(4y –16) 12

or,

Putting the value of x in equation (i), we get 10y – 3(2y + 4) = 28 or, 10y – 6y – 12 = 28 or, 4y = 40 y = 10 Hence the present age of B is 10 years. 11. 5; Let the present age of Bob be x years and that of Abby be y years. Then, x = y – 8 ... (i) Now, four years hence

x6 3 x 14 4 or, 4x + 24 = 3x + 42 or, x = 18 years. Hence Ray’s present age = 18 years

Now,

or, y –

... (ii)

Problem Based on Ages

231

15. 3; Let the present age of Joe’s father be x years. Then, Joe’s age =

2x years 7

2x 2 x 21 3 Joe’s brother’s age = 7 7 Now, ratio of the present age of Joe’s father to

that of Joe’s brother = or,

7x 14 2 x 21 5

x 2 2 x 21 5

or, x = 42 years Joe’s present age = 42

2 = 12 years 7

16. 2; According to the question: Let the present age of Opi be (5x + 5) years and that of Mini be (3x + 5) years. Nikki’s age = 5x + 5 – 5 = 5x ... (i) Nikki’s age as compared to Mini’s = 3x + 5 + 5 = 3x + 10 ... (ii) Now, equating (i) and (ii), we get 3x + 10 = 5x x = 5 years Now, Nikki’s present age = 5x = 5 × 5 = 25 years Alternative Approach: O5 5 ... (i) M5 3 O – N = 5 O – 5 = N ... (ii) N – M = 5 M + 5 = N M – 5 = N – 10 ... (iii) Putting values of O – 5 and M – 5 in (i), we get

2 10 yrs N = 5 25 yrs 17. 2; Let Binny’s present age be x years. Then, Akki’s age = (x – 7) years 16 years hence Binny’s age = (x + 16) years 2 years ago, Akki’s age = x – 7 – 2 = (x –9) years Now, (x + 16) = 2(x – 9) or, x + 16 = 2x – 18 or, x = 34 years Hence Binny’s age = 34 years Akki’s age = 34 – 7 = 27 years Sum of the present ages of Akki and Binny = 34 + 27 = 61 years

18. 3; Five years ago, let Amit’s age be 3x years. Then, Somi’s age = x Now, according to the question, 3x 5 6 7 x 5 12 4

or,

3 x 11 7 x 17 4

or, 12x + 44 = 7x + 119 or, 5x = 75 x = 15 Somi's age 3 yrs ago = 15 + 2 = 17 yrs 19. 1; Let two years ago, A’s age be 5x and B’s age be 9x. Now, according to the question, 3 years ago A’s age = 13 years less than B’s age six years ago or, 5x – 1 = (9x – 4) – 13 or, 5x = 9x – 4 – 13 + 1 = 9x – 16 or, 4x = 16 x=4 Present age of B = 9x + 2 = 9 × 4 + 2 = 38 years 20. 5; Let the present age of A, B and C be x, y and z years respectively. Then, A B C x y z=y+6 x – 8 = 2y – 4 x = 2y + 4 Now,

x 8 y6 5

2y 4 8 or, y 6 5 y2 4 or, y 6 5 or, 5y + 10 = 4y + 24 y = 24 – 10 = 14 years 21. 3; A B Present age 2x 2x – 18 Now, after 6 years

C x

2 x 18 6 3 x 2

4x – 24 = 3x x = 24 yrs B’s age four years ago = 2x – 18 – 4 = 48 – 22 = 26 yrs

Quicker Maths

232 22. 3; After 7 yrs, let their ages be 7x, 6x, 5x and 8x years. Then, sum of their ages at present = (7x + 6x + 5x + 8x) – 7 × 4 = 76 (given) 26x = 76 + 28 = 104 x = 4 C’s age = 5 × 4 – 7 = 13 yrs Note: In exam, don't assume the values in terms of x.

Simply move with terms of the ratio. We have, (7 + 6 + 5 + 18=) 26 (76 + 28 =) 104 yrs 104 5 20 yrs 26 C's present age = 20 – 7 = 13 yrs. This will save your precious time.

C's age after 7 yrs, 5

Chapter 23

Profit and Loss In this chapter, the use of “Rule of Fraction” is dominant. We should understand this rule very well because it is going to be used in almost all the questions.

Ex.3:

The Rule of Fraction says

Soln:

d If our required value is greater than the supplied value, we should multiply the supplied value with a fraction which is more than one. And if our required value is less than the supplied value, we should multiply the supplied value with a fraction which is less than one. (1) If there is a gain of x%, the calculating figures would be 100 and (100 + x). (2) If there is a loss of y%, the calculating figures would be 100 and (100 – y). (3) If the required value is more than the supplied value, our multiplying fractions should be

CP of 110 oranges =

11 100 = 121 10 Profit = 121 – 100 = 21

and % profit =

The profit or loss is generally reckoned as so much per cent on the cost. Gain or loss per cent =

Loss or gain 100 CP

Ex. 1.

A man buys a toy for 25 and sells it for 30. Find his gain per cent.

Soln:

% Gain =

Ex. 2.

Soln:

Gain 5 100 100 20% CP 25 A boy buys a pen for 25 and sells it for 20. Find his loss per cent.

% Loss =

Loss 5 100 100 = 20% CP 25

Profit 21 100 21% ×100 = 100 CP

Quicker Maths: Rewrite the statements as follows: Purchases 11 oranges for 10 Sells 10 oranges for 11 Now, percentage profit or loss is given by:

(4) If the required value is less than the supplied value, our multiplying fractions should be

To find the gain or loss per cent

10 110 = 100 11

SP of 110 oranges =

100 x 100 or (both are greater than 1). 100 100 y

100 100 y or (both are less than 1). 100 x 100 Profit = Selling Price (SP) – Cost Price (CP) Loss = Cost Price (CP) – Selling Price (SP)

If a man purchases 11 oranges for 10 and sells 10 oranges for 11. How much profit or loss does he make? Suppose that the person bought 11 × 10 = 110 oranges.

11 11 10 10 100 21% 10 10 Since the sign is +ve, there is a gain of 21%. The above form of structural adjustment should be remembered. The first line deals with purchase whereas the second line deals with sales. Once you get familiar with the form, you need to write only the figures and not the letters. Ex.4: A man purchases 8 pens for 9 and sells 9 pens for rupees 8. How much profit or loss does he make? Soln: Quicker Maths: Purchases 8 pens for 9 Sells 9 pens for 8 Note:

% profit or loss =

88 9 9 100 99

1700 20.98% 81 Since the sign is -ve, there is a loss of 20.98%.

Quicker Maths

234 Ex.5:

Soln:

A boy buys oranges at 9 for 16 and sells them at 11 for 20. What does he gain or lose per cent? Quicker Maths:

16

or, at

11

20

11x x 300 Now, % profit = 8 100 37.5% x 8

9 20 16 11 3 100 2 % 16 11 11

3 Since the sign is +ve, there is a gain of 2 %. 11

Dishonest dealer using false weight Ex. 6: A dishonest dealer professes to sell his goods at cost price, but he uses a weight of 960 gm for the kg weight. Find his gain per cent. Soln: Suppose goods cost the dealer 1 per kg. He sells for 1 what cost him 0.96. Gain on 0.96 = 1 – 0.96 = 0.04

Gain on 100

0.04 1 100 = 4 0.96 6

1 Gain % = 4 % 6 Direct formula:

Error %gain = True value – Error 100

True weight – False weight 100 or, % gain = False weight 40 1 100 4 % 1000 – 40 6 A dishonest dealer professes to sell his goods at cost price, but he uses a weight of 950 gm for the kg weight. Find his gain per cent. Direct formula:

=

Soln:

Error %gain = True value – Error 100 50 100 5.26% 950 Ex. 8: A grocer sells rice at a profit of 10% and uses a weight which is 20% less. Find his total percentage gain. Soln: Detail method: Suppose he bought at x/kg.

=

110x 100 11x per kg or, at per kg 100 80 8

9

% profit or loss

Ex.7:

80 110x kg Then, he sells at per 100 100

Quicker Method: Total percentage profit =

% profit % less in wt 100 100 % less in wt

10 20 30 100 100 37.5% 100 20 80

1 Ex. 9: A dishonest dealer sells goods at 6 % loss on 4 cost price but uses 14 gm instead of 16 gm. What is his percentage profit or loss? Soln: Detail method: Suppose the cost price is x per kg. 25 100 4 Then, he sells the goods for x 100 15x per kg 16 Now, suppose he bought y kg of goods. Then, his total investment = xy

=

and his total return =

15x 16 15 y = xy 16 14 14

15 xy xy 50 1 100 7 % his % profit = 14 xy 7 7 Direct Formula: If the shopkeeper sells his goods at x% loss on cost price but uses y gm instead of z gm, then his % profit or loss is

z [100 x] 100 as the sign is +ve or -ve. y In the above case, 1 16 % profit or loss = 100 6 100 4 14

Profit and Loss

375 16 1500 1400 100 100 4 14 14 14

50 1 7 % 7 7

235 100 Cost price of 840 gm = 1000 (840) = 84 For (a), selling price of 840 gm = (100 – 4) = 96 Profit = SP – CP = 96 – 84 = 12

1 Since the sign is +ve, there is a profit of 7 % . 7

Note:

See another form of the above question in Ex. 10.

1 Ex. 10: A dishonest dealer sells the goods at 6 % loss 4

Soln:

1 on cost price but uses 12 % less weight. What 2 is his percentage profit or loss? In this case, we use the direct formula as: Profit or loss percentage

1 25 100 4 100 100 4 100 100 1 25 100 12 100 2 2 100 6

375 4 100 100 175 2

Note:

12 100 100 2 14 %. 84 7 7 For (b), selling price of 840 gm = (100 + 4) = 104 Profit = SP – CP = 104 – 84 = 20

% profit =

20 100 17 23 % 84 21 Quicker Method: There is a general formula for such type of questions. See the two cases separately: Case I: If a seller uses ‘X’ gm in place of one kg (1000 gm) to sell his goods and bears a loss of x% on cost price then his actual gain or loss percentage

% profit

100 is (100 x) 100 according as the sign is X +ve or -ve. Case II: If a seller uses ‘X’ gm in place of one kg (1000 gm) to sell his goods and gains a profit of x% on cost price, then his actual gain or loss percentage 100 is (100 x) 100 according as the sign is X +ve or -ve. Combining the two cases, we have 1000 Gain or loss% = (100 x) 100 X according as the sign is +ve or -ve In the above case x = 4 and X = 840 gm. Therefore,

15 100 1 100 100 7 % 14 14 7

1 Since, sign is +ve, there is a profit of 7 % 7 Ex. 9 and Ex. 10 are the same question. In Ex. 9, he uses 14 gm for 16 gm. This implies that he

16 14 25 1 100 12 % less weight. 16 2 2 Thus, we see that any of the direct formula can be used in both the cases. Ex. 11: A seller uses 840 gm in place of one kg to sell his goods. Find his actual % profit or loss (a) when he sells his article on 4% loss on cost price. (b) when he sells his article on 4% gain on cost price. Soln: Detail Method: Suppose the cost price of 1000 gm is 100.

1000 (a) % loss or gain = (100 4) 100 840

uses

=

96 1000 800 100 2 100 100 14 % 840 7 7 7

2 Since the sign is +ve, there is a gain of 14 % 7

1000 (b) % gain = (100 4) 100 840

104 1000 2600 500 17 100 100 23 % 840 21 21 21

Quicker Maths

236 Another Example A seller used 990 gm in place of one kg to sell the rice. Find his actual profit or loss percentage when he sells (a) On 10% loss on cost price. (b) On 10% profit on cost price. Using the above general formula: 1000 (a) % loss or gain = (100 – 10) 100 990 1000 100 1 100 9 % 11 11 11 1 there is a loss of 9 % 11

1000 (b) % gain = (100 + 10) 100 990

1000 100 1 100 11 % 9 9 9

To find the selling price Ex. 12: A man bought a cycle for 250. For how much should he sell it so as to gain 10%? Soln: If CP is 100, the SP is 110. 110 . If CP is 1, the SP is 100 110 250 If CP is 250, the SP is 275 100 Another suggested method (By Rule of Fraction) If he wanted to sell the bicycle at a gain of 10%, the selling price (required value) must be greater than the cost price (supplied value), so we should multiply 250 with a more-than-one value fraction. Since there is a gain, our calculating figures should be 100 and (100 + 10) and 110 . the fraction should be 100 Thus, selling price = 250

110 = 275. 100

OR, As there is a gain, SP must be greater than CP. 110 250 = 275 100 Ex. 13: A man bought a cycle for 560. For how much shall he sell it so as to lose 10%? Soln: As there is a loss, the SP must be less than CP. So, SP = (100 – 10) % of CP

So, SP = (100 + 10) % of CP =

90 560 = 504 100

By Rule of Fraction: Calculating figures are 100 and (100 – 10) Since the required value is less than 1, the multiplying fraction =

90 . 100

90 = 504 100 Once you understand the theorem well, you need to write only the figures and hence you may save a lot of time.

Thus, selling price = 560 Note:

To find the Cost Price Ex. 14: If by selling an article for 390 a shopkeeper gains 20%, find his cost. Soln: If the SP be 120, the CP is 100 If the SP be 390, the CP is

100 390 = 325 120

By Rule of Fraction: Required value is less than the supplied value; therefore 390 should be multiplied by

100 . 100 20

100 = 325 120 Ex. 15: By selling goods for 352.88, I lost 12%. Find the cost price. Soln: CP should be more than SP; so we multiply SP by CP 390

100 100 (a fraction which is more than 1) 100 12 88 100 = 401 88 Goods passing through successive hands Ex.16: A sells a good to B at a profit of 20% and B sells it to C at a profit of 25% . If C pays 225 for it, what was the cost price for A? Soln: During both the transactions there are profits. So our calculating figures would be 120, 125 and 100. A’s cost price is certainly less than C’s selling price.

CP = 352.88 ×

100 100 = 150 120 125 Remark: Since we need a value which is less than the given value, so our multiplying fractions should be less than one. That is why we multiplied 225

Required price = 225

with

100 100 and . 120 125

Profit and Loss

237

Ex.17: A sells a bicycle to B at a profit of 30% and B sells it to C at a loss of 20%. If C pays 520 for it, at what price did A buy? Soln: In the whole transaction there is a gain of 30% and a loss of 20%, so our calculating figures would be 130, 80 and 100. 100 B’s cost price = 520 80 100 100 A’s cost price = 520 = 500 80 130 Alternative method for Ex. 16 & Ex. 17 (1) When there are two successive profits of x% and y%, then the resultant profit per cent xy is given by x y 100 Thus, for Ex.16, the resultant profit

= 20 + 25 +

20 25 50% 100

100 Thus, CP = 225 = 150 150 (2) When there is a profit of x% and loss of y% in a transaction, then the resultant profit or

xy loss per cent is given by x y 100 according to the + ve and the -ve signs respectively. Thus for Ex.17, the resultant profit or loss

= 30 – 20 –

30 20 = 4% profit, because 100

sign is +ve.

520 100 = 500. required price = 104 Note:

xy The second formula x y is obtained 100

xy from the first x y by putting -y for y.. 100 Can you find the reason? Ex.18: By selling a horse for 570, a tradesman would lose 5%. At what price must he sell it to gain 5%? Soln: (100 – 5)% of the CP = 570

(100 + 5)% of the CP =

570 105 = 630 95

If you don’t want to go into details of the method, you may follow the method of fraction. Our calculating figures are (100 – 5), (100 + 5) and 100. Cost price of horse = 570

100 95

100 105 = 630 95 100 Ex.19: A machine is sold for 5060 at a gain of 10%. What would have been the gain or loss per cent if it had been sold for 4370? Soln: Calculating figures are 110 and 100.

Thus, SP = 570

100 CP = 5060 110 = 4600 2nd SP = 4370

loss%

230 100 5% 4600

Ex. 20: I sold a book at a profit of 12% . Had I sold it for 18 more, 18% would have been gained. Find the cost price. Soln: Here, 118% of cost – 112% of cost = 18 6% of cost = 18 18 100 = 300 6 Quicker Maths: Ignoring the intermediate steps, we have a direct formula for such questions.

cost =

Cost =

More gain × 100 Difference in percentage profit

18 100 Cost = 18 – 12 = 300 Ex.21: A man sold a horse at a loss of 7%. Had he been able to sell it at a gain of 9%, it would have fetched 64 more than it did. What was the cost price? Soln: Here, 109 % of cost – 93% of cost = 64 16 % of cost = 64 64 100 = 400 16 By direct formula:

Cost =

64 100 64 100 = 400 9 (–7) 16 Note: Since 7% loss = (-7) % profit

Quicker Maths

238 Ex. 22: A person sells an article at a profit of 10%. If he had bought it at 10% less and sold it for 3 more, he would have gained 25%. Find the cost price. Soln: Let the actual cost price = 100 Actual selling price at 10% profit = 110 Supposed cost price at 10% less = 90 Supposed selling price at 25% gain 125 = 90 = 112.5 100 the difference in the selling prices = 112.5 – 110 = 2.5 If the difference is 2.5, the CP = 100 If the difference is 3, the CP =

100 3 = 120 2.5

By Rule of Fraction: Let the cost of the article be A. 110 Actual selling price A 100

90 Supposed cost price A 100 90 125 Supposed selling price = A 100 100 Therefore, we find a relationship: 110 90 125 A 3 A 100 100 100

------- (*)

90 125 – 110 100 3 or, A 100 100

A

3 100 100 90 125 110 100

If the cost price is A, the supposed selling price 90 80 140 A 55 A 100 100 100 80 140 100 90 or, A 55 100 100 A

55 100 100 11200 9000

55 100 100 = 250 2200 Note: If we write the direct formula, we will have to keep one thing in mind that for x% loss and y% gain our calculating figures will be (100 – x) and (100 + y). Ex. 24: A man buys an article and sells it at a profit of 20% . If he bought it at 20% less and sold it for 75 less, he would have gained 25%. What is the cost price? Soln: Let the actual cost price = 100 Actual selling price at 20% profit = 120 Supposed cost price at 20% less = 80 Supposed selling price at 25% gain

=

125 = 100 100 the difference in selling price = 120 – 100 = 20 If the difference is 20, the CP = 100 If the difference is 75, the CP

=

80

100 75 = 375 20 By the rule of fraction: Let the cost price be A, then 120 80 125 A 75 A 100 100 100

=

------- (*) (*)

3 100 100 = 120 250 Note: In the above example, the relationship given in (*) should be clear. Both sides of the equation are the supposed selling price of the article. With the help of that equation, we get cost price in (*) (*). If you remember (*) (*), you can save much time. But since the type of question varies frequently, you are suggested to proceed after finding the relationship given in (*). Ex. 23: A person bought an article and sold it at a loss of 10%. If he had bought it for 20% less and sold it for 55 more he would have had a profit of 40%. Find the cost price of the article.

=

Soln:

120 100 80 125 or, A 75 100 100 75 100 100 120 100 80 125 75 100 100 = = 375 2000 Ex. 25: A dealer sold a radio at a loss of 2.5%. Had he sold it for 100 more, he would have gained 7.5%. For what value should he sell it in order to

or, A =

1 gain 12 % ? 2

Profit and Loss Soln:

239

Suppose he bought the radio for x. Then selling price at 2.5% loss 100 2.5 97.5x = x 100 100 and selling price at 7.5% gain 100 7.5 107.5x = x = 100 100

From the question,

20x 24 x 100 or, 4x = 2400 x = 600 Thus, cost of the article = 600 Direct Formula: When cost price and selling price are reduced by the same amount (say A) then Cost price

or,

107.5x 97.5x = 100 100 100

=

In this case,

or, 10x = 100 × 100 x = 1000 Therefore, to gain 12.5%, he should sell it for 100 12.5 1000 = 1125 100 Quicker Method: Selling price

=

More rupees (100 % final gain) %gain %loss

100(112.5) = 1125 7.5 2.5 Ex. 26: An article is sold at a profit of 20%. If both the cost price and selling price are 100 less, the profit would be 4% more. Find the cost price. Soln: Suppose the cost price of that article is x.

=

120 The selling price = x 100 New cost price and selling price is (x – 100)

120 – 100 respectively.. and x 100 New profit =

=

120 x 100 – 100 – (x – 100) 120 20 x 100 1 = x 100

New percentage profit 20 x 20x 100 100 % x 100 x 100 We are also given that the new percentage of profit = 20 + 4 = 24%

[Initial profit % Increase in profit %] A Increase in profit%

(20 4) 100 = 600 4 What happens when the cost price and selling price are reduced by different amounts? For that case, we have derived a general formula. Take the following form of questions: “An article is sold at P% profit. If its CP is lowered by c and at the same time its SP is also lowered by s, then percentage of profit increases by p%. Find the cost price of that article.”

Cost price = Note:

Cost Price =

c(P p) 100(s c) p

Ex: (a) An article is sold at 20% profit. If its CP and SP are less by 10 and 5 respectively, the percentage profit increases by 10%. Find the cost price. Soln: Using the above formula: 10(20 10) 100(5 10) 800 = 80 10 10 Ex: (b) An article is sold at 25% profit. If its CP and SP are increased by 20 and 4 respectively, the percentage of profit decreases by 15%. Find the cost price. Soln: We may use the above formula in this question if we put the +ve and -ve signs correctly. For example, in this case, CP and SP are decreased by (-20) and (-4) respectively whereas % profit increases by (15)%.

Now, CP =

20(25 15) 100 4 (20) 15

200 1600 1800 1800 = 120 15 15 15

Quicker Maths

240 Thus, we see that the above question can be asked in so many ways by changing “increase” into “decrease” and “decrease” into “increase”. If you understand the signs used in the above formula, you can solve all these types of questions very easily. 1 Ex. 27: A person sells his table at a profit of 12 % and the 2 1 chair at a loss of 8 % but on the whole he gains 3 25. On the other hand, if he sells the table at a 1 1 loss of 8 % and the chair at a profit of 12 % 3 2 then he neither gains nor loses. Find the cost price of the table. Soln: Suppose the cost price of a table = T and cost price of a chair = C. 1 1 Then; 12 % of T 8 % of C = 25 2 3

In the first case, suppose the overall profit% is x then Table Chair 25 + – 25 % % 2 3 x 25 25 – x 2 3 ratio of cost price of table and chair

x+

25 25 = x : x 3: 2 3 2

25 3 3 or, 25 x 2 2 50 75 3x or, 2x 3 2 x

1 1 and 8 % of T 12 % of C 0 2 3

or,

or, 5x

125 25 x 30 6 %

25 25 T C 2500 -------- (1) 2 3

25 25 T C0 3 2

25 % 25 6 100% 600

Now,

-------- (2)

(1) ÷ 2 + (2) ÷ 3 gives

25 25 T T 1250 4 9

225 100 or, T 1250 36 T = 360 Price of a table = 360

Cost of a table

We see that the method of alligation becomes lengthy because we are not in a condition to use it directly. If we had the % value of profit in the first case, the method of alligation would have been easier. Ex. 28: An article is sold at 20% profit. If its cost price is increased by 50 and at the same time if its selling price is also increased by 30, the percentage of

By Rule of Alligation: Chair

25 % 2

25 % 3 0 25 2

Ratio of cost of table to chair 25 25 1 1 : : 3: 2 = 2 3 2 3

600 3 = 360 3 2

Note:

OR, In second case: Table

75 50 125 2 3 6

25 3

Soln:

1 profit decreases by 3 % . Find the cost price. 3 Suppose the cost price = x 120 6 x = x 100 5 Now, new CP = (x + 50)

Then, SP =

new SP =

120 100 x 30 =

6 5 x 30

Profit and Loss

241

Now, the new % profit 1 2 50 = 20 – 3 16 % % 3 3 3

Another Method: To avoid much calculation we should suppose that the total investment = 10 × 9 = 90

50 6 Thus, 100 % of (x 50) x 30 3 5

Then cost price of 1 article =

90 = 9 10

and selling price of 1 article =

90 = 10 9

350 6 (x 50) x 30 300 5 7 175 6 x 30 or, x 6 3 5

or,

10 9 1 1 100 100 11 % 9 9 9 By Direct Formula (given in theorem):

% profit

175 6 7 30 or, x 3 5 6 1 85 x 30 3 x = 850 Quicker Method: If we think about the changes

or,

% profit =

Ex. 30: I sell 16 articles for the same money as I paid for 20. What is my gain per cent? Soln: I got a profit of 4 articles at the cost of 16 articles. 4 100 25% 16 or, Let the total investment be 16 × 20 = 320

1 only, we find that 3 % of cost price 3

% profit

2 100 16 % of increase in CP – Increase in 3 SP.

or,

10 350 CP 50 30 300 300

CP

350 180 300 = 850 6 10

xy 100% . y

Selling price of 10 articles = 100 9

10 10 100 = 9 9

10 =

100 1 = 11 9 9 1 profit per cent is 11 9 %.

gain on 100 =

320 = 16 20

SP =

320 = 20 16

20 16 100 25% 16 By Direct Formula:

% profit

Ex.29: The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit per cent. Soln: Let the cost price of 1 article be 1. Cost of 10 articles = 10 Selling price of 9 articles = 10

gain on 10 =

CP =

% profit

Theorem: If cost price of x articles is equal to the selling price of y articles, then profit percentage

10 9 1 100 11 % 9 9

10 9

20 16 100 25% 16

Ex. 31: A wholeseller sells 30 pens for the price of 27 pens to a retailer. The retailer sells the pens at the marked price. Find the percent profit/loss of the retailer. 3 100 = 10% 30 Ex. 32: By selling 66 metres of cloth, a person gains the cost of 22 metres. Find his gain % .

Soln:

% loss =

Soln:

% gain =

22 1 100 33 % 66 3

Dealing in two or more parts Ex. 33: If goods be purchased for 450, and one-third be sold at a loss of 10%, at what gain per cent should the remainder be sold so as to gain 20% on the whole transaction?

Quicker Maths

242

Soln:

1 450 Cost of rd of goods = = 150 3 3 The selling price of one-third of goods

= 150

90 = 135 100

120 = 540 100 Hence, the selling price of the remaining twothirds of the goods must be ( 540 – 135) or 405. But the cost price of this two-thirds = 300

The total selling price is to be 450

gain % =

105 100 35% 300

Short-cut suggested method: If we ignore all the intermediate steps we would reach at the following: Let x be the required gain per cent, then 1 2 of (100 10) of (100 x) = whole of 3 3 (100 + 20) 90 2(100 x) 120 3 3 or, 290 +2x = 360

or,

70 = 35% 2 By method of alligation: This question can be solved by the method of alligation very quickly. See Ex 33 in Chapter ALLIGATION. Ex. 34: If goods be purchased for 840, and one-fourth be sold at a loss of 20%, at what gain per cent should the remainder be sold so as to gain 20% on the whole transaction? Soln: Let x be the required per cent, then

By method of alligation: See Ex 34 in chapter ALLIGATION Ex. 35: A man purchases 5 horses and 10 cows for 10000. He sells the horses at 15% profit and the cows at 10% loss. Thus, he gets 375 as profit. Find the cost of 1 horse and 1 cow separately. Soln: Detail Method: Let the cost of 1 horse be x, then total selling price 115 90 5x (10000 – 5x) 10375 100 100 or, 575x + 90 × 10000 – 450x = 10375 × 100 or, 125x = 137500 137500 x 125 = 1100 Therefore, Cost of one horse = 1100 and cost of

one cow =

10000 5 1100 = 450 10

Short-cut Method (By method of alligation): Overall % of profit = Now, Horse 15% profit

x=

1 3 (100 20) (100 x) (100 20) 4 4

or, 20 + 75 + or,

3x 120 4

3x 25 4

x

100 1 33 % 3 3

375 100 = 3.75% 10000

Cow -10% profit 3.75% profit

13.75

11.25

ratio of the amount invested for the horses and that for the cows = 11:9 10000 cost price of 5 horses = 11 9 11 = 5500

cost price of 1 horse = 1100 10000 9 10 = 450 20 Ex. 36: Two-thirds of a consignment was sold at a profit of 6% and the rest at a loss of 3%. If there was an overall profit of 540, find the value of the consignment. Soln: Detail Method: Suppose the value of

and cost price of 1 cow =

consignment was x. Then, profit, i.e., for

2 106 x . 3 100

2 x was sold at 6% 3

Profit and Loss x 97 1 And rd part is sold at . 3 100 3

243 Note:

480 (100 % loss) = (100 15) (100 19) = 200

2x 106 x 97 – x Now, Profit = 3 100 3 100

212x 97x 309x 300x 9x x 300 300 300

Since,

9x = 540 300

540 300 x = = 18,000 9

(2) The direct formula has been derived from the detailed method. Try to find it yourself. Ex. 38:

Soln:

x x 5x profit and x was sold at 24% 3 4 12 profit. Now, profit

Total Pr ofit 100 540 100 2 1 2 1 % profit – % loss 6 3 3 3 3 3 = 18,000 Note: 1. The above formula can be written in the form (General form): If x part is sold at m% profit, y part is sold at n% profit, z part is sold at p% profit and P is earned as overall profit then the value of

=

P 100 xm ny pz In Ex. 36, we used profit = -(loss) in the denominaton. Ex. 37: A person bought two watches for 480. He sold one at a loss of 15% and the other at a gain of 19% and he found that each watch was sold at the same price. Find the cost prices of the two watches. Soln: We are not going to discuss the detailed method. Only some hint regarding it is enough for you. If the CP of the first watch is x, then total consignment

480 (100 % profit) (100 15) (100 19)

480 119 = 280 204 CP of watch sold at gain = 480 – 280 = 200

=

1 1 rd of a commodity is sold at 15% profit, is 3 4 sold at 20% profit and the rest at 24% profit. If a total profit of 62 is earned, then find the value of the commodity. Suppose the value of the commodity was x. Then x x was sold at 15% profit, was sold at 20% 3 4

Quicker Method: Value of Consignment

100 15 100 19 x (480 x) 100 100 Now, solve for x and get the two prices. Direct Formula: CP of watch sold at loss

(1) CP of watch sold at gain

x 15 x 20 5x 24 62 3 100 4 100 12 100

or,

x x x 62 20 20 10

or,

4x 62 20

x

62 20 = 310 4

Quicker Method (Direct Formula): Value of commodity =

62 100 1 1 5 15 20 24 3 4 12

62 100 = 310 5 5 10 Use -ve sign when some part is sold at loss. For example, see the Ex. 39.

Note:

Ex 39:

Soln:

2 rd of a consignment was sold at 6% profit and 3 the rest at a loss of 3%. If there was an overall profit of 540, find the value of the consignment. Value of consignment

540 100 540 100 = 18,000 2 1 4 1 6 (–3) 3 3

Quicker Maths

244 Ex. 40: Nandlal purchased 20 dozen notebooks at 48 per dozen. He sold 8 dozen at 10% profit and the remaining 12 dozen at 20% profit. What is his profit percentage in this transaction? Soln: Cost price of 20 dozen notebooks = 20 × 48 = 960 Selling price of 8 dozen notebooks 110 = 8 × 48 100 Selling price of 12 dozen notebooks 120 = 12 × 48 100

total selling price 2112 + 5

3456 = 5

5568 960 = 5

768 5

= Profit =

5568 5

768 100 profit = 5 960 16% Quicker Method: (Direct formula): Percentage profit

(First part %profit on first part =

second part %profit on second part) Total of two part

Here, total = 20 dozen are sold in two parts; first part = 8 dozen and second part = 12 dozen.

% profit =

8 10 12 20 320 16% 20 20

Reduction in price Ex. 41: A reduction of 10% in the price of sugar enables a person to obtain 25 kg more for 225. What is the reduction price per kg? Also find the original price per kg. Soln: Owing to the fall in price, there is a saving of 45 10% on 225, i.e., . 2 45 For this , a person purchases 25 kg of sugar.. 2 Hence, reduced price per kg = 22.5 ÷ 25 = 0.90 = 90 P Original price = 90P

100 = 1 100 10

Sale of Mixture Ex. 42: A grocer mixes 26 kg of sugar which cost 2 a kg with 30 kg of sugar which cost 3.60 a kg and sells the mixture at 3 a kg. What is his total gain and the profit per cent? Soln: Profit = (26 kg + 30 kg) × 3/kg – [26 kg × 2/ kg + 30 kg × 3.60/kg] = 168 – (52 + 108) = 168 – 160 = 8 8 100 5% 160 Theorem: A man purchases a certain number of articles at x a rupee and the same number at y a rupee. He mixes them together and sells them at z a rupee. Then his gain or loss %

% profit =

2xy – 1 100 according as the sign is z(x y) +ve or -ve. Proof: Let the man purchase A number of articles. At the rate of x articles per rupee, CP of articles A x At the rate of y articles per rupee, CP of A articles

=

=

A y

Total cost price =

A A x y

Now, selling price of 2A articles at the rate of z articles per rupee =

2A z

SP – CP 100 (According to CP +ve or -ve sign respectively)

% profit or loss =

2A A A z x y 100 A A x y x y 2A A z xy 100 2xy 1 100 z(x y) x y A xy Ex. 43: A man purchases a certain number of mangoes at 3 per rupee and the same number at 4 per

Profit and Loss

Soln:

245

rupee. He mixes them together and sells them at 3 per rupee. What is his gain or loss per cent? By the theorem:

2 3 4 Profit or loss per cent = 3(3 4) 1 100

Ex. 45: Oranges are bought at 11 for a rupee and an equal number more at 9 for a rupee. If these are sold at 10 for a rupee, find the loss or gain per cent. Soln: By the theorem:

2 11 9 – 1 100 % profit or loss = 10(11 9)

100 2 24 1 100 14 % 7 7 21 2 Since the sign is +ve, there is a gain of 14 % 7 Ex. 44: A man purchases a certain number of toffees at 25 a rupee and the same number at 20 a rupee. He mixes them together and sells them at 45 for 2 rupees. What does he gain or lose per cent in the transaction? Soln: Suppose the man bought x number of toffees at 25 a rupee. Then,

Profit =

2x x x 2 1 1 45 20 25

4x 9x x 45 100 900 -ve sign shows that there is a loss. % loss

45 = 22.5 2

2xy % profit or loss = z(x y) 1 100 2 25 20 – 1 100 22.5(25 20) 1000 1012.5 12.5 100 – 100 1012.5 1012.5 100 19 –1 % 81 81

Since the sign is -ve there is a loss of 1

(2) If there is x = y = z, there is neither gain nor loss. Do you agree? Ex. 46: If toffees are bought at the rate of 25 for a rupee, how many must be sold for a rupee so as to gain 25%? 125 5 Soln: SP of 25 toffees = 1 = 100 4

5 = 25 4 25 4 20 5

Short-cut Method (Method of Fraction):

By the theorem:

–

xy , there is always loss. 2

No. of toffees sold for 1 =

x 900 100 x 100 100 100 119 % = 9x 900 9x 81 81 100

when z

No. of toffees sold for

=

x = 25, y = 20 and z =

Note:

2 198 – 1 100 100 1% 200 200 Since the sign is -ve there is a loss of 1%. (1) From the above two examples, we find that

19 %. 81

As there is 25% gain so our calculating figure would be 125 and 100. Now, to gain a profit the number of articles sold for one rupee must be less than the number bought for one rupee. Thus, 100 . the multiplying fraction is 125 100 20 required no. of toffees = 25 125 Ex.47: A man buys 5 horses and 7 oxen for 5850. He sells the horses at a profit of 10% and oxen at a profit of 16% and his whole gain is 711. What price does he pay for a horse? Soln: Suppose the man pays x for a horse. Then, we reach at the equation: 10% of 5x + 16% of (5850 – 5x) = 711 5x 16 (5850 5x) 711 or, 10 100 x 4 or, (5850 5x) 711 2 25 or, 25x + 8 (5850 – 5x) = 711 × 50 = 35550 or, 15x = 46,800 – 35,550 = 11,250 x = 750

Quicker Maths

246 Ex. 48: A person bought some oranges at the rate of 5 per rupee. He bought the same number of oranges at the rate of 4 per rupee. He mixes both the types and sells at 9 for 2 rupees. In this business, he bears a loss of 3. Find out how many oranges he bought in all. Soln: Detail Method: Suppose he bought x oranges of each quality. Then, his total investment =

x x = 5 4

Theorem: If a trademan marks his goods at x% above his cost price and allows purchasers a discount xy of y% for cash, then there is x y % 100 profit or loss according to + ve or - ve sign respectively. 1 Here, x = 25% , y = 12 % 2

9x 20

2x 2 4x = 9 9 9x 4x 81x 80x x – total loss 20 9 180 180 x then, = 3 180

xy 1 25 25 x y 100 % 25 12 2 200 %

Total selling price =

180 × 3 = 540 Therefore, he bought 2 × 540 = 1080 oranges in total. Quicker Method: In the above question: If x oranges/rupee and y oranges/rupee are mixed in same numbers and sold at z oranges/rupee then Number of total apples bought loss ruppees 2xyz z(x y) 2xy 3 2 5 4 4.5 4.5(5 4) – 2 5 4

120 4.5 1080 oranges. 40.5 – 40

Tradesman’s discount for cash payment Ex. 49: A tradesman marks his goods at 25% above his cost price and allows purchasers a discount of

Soln:

1 12 % for cash. What profit % does he make? 2 Let the cost price = 100 Marked price = 125 1 5 Discount = 12 % of 125 = 15 2 8 5 3 reduced price = 125 – 15 8 = 109 8 3 3 gain per cent 109 8 100 9 8 %

Note:

3 = 9 % profit 8 Thus, we see that if x = marked percentage above CP y = discount in per cent z = profit in per cent Then, there exists a relationship;

zxy

xy 100

Ex. 50: A trader allows a discount of 5% for cash payment. How much % above cost price must he mark his goods to make a profit of 10%? Soln: If we use the relationship discussed above, we have 10 = x – 5 – or,

5x 100

19x 15 20

15 20 15 15 % 19 19 Ex. 51: A man buys two horses for 1350. He sells one so as to lose 6% and the other so as to gain 7.5%. On the whole he neither gains nor loses. What does each horse cost? Soln: Loss on one horse = gain on the other 6 % of the cost of first horse = 7.5 % of the cost of the second.

x

Cost of first horse 7.5% 15 5 Cost of second horse 6% 12 4 Dividing 1350 in the ratio of 5 : 4, Cost of first horse = 750 Cost of the second = 600 By the method of Alligation: (See Ex 35 in Chapter ALLIGATION)

Profit and Loss

247

Direct Formula: Cost of first horse

CP of both % loss or gain on 2nd % loss or gain on 1st % loss or gain on 2nd Cost of second horse

CP of both % loss or gain on 1st % loss or gain on first % loss or gain on 2nd In this case: Cost of 1st horse =

1350 7.5 = 750 6 7.5

Cost of 2nd horse =

1350 6 = 600 6 7.5

Some More Uses of Rule of Fraction Ex. 52: Manju Sells an article to Anju at a profit of 25%. Anju sells it to Sonia at a gain of 10% and Sonia sells to Bobby at a profit of 5%. If Sonia sells it for 231, find the cost price at which Manju bought the article. Soln:

100 Sonia bought for 231 100 5 100 100 Anju bought for 231 105 100 10 100 100 100 Manju bought for 231 105 110 100 25

= 160 Ex. 53: Satish marks his goods 25% above cost price but allows 12.5% discount for cash payment. If he sells the article for 875, find his cost price. Soln:

100 Marked price = 875 100 12.5 100 = 875 87.5 100 100 = 800 Cost price = 875 87.5 100 25

Ex. 54: If oranges are bought at the rate of 30 for a rupee, how many must be sold for a rupee in order to gain 25%? Soln: He must sell less than 30 oranges in order to gain. Hence, required number of oranges

100 24 = 30 100 25

Ex. 55: By selling oranges at 32 a rupee, a man loses 40%. How many for a rupee should he sell in order to gain 20%? Soln: The man bought less than 32 oranges for a rupee. 100 40 60 32 Therefore, he bought 32 100 100 oranges for a rupee. 60 oranges for a He must sell less than 32 100 rupee. So, the required number of oranges 60 100 60 100 32 16 = 32 100 100 20 100 120 Ex. 56: If a man sells two horses for 3910 each, gaining 15% on one and losing 15% on the other, find his total gain or loss. Soln: By the theorem, there is always loss in this case

(15) 2 2.25% 100 Now, we see that for SP (100 – 2.25) there is loss of 2.25 and the per cent value is given by

2.25 3910 = 90 97.75 Total loss over two horses = 2 × 90 = 180 Ex. 57: By selling an article for 19.50 a dealer makes a profit of 30%. By how much should he increase his selling price so as to make a profit of 40%? 100 Soln: Cost price = 19.50 100 30

When SP is 3910, loss =

100 = 19.50 130 100 100 40 New selling price = 19.50 130 100 100 140 = 21 = 19.50 130 100

increase in SP = 21 – 19.5 = 1.5 Ex. 58: A man bought a certain quantity of rice at the rate of 150 per quintal. 10% of the rice was spoiled. At what price should he sell the remaining to gain 20% of his outlay? Soln: “10% of the rice is spoiled” may be considered as if he bought the rice at 10% loss. CP per quintal 100 100 = 150 = 150 100 10 90

Quicker Maths

248 Then,

100 100 20 120 SP = = 150 = 200 90 100 90 Ex. 59: A person sold his watch for 144, and got a percentage of profit equal to the cost price. Find the cost of the watch. Soln: Let the cost of the watch = x 100 x 144 Then, x 100 or, x 2 100 x 14400 0 or, (x + 180) (x – 80) = 0 x = –180 or 80 The only +ve value should be our answer, so cost of watch = 80. Note: In such questions, you are suggested to move from the given choices. Ex. 60: What profit per cent is made by selling an article at a certain price, if by selling at 2/3rd of that price there would be a loss of 20%? 2 Soln: rd of the selling price = (100 – 20)% of the 3 cost price 80 3 % of the C.P. = 120% or, Selling price = 2 of the C.P. 20% profit. Ex. 61: A sells a pen to B at a gain of 20% and B sells it to C at a gain of 10% and C sells it to D at a gain of 12.5%. If D pays 14.85, what did it cost to A? Soln: By the rule of fraction: 100 100 100 A’s cost = 14.85 = 10 112.5 110 120 9 Ex. 62: Suresh purchased a horse at th of its selling 10 price and sold it at 8% more than its selling price. Find his gain per cent. 9 Soln: Cost price = 10 108 27 Selling price = 100 25

27 9 25 10 100 % profit = 9 10

270 225 10 100 20 250 9 20% profit.

Ex. 63: The marked price of a radio is 480. The shopkeeper allows a discount of 10% and gains 8%. If no discount is allowed, find his gain per cent. Soln:

100 10 = 432 Selling price = 480 100 100 = 400 Cost price = 432 100 8 If there is no discount, SP = 480

480 – 400 100 20% 400 OR If we recall the relationship

% profit =

xy 100 Where, z = % profit = 8% x = % higher mark y = % discount = 10% z=x–y–

We have; 8 = x – 10 –

10x 100

9x 18 10 x = 20% Hence, the shopkeeper marks 20% higher. If he gives no discount his gain is the same as he marks higher. Therefore, % gain = 20%. Ex. 64: A dealer bought a horse at 20% discount on its original price. He sold it at a 40% increase on the original price. What percentage of profit did he get? Soln: Let the original C.P. = 100 Dealer’s C.P. = 100 – 20% of 100 = 80 Dealer’s S.P. = 100 + 40% of 100 = 140 or,

140 – 80 100 = 75% 80 If we ignore the intermediate steps we have a direct formula as: Dealer’s profit % =

(100 40) – (100 – 10) 100 75% (100 – 20)

Other form of the above example: Ex. 65: A trader bought a car at 20% discount on its original price. He sold it at a 40% increase on the price he bought it. What percentage of profit did he make on the original price?

Profit and Loss Soln:

249

Let the original price be 100. 80 = 80 C.P. = 100 100 S.P. = 80 + 40% of 80 = 112 % profit on original price

112 – 100 100 12% 100 OR Using the direct formula: =

40 20 = 12% 100 Ex. 66: There would be 10% loss if rice is sold at 5.40 per kg. At what price per kg should it be sold to earn a profit of 20%? Soln: By the rule of fraction: % profit = 40 – 20 –

100 100 20 S.P. = 5.4 100 10 100 120 = 7.2/kg. = 5.4 90 Ex. 67: A horse worth 9000 is sold by A to B at 10% loss. B sells the horse back to A at 10% gain. Who gains and who loses? Also find the values. Soln:

90 = 8100 A sells to B for 9000 100 110 = 8910 Again, B sells to A for 8100 100 Thus, A loses (8910 – 8100) = 810 In this whole transaction, A’s investment is only 9000 (the cost of the horse) because the horse returned to his hand.

810 100 9% A’s % loss = 9000 B gains 810 (the same as A loses) and his investment in this transaction is 8100. 810 100 10% 8100 Quicker Maths (direct formula): In such case, the first buyer bears loss and his % of loss is given by

B’s % gain =

% gain (100 – % loss) . 100 In this case, A’s loss%

10(100 10) = 9% 100

9 = 810 and loss amount = 9000 100 Ex. 68: A man purchased two cows for 500. He sells the first at 12% loss and the second at 8% gain. In this bargain he neither gains nor loses. Find the selling price of each cow. Soln: If you recall, you will find that CP of first cow

500 8 = 200 12 8

100 12 SP of first cow = 200 = 176 100 And CP of second cow =

500 12 = 300 12 8

108 SP of second cow = 300 = 324 100 Ex. 69: A milkman buys some milk. If he sells it at 5 a litre, he loses 200, but when he sells it at 6 a litre, he gains 150. How much milk did he purchase? Soln: Difference in selling price = 6/litre – 5/litre = 1/litre If he increases the SP by 1/litre, he gets 200+ 150 = 350 more.

350 350 litres milk 1/ litre Quicker Maths (direct formula): Quantity of milk he purchased

=

Difference of amount 150 – (–120) Difference of rate 6–5

350 = 350 litres. 1 Ex. 70: An article is marked for sale at 275. The shopkeeper allows a discount of 5% on the marked price. His net profit is 4.5%. What did the shopkeeper pay for the article? Soln: We know that if the shopkeeper marked x% higher then

4.5 = x – 5 –

5x x 10% 100

Therefore, cost price = 275

100 = 250 100 10

Quicker Maths

250 Ex.71: 9 kg of rice cost as much as 4 kg of sugar; 14 kg of sugar cost as much as 1.5 kg of tea; 2 kg of tea cost as much as 5 kg of coffee; find the cost of 11 kg of coffee, if 2.5 kg of rice cost 12.50. Soln: 2.5 kg of rice cost 12.50

9 kg of rice cost

12.50 × 9 = 45 2.5

side column. And so on. That is why the last information (n kg of milks cost A) is written at the top. Step II: Mark the side of question mark (?). It is in the rightside column. So, the figures in the left-side column will go in the numerator and the figures in the right side column will go in the denominator.

?

Cost of 9 kg of rice = Cost of 4 kg of sugar = 45

Cost of 14 kg of sugar =

45 14 = Cost of 4

1.5 kg of tea

Cost of 2 kg of tea =

45 14 2 = 210 = Cost 4 1.5

of 5 kg of coffee

Cost of 11 kg of coffee =

210 11 = 462 5

By the Rule of Column This type of question creates confusion and leads to unsuccessful attempt. A simple method has been derived which is easy to understand and apply. As per its name, the whole information is arranged in columns. Once you learn the method of arrangement, your problem will be solved within seconds. The following two points should be taken care of while arranging the information in columns. (It is easy to understand the method with the help of an example.) Take an example x kg of milk costs as much as y kg of rice; z kg of rice costs as much as p kg of pulse; w kg of pulse costs as much as t kg of wheat; u kg of wheat costs as much as v kg of edible oil. Find the cost of m kg of edible oil if n kg of milk costs A. Step I: Arrange the information like; A = n kg milk x kg milk = y kg rice z kg rice = p kg pulse w kg pulse = t kg wheat u kg wheat = v kg edible oil m kg edible oil = ? Note: While arranging the data, the first point to be marked is that the first commodity in the rightside column should be the same as the second commodity in the left-side column. Similarly, the second commodity in the right-side column should be the same as the third commodity in the left-

Note:

Axzwum n y p t v

The second remarkable point is the position of question mark (?). Our numerator and denominator depend on it, and hence the required answer. Suppose the above example is changed. Instead of the last given sentence, we are given

“Find the cost of n kg of milk if k kg of edible oil costs B”. Then our arrangement will be (taking the first point into consideration) ? = n kg milk x kg milk = y kg rice z kg rice = p kg pulse w kg pulse = t kg wheat u kg wheat = v kg edible oil m kg edible oil = B We see that the question mark (?) is in the leftside column, so the right side is our numerator and the left side is our denominator.

B v t p y n required answer = ? m u w z x Hope you have understood the method. Now apply it to the above example. Soln (Ex. 71): 12.5 = 2.5 kg rice 9 kg rice = 4 kg sugar 14 kg sugar = 1.5 kg tea 2 kg tea = 5 kg coffee 11 kg coffee =?

12.5 9 14 2 11 = 462 2.5 4 1.5 5 Note: Solve the same question if the last sentence is changed to ''Find the cost of 2.5 kg of rice if the cost of 11 kg of coffee is 462". Ex. 72: A fruit merchant makes a profit of 25% by selling mangoes at a certain price. If he charges 1 more on each mango, he would gain 50%. Find what price per mango did he sell at first. Also find the cost price per mango. ?

Profit and Loss Soln:

251

Suppose the cost price of a mango be x.

Soln:

100 25 = Then, first selling price = x 100

5x 4 If he charges 1 more and gets 50% profit then there exists a relationship: 5x 100 50 3x 1 x 100 2 4

3x 5x 1 2 4 x= 4 Cost price/mange = 4

or,

125 = 5 and first selling price = 4 100 Quicker Maths (direct formula):

Cost price

100 More charge % Difference in profit

and Selling price

More charge (100 % first profit) % Difference in profit

Thus in this case

100 1 1 125 C.P. = 4, S.P. = 5 50 25 50 – 25 Ex. 73: A fruit merchant makes a profit of 20% by selling a commodity at a certain price. If he charges 3 more on each commodity, he would gain 50%. Find the cost price and first selling price of that commodity. Soln: By Direct Formula: 100 3 3(120) = 10, S.P. = 12 50 20 50 20 Ex. 74: A salaried employee sticks to save 10% of his income every year. If his salary increases by 25% and he still sticks to his decision of his saving habit of 10%, by what per cent has his saving increased? Soln: There should be no hesitation in saying that his saving will be increased by as many per cent as his salary is increased by. So the required answer is 25%. But what happens when his saving % is changed? See the following example: Ex. 75: A person saves 10% of his income. If his income increases by 20% and he decides to save 15% of his income, by what per cent has his saving increased?

By Quicker Maths (direct formula): % increase in saving

(100 20)15 10 100 80% 10 If he sticks to his previous saving habit of 10% then by the direct formula: =

Note:

120 10 10 100 10 = 20%, which is the same as % increase in income. Theorem: When each of the two commodities is sold at the same price, and a profit of P% in made on the first and a loss of L% is made on the second, then the percentage gain or loss % increase in saving =

100(P L) 2PL ( 100 P) ( 100 L) according to the +ve or

-ve sign. Proof: Let each commodity be sold at A. A profit of P% is made on the first, then cost price

100 of the first commodity = A 100 P A loss of L% is made on the second, then cost price

100 of the second commodity = A 100 L 100 100 A Total C.P. = A 100 P 100 L 100(100 L) 100(100 P) = A (100 P)(100 L)

C.P.

100A[(100 L) (100 P)] (100 P)(100 L)

Total S.P. = 2A

% profit or loss =

SP CP 100 CP

100A [(100 L) (100 P)] (100 P)(100 L) 100 100A[(100 L) (100 P)] (100 P)(100 L)

2(100 P)(100 L) 100[(100 L) (100 P)] 100 100[(100 L) (100 P)]

2A

Quicker Maths

252

2 1002 200P 200L 2PL

Note:

2 1002 100L 100P (100 P) (100 L)

100P – 100L 2PL 100(P L) 2PL (100 P) (100 L) (100 P) (100 L)

In the special case when P = L, we have

Ex. 78: A man sells two horses for 1710. The cost price of the first is equal to the selling price of the second. If the first is sold at 10% loss and the second at 25% gain, what is his total gain or loss (in rupees)? Soln: We suppose that the cost price of the first horse is 100. Then we arrange our values in a tabular form: 1st horse 2nd horse Total

100 0 2P2 P2 200 100 Since the sign is -ve, there is always loss and the (% value) 2 . value is given as 100 Ex.76: Each of the two horses is sold for 720. The first one is sold at 25% profit and the other one at 25% loss. What is the % loss or gain in this deal? Soln: Total selling price of two horses = 2 × 720 = 1,440 The CP of first horse = 720

100 = 576 125

100 = 960 75 Total CP of two horses = 576 + 960 = 1,536 Therefore, loss = 1,536 – 1,440 = 96

CP 100

100 100 80 125

180

90 90 SP 100 100

100

190

CP : SP = 180 : 190 = 18 : 19 19 18 1710 = 90 19 Note: We suggest you to solve such lengthy questions by making a tabular arrangement like the above one. This gives a quick solution without any confusion. Direct Formula: If you need the direct formula for this question, see the following: Pr ofit

The CP of second horse = 720

96 100 6.25% 1536 Direct Formula: (See theorem; note) In this type of question where SP is given and profit and loss percentage are same, there is always loss and the % loss

(25)2 625 6.25% 100 100 Note: The above example is a special case when percentage values of loss and gain are the same. But what happens when they are different? See in the following example. Ex.77: Each of the two cars is sold at the same price. A profit of 10% is made on the first and a loss of 7% is made on the second. What is the combined loss or gain? Soln: By the theorem: 100(10 7) 2 10 7 160 % gain as the 200 10 7 203 sign is +ve. Note: You may notice that Ex.77 is a special case of Ex.76. % loss =

Profit

100 (100 10) 100 100 25 = 1710 100 (100 10)

90 – 80 1710 = 90 190 Note: In the above formula, 10% loss is represented as (100 – 10) and 25% profit is represented as (100 + 25). Also, if we find the value -ve, we may conclude that there is a loss. Ex. 79: A dealer sells a table for 400, making a profit of 25%. He sells another table at a loss of 10%, and on the whole he makes neither profit nor loss. What did the second table cost him? =

Soln:

25 = 80 Profit on the first table = 400 125

he loses 80 on the second table (Since there is neither profit nor loss)

80 100 = 800 10 Direct Formula: In the case, when there is neither profit nor loss and selling price of first is given, then Cost price of second table =

100 25 cost price of second = 400 = 800 125 10

Profit and Loss

253

Ex. 80: Rakesh calculates his profit percentage on the selling price whereas Ramesh calculates his on the cost price. They find that the difference of their profits is 100. If the selling price of both of them are the same and both of them get 25% profit, find their selling price. Soln: Suppose the selling price for both of them is x.

100 25 3 x. Now, cost price of Rakesh = x 100 4 100 4 x and cost price of Ramesh x 100 25 5

3 x Rakesh’s profit = x x 4 4 4 x Ramesh’s profit = x x 5 5 Now, difference of their profits x x = = 100 (given) 4 5 x or, = 100 20 x = 2000 Thus, selling price = 2000. Quicker Method (Direct Formula) Selling price =

Diff in profit 100 (100 25) (25) 2

100 100 125 = 2000 25 25 What happens when % profits are different? In that case use the following formula: If Rakesh gets x% profit and Ramesh gets y% profit then

Note:

Selling price =

Diff in profit 100 (100 y) (100)2 (100 y)(100 x)

Please note that when % profit is calculated over cost price we use (100 + y) and when % profit is calculated over selling price we use (100 – x). If we put x = y = 25 in the above general formula, we can get the previously-used formula. Ex. 81: If a discount of 10% is given on the marked price of an article, the shopkeeper gets a profit of 20%. Find his % profit if he offers a discount of 20% on the same article.

Soln:

Detail Method: Suppose the marked price = 100 Then selling price at 10% discount = (100 – 10) = 90 Since he gets 20% profit, his cost price

100 = 75 = 90 120 Now, at 20% discount, the selling price = (100 – 20) = 80 Thus, his % profit

80 75 500 20 2 100 6 % 75 75 3 3 Quicker Method (Direct Formula): Required % profit =

= (100 + % first profit)

100 % 2nd discount – 100 100 % 1st discount

100 20 (100 20) – 100 100 10 320 20 2 80 120 100 100 6 % 90 3 3 3 Ex. 82: A farmer sold a cow and a calf for 760 and got a profit of 10% on the cow and 25% on the calf. If he sells the cow and the calf for 767.50 and gets a profit of 25% on the cow and 10% on the calf, find the individual cost price of the cow and the calf. Soln: Quicker Method: (i) For cost of cow: Cow Calf (1) 110% + 125% = 760 (2) 125% + 110% = 767.5 Cost of cow =

125% of 767.5 110% of 760 (125%)2 (110%)2

5 11 767.5 – 760 10 4 (1.25) 2 (1.1) 2

959.375 836 (1.25 1.1)(1.25 1.1)

123.375 = 350 2.35 0.15 (ii) For cost of calf: Cow Calf (1) SP 110% + 125% = 760 (2) SP 125% + 110% = 767.5

Quicker Maths

254

Cost of calf =

125% of 760 – 110% of 767.5 (125%)2 (110) 2

950 – 844.25 = 300 2.35 0.15 Ex. 83: A profit of 20% is made on goods when a discount of 10% is given on the marked price. What profit per cent will be made when a discount of 20% is given on the marked price? Soln: Detail Method: Suppose the cost price of the goods is 100. Then, selling price in the first case =

120 = 120 = 100 100 Therefore, marked price

100 = = 120 100 10

400 3 Now, selling price in the second case =

400 100 20 320 = 3 100 3

Therefore, % profit =

320 100 ( CP 100) 3

20 2 6 % 3 3 Quicker Method: In such cases: % profit

100 % II discount = (100 + % profit) 100 % I discount 100

80 320 20 2 – 100 100 6 % 90 3 3 3 Ex. 84: What will be the percentage profit after selling an 120

Soln:

1 article at a certain price if there is a loss of 12 % 2 when the article is sold at half of the previous selling price? Detailed Method: Suppose the previous selling price = x Now, the later selling price =

x 2

1 x There is a loss of 12 % when selling price = 2 2

x 100 100x 4x 2 100 12.5 175 7 Now, when selling price is x, % profit 4x x 7 100 7x 4x 100 3 100 75% 4x 4x 4 7 Quicker Method: Required % profit = 100 – 2 × % loss 1 100 – 2 12 100 25 75% 2 Ex. 85: What will be the percentage profit after selling an article at a certain price if there is a loss of 45% when the article is sold at half of the previous selling price? Soln: Quicker Method: % profit = 100 – 2 × % loss = 100 – 2 × 45 = 10% Note: The more general formula for the Ex. 85 may be like: 1 When the second selling price is of the original x selling price, then % profit = x (100 – % loss) – 100 This formula is the same as used in Ex. 84 and Ex. 85. In these two cases % profit = 2(100 – % loss) – 100 = 200 – 2 × % loss – 100 = [100 – 2 × % loss] which are the same as used in Ex. 84 & Ex. 85 Ex. 86: What will be % profit after selling an article at a certain price if there is loss of 45% when the 1 article is sold at rd of previous selling price? 3 Soln: By the general formula given in note of Ex. 85 % profit = 3(100 – % loss) – 100 = 3 × (100 – 45) – 100 = 3 × 55 – 100 = 65% Ex. 87: A horse and a cow were sold for 540, making a profit of 25% on the horse and 20% on the cow. By selling for 538, the profit would be 20% on the horse and 25% on the cow. Find the cost of each. Soln: Detailed Method: Suppose the cost price of a cow and a horse are C and H respectively. Then, selling price of both = 125% of H + 120% of C = 540 Cost price =

or,

5 6 H C 540 4 5

Profit and Loss

255

or, 25H + 24C = 540 × 20 ....(i) Total selling price in the second case = 120% of H + 125% of C = 538

6 5 H C 538 5 4 or, 24 H + 25 C = 538 × 20 ...(ii) Performing (1) × 25 – (2) × 24, we have

10 100 = 200 5 Direct Formula: Cost price CP

or,

(25) 2 H – (24)2 H (540 20 25) (538 20 24)

or, 49 H = 11760 H = 240 If we put the value of H in (1) we get; 24 C = 540 × 20 – 25 × 240 = 4800 C = 200 Cost of a horse is 240 and that of a cow is 200. Quicker Method: In the above case, when the % of profit interchange in the two cases: H+C= =

540 538 125% 120%

540 538 1078 440 1.25 1.20 2.45

and H – C =

=

Ex. 89: I bought two calculators for 480. I sold one at a loss of 15% and the other at a gain of 19% and then I found that each calculator was sold at the same price. Find the cost of the calculator sold at a loss. Soln: Let the CP of the calculator which was sold at 15% loss be x then 100 15 100 19 x (480 x) 100 100 or, 85x = 480 × 119 – 119x or, 204x = 480 × 119 480 199 = 280 x= 204 Quicker Method (Direct Formula): Cost of the calculator sold at 15% loss

480(100 % profit) = (100 % loss) (100 % profit)

540 – 538 125% – 120%

2 2 200 40 1.25 1.20 0.05 5 Now, solve the above two easier equations by =

adding and subtracting and get H =

100 diff. in SP 100 10 = 200 % diff. in profit 5

480 = 240 2

400 and C = = 200. 2 Ex. 88: 5% more is gained by selling a cow for 1010 than by selling it for 1000. Find the cost price of the cow. Soln: Suppose the cost price = x 1000 x 1010 x 100 5 100 Then x x 100 [1010 x 1000 x] 5 or, x 100 (10) 5 or, x x = 200 Quicker Method: 5% of cost price = (1010 - 1000) = 10

480 119 480 119 = 280 (100 15)(100 19) 204

and cost of the calculator sold at 19% profit

480(100 % loss) (100 % loss) (100 % profit)

480 85 = 200 204 Ex. 90: I buy two tables for 1,350. I sell one so as to lose

Soln:

1 6% and the other so as to gain 7 % . On the 2 whole I neither lose nor gain. What did each table cost? Let the first table costs x. 94 107.5 (1350 x) 1350 Then, x 100 100 107.5 94 107.5 100 1350 or, x 100 100

1350 7.5 1350 7.5 = 750 107.5 94 13.5 Thus, the prices of tables are 750 and (1300 – 750 =) 600 x

Quicker Maths

256 Quicker Method (Direct Formula):

Therefore costs of the tables are in the ratio 7.5 : 6, 5 : 4

1350 7.5 1350 6 and 7.5 6 7.5 6 = 750 and 600. This can also be solved by the method of Alligation. We have overall profit = 0% Then I II Price of tables =

Note:

-6%

1350 Cost of tables = 5 4 5 = 750 and

1350 4 = 600 9

7.5%

0 7.5%

6%

EXERCISES 1. A book costing 15 P was sold for 18 P. What is the gain or loss per cent? 2. If oranges are bought at 11 for 10 P and sold at 10 for 11P, what is the gain or loss per cent? 3. A dishonest dealer professes to sell his goods at cost price but uses a weight of 875 grams for the kilogram weight. His gain per cent is _________. 4. A man buys milk at 60 P per litre, adds one-third of water to it and sells the mixture at 72 P per litre. The profit per cent is _________. 5. A watch costing 120 was sold at a loss of 15%. The selling price is _________ . 6. If mangoes are bought at 15 a rupee, how many must be sold for a rupee to gain 25%? 7. Find the cost price if, by selling goods for 279, a merchant loses 7 per cent. 8. A man sells two watches for 99 each. On one he gained 10% and on the other he lost 10%. His gain or loss per cent is _________. 9. By selling goods for 153, a man loses 10% . For how much should he sell them to gain 20%? 10. By selling goods for 240, a merchant gains 25%. How much per cent would he gain by selling it for 216? 11. What profit per cent is made by selling an article at a certain price, if by selling at two-third of that price there would be a loss of 20%? 12. By selling oranges at 32 a rupee, a man loses 40%. How many a rupee must he sell to gain 20 p.c.? 13. The cost price of 16 articles is equal to the selling price of 12 articles. The gain or loss per cent is _________. 14. By selling 33 metres of cloth, I gain the selling price of 11 metres. The gain per cent is_________ .

15. 5% more is gained by selling a cow for 350 than by selling it for 340. The cost price of the cow is _________ . 16. A man buys apples at a certain price per dozen and sells them at eight times that price per hundred. His gain or loss per cent is _________. 17. A shopkeeper marks his goods 20 per cent above cost price, but allows 10% discount for cash. The net profit per cent is _________. 18. A shopkeeper bought a table marked at 200 at successive discounts of 10% and 15% respectively. He spent 7 on transport and sold the table for 208. Find his profit per cent. 19. A merchant sold his goods for 75 at a profit per cent equal to the cost price. His cost price is _________. 20. I purchased a box full of pencils at the rate of 7 for 5 and sold the whole box at the rate of 9 for 8. In this process I gained 44. How many pencils were contained in the box? 21. A dishonest dealer professes to sell his goods at a profit of 20% and also weighs 800 grams in place of a kg. Find his actual gain % . 22. A merchant professes to sell his goods at a loss of 10% , but weighs 750 gm in place of a kg. Find his real loss or gain per cent. 23. By selling salt at Re. 1 a kg, a man gains 10%. By how much must he raise the price so as to gain 21%? 24. A milkman buys some milk contained in 10 vessels of equal size. If he sells his milk at 5 a litre, he loses 200; while selling it at 6 a litre, he would gain 150 on the whole. Find the number of litres contained in each cask.

Profit and Loss

257

25. A watch passes through three hands and each gains 25% . If the third sells it for 250, what did the first pay for it? 26. If by selling an article for 60, a person loses

1 of his 7

outlay (cost), what would he have gained or lost per cent by selling it for 77? 27. I sold a book at a profit of 7%. Had I sold it for 7.50 more, 22% would have been gained. Find the cost price. 28. A reduction of 40 per cent in the price of bananas would enable a man to obtain 64 more for 40. What is the reduced price per dozen?

3 29. A man purchased an article at th of the list price and 4 sold at half more than the list price. What was his gain per cent? 30. I lose 9 per cent by selling pencils at the rate of 15 a rupee. How many for a rupee must I sell them to gain 5 per cent? 31. Goods are sold so that when 4 per cent is taken off the list price, a profit of 20% is made. How much per cent is the list price more than the cost price? 32. A watch was sold at a loss of 10 per cent. If it were sold for 70 more, there would have been a gain of 4 per cent. What is the C.P. of the watch? 33. A man sells an article at 5% profit. If he had bought it at 5% less and sold it for 1 less, he would have gained 10%. Find the cost price. 34. A sold an article at 10 per cent loss on the cost price. He had bought it at a discount of 20 per cent on the labelled price. What would have been the percentage loss had he bought it at the labelled price? 35. A bakery bakes cake with the expectation that it will earn a profit of 40% by selling each cake at marked price. But during the delivery to showroom 16% of the cakes were completely damaged and hence could not be sold. 24% of the cakes were slightly damaged and hence could be sold at 80% of the cost price. The remaining 60% of the cakes were sold at marked price. What is the percentage profit in the whole consignment? 36. A shopkeeper bought 84 identical shirts priced at 240 each. He spent a total of 3200 on transportation and packaging. He put the label of marked price of 420 on each shirt. He offered a discount of 15% on each shirt at the marked price. What is the total profit of the shopkeeper in the whole transaction ? 37. A merchant buys two items for 7500. He sells one item at a profit of 16% and the other item at 14% loss.

In the deal he makes neither any profit nor any loss. What is the difference between the selling price of both the items? (in ) 38. An item was bought for X and sold for Y, thereby earning a profit of 20%. Had the value of X been 15% less and the value of Y 76 less, a profit of 30% would have been earned. What was the value of ‘X’? 39. A bought a certain quantity of oranges at a total cost 1 of those oranges at 20% loss. 3 If A earns an overall profit of 10%, at what percentage profit did A sell the rest of the oranges? A trader has 600 kg of rice, a part of which he sells at 15% profit and the remaining quantity at 20% loss. On the whole, he incurs an overall loss of 6%. What is the quantity of rice he sold at 20% loss ? Two mobile phones were purchased at the same price. One was sold at a profit of 20% and another was sold at a price which was 1520 less than the price at which the first was sold. If the overall profit earned by selling both the mobile phones was 1%, what was the cost price of one mobile phone? The cost price of article A is 100 more than the cost price of article B. Article A was sold at 40% profit and article B was sold at 40% loss. If the overall profit earned after selling both the articles is 5%, then what is the cost price of article B? A trader sells two bullocks for 8,400 each, neither losing nor gaining in total. If he sold one of the bullocks at a gain of 20%, the other is sold at a loss of _________. The percentage profit earned when an article is sold for 546 is double the percentage profit earned when the same article is sold for 483. If the marked price of the article is 40% above the cost price, then what is the marked price of the article? The cost price of two beds are equal. One bed is sold at a profit of 30% and the other one for 5504 less than the first one. If the overall profit earned after selling both the beds is 14%, what is the cost price of each bed? A man sold two articles – A (at a profit of 40%) and B (at a loss of 20%). He earned a total profit of 8 in the whole deal. If article A costs 140 less than article B, what is the price of article B? An article was purchased for 78,350. Its price was marked up by 30%. It was sold at a discount of 20% on the marked-up price. What was the profit per cent on the cost price?

of 1200. He sold

40.

41.

42.

43.

44.

45.

46.

47.

Quicker Maths

258 48. A dealer marked the price of an item 40% above the cost price. He allowed two successive discounts of 20% and 25% to a particular customer. As a result he incurred a loss of 448. At what price did he sell the item to the said customer? 49. The marked price of A is 1600 more than its cost price. When a discount of 500 is allowed a profit of

25% is earned. At what price should A be sold to earn a 30% profit? 50. The ratio of the cost price to the selling price of an article is 5 : 6. if 20% discount is offered on the marked price of the article then the marked price is what per cent more than the cost price?

Answers 1. % gain 2. 11 10

7. By rule of fraction:

18 15 100 = 20% 15

100 Cost price = 279 = 300 100 7

10 11

8. There is always a loss in such case and the loss %

11 11 10 10 100% 10 10 = 21% profit, since sign is positive. 3. % gain

True wt – False wt 100 False wt

1000 875 100 2 100 14 % 875 7 7

4. When he added

1 rd of water, the cost of one litre of 3

impure milk

=

(10) 2 % 1% 100

100 120 = 204 9. By rule of fraction: 153 90 100 100 = 192 10. CP = 240 125

216 192 25 1 100 12 % 192 2 2 11. Let the cost price be 100.

% profit

3 = 60P 45P -------- (*) 4

2 rd of original SP = 100 – 20% of 100 = 80 3

72 – 45 % profit = 45 100 = 60%

Original SP =

Note (*) Quantity of milk = 1

1 4 litre 3 3

4 litre cost 60 P. By rule of fraction, 1 litre will cost 3 less than 60; hence we multiplied 60 by less-than-one fraction i.e. 3 . 4 5. By rule of fraction: 85 = 102 SP = 120 100 100 12 6. By rule of fraction: 15 125

80 3 = 120 2

120 – 100 100 = 20% 100 Quicker Formula:

% profit =

100 20 – 100 % % profit or loss = profit or loss 2 3 according to +ve or negative sign.

80 – 100 20% profit 2 3 12. By the rule of fraction: He must have purchased less number of oranges for a rupee, as he bears a loss. Therefore, no. of oranges

Profit and Loss

259

60 purchased for a rupee 32 100 Now, to gain 20%, he must sell less number of oranges for a rupee. And that number is = 60 100 = 16 32 100 120 13. Suppose he invested 16 × 12. Then, CP of 1 article = and SP of 1 article =

16 12 = 12 16

16 12 = 16 12

16 12 100 1 100 33 % 12 3 3 Quicker Method:

% profit =

No. of purchased goods – No. of sold goods % profit 100 No. of sold goods 16 12 1 100 33 % 12 3 14. Suppose the S.P. per metre = 1 Then, S.P. of 33 metres = 33 Profit = 11 C.P. of 33 metres = 33 - 11= 22 In this case,

11 % Profit = 22 100 = 50% 11 100 50% 33 11 Note: For the above two questions never use the detailed method. Remember the direct formula and its usage. 15. Difference in 5% profit = Diff. in 10 profit Quicker Method: % profit =

10 100 = 200 5 Quicker Method (direct formula):

100% =

100(Diff in S.P.) 100(350 340) Cost price Diff in profit % 5 = 200 16. Let C.P. = x/dozen = 8x/hundred

100x per hundred and SP 12

% profit 100x 12 100 96x 100x 100 12 4% 100x 12 100x 12 -ve sign shows that there is a loss of 4%. Quicker Method (direct formula): % profit or loss = 8 × dozen - Hundred = 96 - 100 = -4% Since sign is -ve, there is a loss of 4%. 17. Use the Direct Formula: 8x

20 10 = 20 - 10 - 2 = 8% 100 18. Single equivalent discount % profit = 20 - 10 -

10 15 23.5% 100 CP for the shopkeeper = 200 - 23.5% of 200 = 153 Total cost = 153 + 7 = 160 = 10 + 15 -

208 – 160 100 = 30% 160 Quicker Method: Profit % =

90 85 = 153 C.P. = 200 100 100 Total cost = 153 + 7 = 160 Required % profit =

208 160 100 = 30% 160

19. x + x% of x = 75 or, x 2 100 x 7500 0 or, (x – 50) (x + 150) = 0 x = 50 or – 150 neglecting the -ve value, the CP = 50. 20. CP of 7 pencils is 5. SP of 7 pencils is

8 56 7 = 9 9

Profit on 7 pencils =

56 – 45 11 = 9 9

79 44 = 252 11 21. Let CP of 1000 gm = 100 SP of 800 gm = 100 + 20% of 100 = 120

Total pencils =

or, SP of 1000 gm =

120 1000 = 150 800

Quicker Maths

260

150 – 100 100 50% 100 Quicker Method: True weight % profit = (100 + % profit) False weight – 100

% profit =

1000 120 – 100 50% 800 22. Let CP of 1000 gm = 100 SP of 750 gm = 90 (as there is 10% loss) or, SP of 1000 gm =

90 1000 = 120 750

120 100 100 = 20% 100 Quicker Method:

% profit

1000 – 100 = 20% % profit or loss = (100 – 10) 750 Since sign is +ve, there is profit of 20%.

100 121 = 1.10 23. By Rule of fraction: SP = 1 110 100

he must raise the price by 0.1 rupee or 10 paise. 24. Suppose he has x litre of milk in total. Thus, we have 5x + 200 = 6x - 150 or, x (6 - 5) = 200 + 150 x = 350 litres. Each vessel contains

350 = 35 litres 10

Quicker Method:

Total quantity of milk

Difference in Amount Difference in Rates

150 (–200) 350 litres 65 Note: Difference in amount = Gain + loss = 150 + 200 = 350 25. By Rule of fraction:

100 100 100 First Purchased for 250 125 125 125 4 4 4 250 = 128 5 5 5 26. Cost Price =

Selling price 60 7 = 70 1 6 1 7

77 70 100 10% 70 27. Quicker Method: 7.5 100 7.5 100 Cost Price = Difference in % profit 22 – 7 = 50 28. He purchases 64 bananas more for 40% of 40 or, 16.

% profit

Reduced price per dozen =

16 12 = 3 64

29. Let the listed price be = 100 Then, CP =

3 3 100 = 75 and SP = 100 = 150 4 2

150 75 100 = 100% 75 Direct Method:

% profit =

1 3 1 – 2 4 100 100% % profit 3 4 30. By the rule of fraction:

100 9 for a rupee. He purchased 15 100 91 100 13 Now to gain 5%, he must sell 15 100 105 for a rupee 31. Let the CP be = 100 Actual SP = 100 + 20% of 100 = 120

Marked Price = 120 × Marked price is

100 120 100 = 125 100 4 96

125 100 100 25% more than 100

the cost price. Quicker Maths (Direct Formula): We may use the formula: z = x – y – Where, z = % profit = 20% x = marked % above CP = ? y = discount = 4% or, 20 = x – 4 – or, x =

4x 100

24 100 = 25% 96

xy 100

Profit and Loss

261

32. Try it. Follow the rule as in Q. 27 33. Let cost price = x 95 110 105 x 1 Then we have, x 100 100 100 or, x

100 100 200 105 100 – 95 110

Cost price = 200 34. Let the labelled price be 100. 80 Then cost price = 100 100 = 80 80 90 Selling price = = 72 100 If he had bought it at the labelled price, loss = 100 – 72 = 28 28 100 = 28% 100 Quicker Method:

Reqd % loss =

20 10 = 28% 100 35. Let the number of cakes be 100. Let the cost price of each cake be 100. Then, total cost price = (100 × 100) = 10000

Reqd % loss = 20 + 10 –

Now, marked price of each cake =

100 140 = 140 100

Now, selling price of 24 cakes 100 80 = 1920 100 And selling price of 60 cakes = 60 × 140 = 8400 Total selling price = 8400 + 1920 = 10320 Profit = 10320 – 10000 = 320

= 24 ×

320 100 = 3.2% 10000 36. Total actual cost = (84 × 240 + 3200) = (20160 + 3200) = 23360

Reqd % profit =

420 85 = 357 100 SP of 84 shirts = (84 × 357) = 29988 Profit = (29988 – 23360) = 6628 37. Let the cost price of the first item be x. Then the cost price of the other item will be (7500 – x). (7500 – x ) 86 7500 Now, x 116 100 100 or, 116x – 86x = 7500(100 – 86) or, 30x = 7500 × 14

And cost price of the other item = (7500 – 3500) = 4000 Difference in selling prices of both items 3500 116 – 4000 86 = 4060 – 3440 = 620 100 100 Quicker Method (Alligation Method) :

=

CP I = 7500 7 3500 15 7500 8 4000 15 Reqd diff = 116% of 3500 – 86% of 400 = 4060 – 3440 = 620 38. The price of the item is X. And SP = Y. Given, Y = 1.2X If the cost price of the item is 15% less then CP = 0.85 × X = 0.85X According to the question,

and CP II =

130 = 1.2X – 76 100 or, 11.05X = 12X – 760 or, 0.95X = 760

0.85X ×

760 = 800 0.95 Cost price of the item = 800 Quicker Method:

X=

SP of each shirt =

x=

7500 14 = 3500 30

From question, 120 – 110.5 76 or, 9.5 76 100

76 100 = 9.5

800

1 1200 of oranges = 20% of = 80 3 3 Profit = 10% of 1200 = 120 To get final profit of 120 he should sell rest of the oranges of 800 to ain profit of 120 + 80 = 200

39. Loss value on

Quicker Maths

262 200 100 = 25% 800 40. Let the quantity of rice sold at 20% loss be x kg. Quantity of rice sold at 15% gain = (600 – x) kg Now, according to the question,

% profit on rest of the oranges =

115 x 80 600 94 100 100 100 115 × 600 – 115x + 80x = 600 × 94 69000 – 35x = 56400 35x = 69000 – 56400 35x = 12600

38% of CP = 1520 100 = 4000 CP = 1520 38

42. Quicker Method (By alligation method):

(600 – x) ×

12600 = 360 kg 35 Quicker Method (Alligation Method):

x=

= 9 : 7 is ratio of CP 9 – 7 = 2 100 Cost of B = 7 350 43. Total selling price of two bullocks = 8400 + 8400 = 16800 100 = 7000 120 According to the question, there is no profit or loss. Cost price of the second bullock = 16800 – 7000 = 9800 Selling price of the second bullock = 8400 Loss = 9800 – 8400 = 1400 Percentage loss on the second bullock

Cost price of the first bullock = 8400

1st : 2nd = 2 : 3 600 2nd part = 2 3 3 = 360 kg 41. Let the cost price of each mobile phone be x. Then, selling price of the first mobile phone = 1.2x Now, according to the question, Selling price of the second = 1.2x – 1520 Total selling price of both mobile phones = 1.2x + 1.2x – 1520 = 2.4x – 1520 Cost price of both mobile phones = x + x = 2x 1 = 2.4x – 1520 100 or, 2x + 0.02x = 2.4x – 1520 or, 2.4x – 2.02x = 1520 or, 0.38x = 1520

Now, 2x + 2x ×

1520 100 = 4000 38 Quicker Method (By Alligation):

x=

Now, given that 120% of CP – 82% of CP = 1520

1400 100 2 100 % 14 % 9800 7 7 Quicker Approach: Suppose selling prices are 120 for each. The cost price of 1st is 100 (as it was sold at 20% profit). Since, there is neither profit nor loss, cost price of 2nd = 2 × 120 – 100 = 140

required loss on 2nd =

140 120 100 140

20 100 100 14 2 % 140 7 7 44. Let the cost price be x. According to the question,

2( 483 x ) 546 x 100 100 = x x or, 546 – x = 2(483 – x) or, x = 966 – 546 = 420 MP = 140% of 420 = 588 Note: For the same article if percentage profit is double, then profit is also double. Keeping this in mind, we can write our first step directly as: 546 – x = 2 (483 – x).

Profit and Loss

263

45. Let the cost of each bed be x. Now, according to the question, 2x 114 x 130 x 130 5504 100 100 100 or, 2.28x = 2.6x – 5504 or, 0.32x = 5504 5504 100 = 172 × 100 = 17200 32 Quicker Method: I bed II bed CP 100 100 SP 130 (228 – 130 =) 98 (SP of both = 200 + 28 = 228) Now 130 – 98 = 32 5504

x=

5504 100 = 17200 32 46. 40% of (x –140) – 20% of x = 8 20% of x –56 = 8 20% of x = 64

100

x 64 5 x = 64 × 5 = 320 47. Cost price = 78350

Marked price = 78350

130 = 101855 100

80 = 81484 100 Profit = 81484 – 78350 = 3134 Selling price = 101855

Reqd % profit =

3134 100 4% 78350

Quicker Method (Using compounding): ( 30) ( 20) % profit = 30% 20% 100 = 10 – 6 = 4% 48. Let the cost price be 100. Then, Loss = 100 – 84 = 16 Loss = 100 – 84 = 16 448 448 SP = 84 16 84 = 28 × 84= 2352 49. Let CP be x Then MP = x + 1600 x 125 Selling price = x + 1600 – 500 = 100 5x x + 1100 = 4 x = 4400 130 New selling price, 4400 × = 5720 100 50. Let CP be 5x. SP = 6x 100 6x 5 15x Now, MP = 6x = 80 4 2 Reqd % more 15x – 5x 2 100 5x 100 50% 5x 2 5x Quicker Approach: CP SP MP 5 6

120 100 = 150 80

100

MP is 50% more than CP.

120

264

Quicker Maths

Chapter 24

Simple Interest Interest is the money paid by the borrower to the lender for the use of money lent. The sum lent is called the principal. Interest is usually calculated at the rate of so many rupees for every 100 of the money lent for a year. This is called the rate per cent per annum. ‘Per annum’ means for a year. The words ‘per annum’ are sometimes omitted. Thus, 6 p.c. means that 6 is the interest on 100 in one year. The sum of the principal and interest is called the amount. The interest is usually paid yearly, half-yearly or quarterly as agreed upon. Interest is of two kinds, Simple and Compound. When interest is calculated on the original principal for any length of time it is called simple interest. Compound interest is defined in the next chapter. To find Simple Interest, multiply the principal by the number of years and by the rate per cent and divide the result by 100. This may be remembered in the symbolic form

ptr 100 Where I = interest, p = principal, t = number of years, r = rate % Ex.1. Find the simple interest on 400 for 5 years at 6 per cent. SI =

Soln.

SI =

400 5 6 = 120 100

Soln:

1225 2 15 1 147 = = 4.59 (nearly) 4 5 4 100 32 Note: 73, 146, 219 and 292 days are respectively =

1 2 3 4 , , and of a year.. 5 5 5 5 The interest I on principal P for d days at r p.c. is given by d 1 2r Pd 365 100 73000 Hence, we deduce the following rule for calculating the total interest on different principals for different number of days, the rate of interest being the same in each case. Rule. Multiply each principal by its own number of days, and find the sum of these products. Multiply this sum by twice the rate and divide the product by 73000. The four quantities involved in questions connected with interest are P, t, r and I. If any three of these be given, the fourth can be found. I Pr

To find principal Since I =

3 3rd to July 27th at 3 % per annum. 4

Ptr 100

P

100 I tr

Ex.3:What sum of money will produce 143 interest in

Interest for a number of days When the time is given in days or in years and days, 365 days are reckoned to a year. But when the time is given in months and days, 12 months are reckoned to a year and 30 days to the month. The day on which the money is paid back should be included but not the day on which it is borrowed, ie, in counting, the first day is omitted. Ex.2: Find the simple interest on 306. 25 from March

1 146 15 1 Interest = 306 4 365 4 100

1 1 years at 2 p.c. simple interest? 4 2 Let the required sum be P. Then 3

Soln:

100 143 = P = 31 21 4 2 =

100 143 4 2 = 1760 13 5

Quicker Maths

266

To find rate % Since I

t2

Pr t 100

r

100 I Pt

Ex.4: A sum of 468.75 was lent out at simple interest and at the end of 1 year 8 months the total amount was 500. Find the rate of interest per cent per annum. Soln:

2 5 or years 3 3 I = (500 – 468.75) = 31.25

Here, P = 468.75, t = 1

rate p.c.

100 31.25 5 468.75 3

100 Ex. 5:

Soln:

3125 3 = 4% 46875 5

A lent 600 to B for 2 years, and 150 to C for 4 years and received altogether 90 from both as interest. Find the rate of interest, simple interest being calculated. 600 for 2 years = 1200 for 1 year and 150 for 4 years = 600 for 1 year Int. = 90

Rate

90 100 5%. 1800 1

To find Time Since I =

Ptr 100

100 I t Pr

Ex.6:

In what time will 8500 amount to 15767.50 at

Soln:

1 per cent per annum? 2 Here, interest = 15767.50 – 8500 = 7267.50 4

t

7267.50 100 19 years 8500 4.5

Miscellaneous Examples Ex.7:

Soln:

1 The simple interest on a sum of money is th of 9 the principal, and the number of years is equal to the rate per cent per annum. Find the rate per cent. Let Principal = P, time = t years, rate = t Ptt P Then, 100 9

t

100 9

10 1 3 3 3

1 rate 3 % 3 Direct formula: 1 10 1 3 % 9 3 3 What annual payment will discharge a debt of 770 due in 5 years, the rate of interest being 5% per annum? Let the annual payment be P. The amount of P in 4 years at 5% Rate = time = 100 Ex.8:

Soln:

100 4 5P 120 P 100 100

''

''

''

3 yrs

'' ''

115 P 100

''

''

''

2 yrs '' ''

110 P 100

105 P 100 These four amounts together with the last annual payment of P will discharge the debt of 770. ''

''

''

1 yrs '' ''

=

120 P 115 P 110 P 105 P 100 100 100 100 P 770

550P 770 100

770 100 140 550 Hence, annual payment = 140 Theorem: The annual payment that will discharge a debt of A due in t years at the rate of interest r% per P

100A . rt(t 1 ) 100t 2 Proof: Let the annual payment be x rupees. The amount of x in (t – 1) yrs at r% annum is

=

100 (t 1)r x 100

Simple Interest The amount of

267 x in (t – 2) yrs at r%

100 (t 2)r x 100 ----------------------------------------------------------------

The amount of x in 2 yrs at r% =

100 2r x 100

100 1x x 100 These (t – 1) amounts together with the last annual payment of x will discharge the debt of A. 100 (t 1)r 100 (t 2)r x x... 100 100 100 r xxA 100 or, x [{100 + (t – 1) r} + {100 + (t – 2)r} + ... {100 + r} + {100}] = 100A The amount of x in 1 year at r% =

r (t 1)(t) or, x 100t 100A 2

x

Note:

100A r(t 1)(t) 100t 2

m(m 1) 2 Using the above theorem: 1 + 2 + 3 + .... + m

100 770 770 100 5(4)(5) 550 5 100 2 = 410 What annual payment will discharge a debt of 848 in 4 yrs at 4% per annum? By the theorem: Annual payment =

Ex.9: Soln:

848 100 4(3)(4) = 200 4 100 2 Ex. 10: The annual payment of 80 in 5 yrs at 5% per annum simple interest will discharge a debt of _______ . Soln: Putting the values in the above formula: Annual payment =

80

A 100 5(4)(5) 5 100 2

80 550 = 440 100 Ex.11: The rate of interest for the first 2 yrs is 3% per annum, for the next 3 years is 8% per annum and for the period beyond 5 years 10% per annum. If a man gets 1520 as a simple interest for 6 years, how much money did he deposit? Soln: Let his deposit be = 100 Interest for first 2 yrs = 6 Interest for next 3 yrs = 24 Interest for the last year = 10 Total interest = 40 When interest is 40, deposited amount is 100. when interest is 1520, deposited amount or, A

100 1250 = 3800 40 Direct formula:

Interest 100 Prinicipal t r t r t r ... 1 1 2 2 3 3

1520 100 1520 100 = 3800. 40 2 3 3 8 1 10 Ex.12: A sum of money doubles itself in 10 years at simple interest. What is the rate of interest? Soln: Let the sum be 100. After 10 years it becomes 200. Interest = 200 – 100 = 100

100 I 100 100 Then, rate P t 100 10 10% Direct formula: Time × Rate = 100 (Multiple number of principal – 1) or, Rate 100

Multiple number of principal 1 time

100(2 1) 10% 10 Ex.13: A sum of money trebles itself in 20 yrs at SI. Find the rate of interest. Using the above formula: rate =

Soln: Note:

100(3 1) 10% 20 A generalised form can be shown as: If a sum of money becomes ‘x’ times in ‘t’ years at SI, the rate of interest is given by Rate

100 (x 1 ) %. t

Quicker Maths

268 Ex.14: In what time does a sum of money become four times at the simple interest rate of 5% per annum? Soln: Using the above formula,

Time

100(Multiple number of principal – 1) Rate

100(4 1) 60 yrs 5 Ex.15: Divide 2379 into three parts so that their amounts after 2, 3 and 4 years respectively may be equal, the rate of interest being 5% per annum.

Soln:

Amount of 1st part = ''

110 × 1st part 100

'' 2nd part =

115 × 2nd part 100

120 × 3rd part 100 According to the question, these amounts are equal 110 × 1st part = 115 × 2nd part = 120 × 3rd part ''

'' 3rd part =

1st part : 2nd part : 3rd part =

1 1 1 : : 110 115 120

= 276 : 264 : 253 Hence, dividing 2379 into three parts in the ratio 276 : 264 : 253, we have 1st part = 828, 2nd part = 792, 3rd part = 759. Ex.16: A certain sum of money amounts to 756 in 2 yrs and to 873 in 3.5 yrs. Find the sum and the rate of interest. Soln: P + SI for 3.5 yrs = 873 P + SI for 2 yrs = 756 On subtracting, SI for 1.5 yrs = 117

117 2 = 156 1.5 P = 756 – 156 = 600 and

Therefore, SI for 2 yrs =

100 156 13% per annum 600 2 Ex.17: A sum was put at SI at a certain rate for 2 yrs. Had it been put at 3 % higher rate, it would have fetched 300 more. Find the sum. Soln: Let the sum be x and the original rate be y% per annum. Then, new rate = (y + 3)% per annum. rate =

x (y 3) 2 x (y) 2 300 100 100 xy + 3x – xy = 15,000 or, x = 5000 Thus, the sum = 5000

Quicker Method: Direct Formula

Sum

More Interest 100 300 100 5000 Time More rate 23

Ex.18: A sum of money doubles itself in 7 yrs. In how many years will it become fourfold? Soln:

Rate

100(2 1) 100 7 7

100(4 1) = 21 years 100 7 Other Method: This question can be solved without writing anything. Think like. Doubles in 7 years Trebles in 14 years 4 times in 21 years 5 times in 28 years and so on. Ex.19: 4000 is divided into two parts such that if one part be invested at 3% and the other at 5%, the annual interest from both the investments is 144. Find each part. Soln: Let the amount lent at 3% rate be x, then 3% of x + 5% of (4000 – x) = 144 or, 3x + 5 × 4000 – 5x = 14400 or, 2x = 5600 x = 2800 Thus, the two amounts are 2800 and (4000 – 2800) or 1200. By the Method of Alligation: See example 26 in ALLIGATION chapter. Ex.20: At a certain rate of simple interest 800 amounted to 920 in 3 years. If the rate of interest be increased by 3%, what will be the amount after 3 years?

Time

Soln:

First rate of interest =

120 100 5% 800 3

New rate = 5 + 3 = 8%

800 3 8 = 192 100 New amount = 800 + 192 = 992. Ex. 21: The simple interest on a sum of money will be 300 after 5 years. In the next 5 yrs principal is trebled, what will be the total interest at the end of the 10th year? Soln: Simple interest for 5 years = 300 Now, when principal is trebled, the simple interest for 5 years will also treble the simple interest on

New interest =

Simple Interest original principal for the same period. Thus, SI for last 5 years when principal is trebled. = 3 × 300 = 900 Total SI for 10 yrs = 300 + 900 = 1200 Theorem: A sum of X is lent out in n parts in such a way that the interest on first part at r1% for t1 yrs, the interest on second part at r2% for t2 years, the interest on third part at r3% for t3 years, and so on, are equal, the ratio in which the sum was divided in n parts is given by 1 1 1 1 : : : .... . r1 t1 r2 t2 r3 t3 rn tn Proof: Let the sum be divided into S1 , S2 , ....Sn . Then,

Int 100 S1 r1 t1 Int 100 S2 r2 t 2 Int 100 S3 r3 t 3 .. .. . . Int 100 Sn rn t n [Since interests of all parts are equal] S1 : S2 : S3 : ... :Sn

Ex 23: A certain sum of money amounted to 575 at 5% in a time in which 750 amounted to 840 at 4%. If the rate of interest is simple, find the sum. Soln: Interest = 840 – 750 = 90

90 100 3 yrs 750 4 Now, by the formula:

Time =

Sum Note:

100 Amount 100 575 100 rt 100 3 5 = 500

There is a direct relationship between the principal and the amount and is given by

Sum

100 Amount 100 rt

Ex.24: A certain sum of money amounts to 2613 in 6 yrs at 5% per annum. In how many years will it amount to 3015 at the same rate? Soln: Use the formula:

Pr incipal

100 Amount 100 2613 100 rt 100 30

100 2613 = 2010 130 Again by using the same formula:

100 3105 100 5t

Int 100 Int 100 Int 100 Int 100 : : ... : r1 t1 r2 t 2 r3 t 3 rn t n

2010

1 1 1 1 : : : ... : r1 t1 r2 t 2 r3 t 3 rn t n

or, 100 + 5t =

Ex.22: A sum of 2600 is lent out in two parts in such a way that the interest on one part at 10% for 5 years is equal to that on another part at 9% for 6 years. Find the two sums. Soln: Each Interest =

1st part 5 10 2nd part 6 9 100 100

or,

1st part 6 9 27 27 : 25 2nd part 5 10 25

2600 27 = 1350 27 25 and 2nd part = 2600 – 1350 = 1250 If we use the above theorem, or, 1st part

Note:

269

S1 : S2

1 1 : 54 : 50 27 : 25 50 54

100 3015 2010

1 100 3015 – 100 2010 t 5 2010

100 (3015 – 2010) 100 1005 =10 years 2010 2010 5 Ex.25: A person lent a certain sum of money at 4% simple interest; and in 8 years the interest amounted to 340 less than the sum lent. Find the sum lent. Soln: Let the sum be x. x 8 4 32x Interest = 100 100 32x 68x x 100 100 68x When interest is less, the sum is x. 100 when interest is 340 less, the sum is x 100 340 = 500 68x

Quicker Maths

270 Direct Formula:

Soln:

100 100 340 340 = 500 100 8 4 68 Ex.26: The simple interest on 1650 will be less than the interest on 1800 at 4% simple interest by 30. Find the time. Soln: We may consider that (1800 – 1650) gives interest of 30 at 4% per annum.

1 1 1 : : 100 2 5 100 3 5 100 4 5

Sum

30 100 5 yrs 150 4 Ex.27: Arun and Ramu are friends. Arun borrowed a sum of 400 at 5% per annum simple interest from Ramu. He returns the amount with interest after 2 yrs. Ramu returns to Arun 2% of the total amount returned. How much did Arun receive? Soln: After 2 yrs, amount returned to Ramu Time

400 5 2 = 440 100 Amount returned to Arun = 2% of 440 = 8.80 Ex. 28: A man invests an amount of 15,860 in the names of his three sons A, B, and C in such a way that they get the same amount after 2, 3, and 4 years respectively. If the rate of simple interest is 5% then find the ratio in which the amount was invested for A, B and C? Theorem: When different amounts mature to the same amount at simple rate of interest, the ratio of the amounts invested are in inverse ratio of (100 + time × rate). That is, the ratio in which the amounts are invested is 400

1 1 1 : : 110 115 120 Ex. 29: A sum of money doubles itself in 4 yrs at a simple interest. In how many yrs will it amount to 8 times itself? Soln: Doubles in 4 yrs 3 times in 4 × 2 = 8 yrs 4 times in 4 × 3 = 12 yrs 8 times in 4 × 7 = 28 yrs Thus direct formula: x times in = No. of yrs to double (x – 1) 8 times in = 4 (8 – 1) = 4 × 7 = 28 yrs Ex.30: Two equal amounts of money are deposited in two banks each at 15% per annum for 3.5 yrs and 5 yrs respectively. If the difference between their interests is 144, find each sum. Soln: Let the sum be x, then

x 15 5 x 15 7 144 100 200 or, 150x – 105x = 144 × 200

144 100 = 640 45 Direct formula: Two equal amounts of money are deposited at r1 % and r2% for t1 and t2 yrs respectively. If the difference between their interests is Id then Sum x

I d 100 = r t –r t 1 1 2 2

1 1 1 1 : : 100 r1t1 100 r2 t2 100 r3t3 : ... : 100 rn tn Proof: We know that Sum

100 Amount 100 rt

of

r1 , r2 , r3 , ...., rn

t 1 , t 2 , t 3 , ... t n yrs

for

the

respectively.

time

144 100 144 100 = 640 15 5 15 3.5 22.5 Ex.31: The difference between the interest received from two different banks on 500 for 2 yrs is 2.5. Find the difference between their rates.

Then

S1 : S 2 : S3 : ... : S n

Thus, in this case: Sum =

Let the sums invested be S1 , S 2 , S3 , ... S n , at the rate

Therefore, the required ratio is this case is

100 A 100 A 100 A 100 A : : : ... : 100 r1 t1 100 r2 t 2 100 r3 t 3 100 rn t n

[Since the amount (A) is the same for all] 1 1 1 1 : : : ... : 100 r1 t1 100 r2 t 2 100 r3 t 3 100 rn t n

Soln:

I1

500 2 r1 10 r1 100

I2

500 2 r2 10 r2 100

I1 – I 2 10r1 10r2 2.5 or, r1 r2

2.5 0.25% 10

Simple Interest

271

By Direct formula (Used in previous example): When t1 = t2,

I 100 2.5 100 (r1 – r2 ) d 0.25% Sum t 500 2 Ex. 32: The simple interest on a certain sum of money at 4% per annum for 4 yrs is 80 more than the interest on the same sum for 3 yrs at 5% per annum. Find the sum. Soln: Let the sum be x. Then, at 4% rate for 4 yrs the simple interest

x44 4x = 100 25 At 5% rate for 3 yrs the simple interest

x 53 = 100 Now, we have,

3x 20

16x 15x 80 100 x = 8000 Quicker Method: For this type of question or,

Difference 100 80 100 r1 t1 – r2 t 2 4 4 35

= 8000 Ex. 33: A sum of money at simple interest amounts to 600 in 4 years and 650 in 6 years. Find the rate of interest per annum. Soln: Suppose the rate of interest = r% and the sum = A Now, A or, A

Ar4 600; 100

Ar 600 25

r or, A 1 600 25 And, A +

-------- (1)

(25 r) 2 12 50 3r 13 or, (50 + 2r) × 13 = (50 + 3r) × 12 or, 650 + 26r = 600 + 36r; or, 10r = 50 r = 5% Direct Formula: If a sum amounts to A1 in t1 years and A2 in t2 years at simple rate of interest, or,

100[A2 – A1 ] then rate per annum (A t – A t ) 1 2 2 1 In the above case,

100[650 – 600] 100 50 5% 6 600 4 650 1000 Ex. 34: Ramesh borrows 7000 from a bank at SI. After 3 yrs he paid 3000 to the bank and at the end of 5 yrs from the date of borrowing he paid 5450 to the bank to settle the account. Find the rate of interest. Soln: Any sum that is paid back to the bank before the last instalment is deducted from the principal and not from the interest. Thus, Total interest = Interest on 7000 for 3 yrs + Interest on ( 7000 – 3000 =) 4000 for 2 yrs. Or, (5450 + 3000 – 7000) 7000 3 r 4000 2 r 100 100 or, 1450 = 210r + 80r

1450 5%. 290 Ex. 35: Some amount out of 7000 was lent at 6% per annum and the remaining at 4% per annum. If the total simple interest from both the fractions in 5 yrs was 1600, find the sum lent at 6% per annum. Soln: Suppose x was lent at 6% per annum. r

Thus,

Ar6 650; 100

3r or, A 1 650 50

r 25 600 ; 3r 650 1 50 1

r

4x 3x – 80 25 20

Sum

Dividing (1) by (2), we have

or, -------- (2)

x 6 5 (7000 – x) 4 5 1600 100 100

3x 7000 x 1600 10 5

3x 14,000 2x 1600 10 x = 16000 – 14000 = 2000

or,

Quicker Maths

272

ratio of two amounts = 2 : 5

By Method of Alligation: Overall rate of interest

1600 100 32 % 5 7000 7 6%

amount lent at 6%

000 2 = 2000 7

4%

32 % 7 4% 7

10% 7

EXERCISES 1. The simple interest on a certain sum for 3 years at 14% per annum is 235.20.The sum is _______. 2. If 64 amounts to 83.20 in 2 years, what will 86 amount to in 4 years at the same rate per cent per annum? 3. A sum of money amounts to 850 in 3 years and 925 in 4 years. What is the sum? 4. A sum amounts to 702 in 2 years and 783 in 3 years. The rate per cent is _______. 5. A money-lender finds that due to a fall in the rate of

1 interest from 13% to 12 %, his yearly income 2 diminishes by 104. What is his capital? 6. If the amount of 360 in 3 years is 511.20, what will be the amount of 700 in 5 years? 7. A sum of 2540 is lent out in two parts, one at 12% and 1 the other at 12 %. If the total annual income is 2 312.42 the money lent at 12% is _______. 8. A sum of 2600 is lent out in two parts in such a way that the interest on one part at 10% for 5 years is equal to that on the other part at 9% for 6 years. The sum lent out at 10% is __________. 1 th of the 16 principal and the number of years is equal to the rate per cent per annum.The rate per cent per annum is ____________. 10. The simple interest on a certain sum at a certain rate is 9. The simple interest on a sum of money is

9 th of the sum. If the number representing rate per 16 cent and time in years be equal, then the time is _________.

11. A sum of money will double itself in 16 years at simple interest at an yearly rate of __________. 12. A sum of money put at simple interest trebles itself in 15 years. The rate per cent per annum is ____________. 13. At a certain rate of simple interest, a certain sum doubles itself in 10 years. It will treble itself in __________. 14. 800 amounts to 920 in 3 years at simple interest. If the interest rate is increased by 3%, to how much would it amount? 15. A lent 600 to B for 2 years and 150 to C for 4 years and received altogether from both 90 as simple interest. The rate of interest is ________.

1 1 rd of his capital at 7% , th at 8% 3 4 and the remainder at 10% . If his annual income is 561, the capital is ________. The simple interest at x% for x years will be x on a sum of ______. If the interest on 1200 be more than the interest on 1000 by 50 in 3 years, the rate per cent is _______. A sum was put at simple interest at a certain rate for 2 years. Had it been put at 1% higher rate, it would have fetched 24 more. The sum is __________. A sum of money becomes 8/5 of itself in 5 years at a certain rate of interest. The rate per cent per annum is _________. A man lends 10000 in four parts. If he gets 8% on

16. A man invested

17. 18. 19.

20.

21.

1 1 2000, 7 % on 4000 and 8 % on 1400, what per 2 2 cent must he get for the remainder, if the average interest is 8.13%? 22. The simple interest on a sum of money at 8% per annum for 6 years is half the sum.The sum is ____.

Simple Interest

273

23. The difference between the interest received from two different banks on 500 for 2 years is 2.50. The difference between their rates is ________. 24. Two equal amounts of money are deposited in two banks,

25.

26.

27.

28.

29.

1 each at 15% per annum, for 3 % years and 5 years 2 respectively. If the difference between their interests is 144,each sum is _________. The rate of interest on a sum of money is 4% per annum for the first 2 years, 6% per annum for the next 4 years and 8% per annum for the period beyond 6 years. If the simple interest accrued by the sum for a total period of 9 years is 1120,what is the sum? A sum of 16800 is divided into two parts. One part is lent at the simple rate of interest 6% per annum and the other at 8% per annum. After 2 years the total sum received is 19000. The sum lent at the rate of 6% simple interest is _________. The sum invested in Scheme B is thrice the sum invested in Scheme A. The investment in Scheme A is made for 4 years at 8% p.a. simple interest and in Scheme B for 2 years at 13% p.a. simple interest. The total interest earned from both the schemes is 1320. How much amount was invested in Scheme A? 16000 was invested for three years, partly in scheme A at the rate of 5% simple interest per annum and partly in scheme B at the rate of 8% simple interest per annum. The total interest received at the end was 3480. What amount of money was invested in scheme A? A took a certain sum as loan from bank at a rate of 8% simple interest per annum. A lends the same amount to B at 12% simple interest per annum. If at the end of

30.

31.

32.

33.

34.

five years, A made a profit of 800 from the deal, what was the original sum? The interest earned when P is invested for five years in a scheme offering 12% pa simple interest is more than the interest earned when the same sum ( P) is invested for two years in another scheme offering 8% pa simple interest, by 1100. What is the value of P? Mr Phanse invests an amount of 24200 at the rate of 4 pcpa for 6 years to obtain a simple interest. Later he invests the principal amount as well as the amount obtained as simple interest for another 4 years at the same rate of interest. What amount of simple interest will be obtained at the end of the last 4 years? According to a new plan rolled out by HISP Bank, the rate of simple interest on the sum of money is 8% pa for the first two years, 10% pa for the next three years and 6% pa for the period beyond the first five years. The simple interest accrued on a sum for a period of eight years is 12,800. Find the sum. Ravi invested P in a scheme A offering simple interest at 10% pa for two years. He invested the whole amount he received from scheme A in another scheme B offering simple interest at 12% pa for five years. If the difference between the interests earned from schemes A and B was 1300, what is the value of P? Nikhilesh invested certain amount in three different schemes A, B and C with the rate of interest 10 p.c.p.a., 12 p.c.p.a. and 15 p.c.p.a. respectively. If the total interest accrued in one year was 3200 and the amount invested in scheme C was 150% of the amount invested in scheme A and 240% of the amount invested in scheme B, what was the amount invested in scheme B?

Solutions 1. Principal

235.20 100 = 560 3 14

19.20 100 15% 2. Rate of interest = 64 2 86 15 4 Interest on 86 = 51.6 100 Amount = 86 + 51.6 = 137.6

Quicker Maths: Interest on 64 for 2yrs is 19.2. Hence, by Rule of

86 4 fraction, interest on 86 for 4 yrs is 19.2 64 2 = 51.6 Amount = 86 + 51.6 = 137.6 3. P + SI for 4 yrs = 925 P + SI for 3 yrs = 850 On subtracting, SI for 1 yr = 75 SI for 3 yrs = 3 × 75 = 225 P = 850 – 225 = 625

Quicker Maths

274 4. Follow the same rule as in Q. 3. You get the sum, interest and time. Find the rate of interest. 1 5. decreases on 100 2

104 decreases on

100 104 100 2 4 1 2

= 20800 6. Same as Q. 2. 7. Solve it by the method of alligation.

312.42 100 12.3% Overall rate of interest 2540 Now, I part II part

12%

12.5%

13. It doubles in 10 yrs. Then trebles in 20 yrs.

120 100 = 5% 800 3 Now, the new rate becomes 8%. Then interest

14. Rate of interest =

800 8 3 = 192 100 Amount = 800 + 192 = 992 Quicker Method: =

800 3 3 = 72 100 Increased amount = 920 + 72 = 992 15. Let the rate of interest be r% per annum. Increase in interest =

600 2 r 150 4 r 90 100 100 or, 12r + 6r = 90 r = 5% Note: Solve it by method of ‘Alligation’ 16. Let the capital be 120. Then, total interest Then,

12.3% 0.2

0.3

Therefore, the sum will be divided into the ratio of 0.2 : 0.3 or 2 : 3 2540 2 = 1016 Then, sum lent at 12% = 5

= 7% of 40 + 8% of 30 + 10% of 50 = 2.8 + 2.4 + 5 = 10.2

Ratio of two parts = r2 t 2 : r1 t 1 54 : 50 27 : 25

2600 27 = 1350 52 9. Let the rate of interest = r% times = r yrs S Sr r Now, 16 100 100 2 or, r 16

Sum lent out at 10% =

25 1 r 4 64% 10. Same as Q. 9 11. Quicker Maths: Rate of interest =

120 120 8% of 10% of remainder 3 4

= 7% of

1 2540 3 = 1524 and sum lent at 12 % 2 5 8. Quicker Method:

actual capital =

120 561 6600 10.2

Quicker Method: Capital =

561 1 1 5 7% of 8% of 10% of 12 3 4

561 100 561 100 12 = 6600 7 8 25 28 24 50 3 4 6

x 100 100 = xx x 18. Let the rate of interest be x%. 17.

100(2 1) 1 12 % 16 2

12. Quicker Maths:

100(3 1) 200 1 13 % Rate of interest = 15 15 3

Then,

1200 3x 1000 3x 50 100 100

6x = 50 1 x= 8 % 3

Simple Interest Quicker Maths:

Rate

Difference in Interest 100 Time (Difference in Principal)

50 100 25 1 8 % 3(1200 1000) 3 3

Note: The above-used formula is simply based on Rate =

Interest×100 Time× Principals

19. Let the sum be P and rate be r%. Then

P 2 r P(r 1) 2 24 100 100 or, 2Pr = 2Pr + 2P – 2400 or, 2P = 2400 P = 1200 Thus, the sum is 1200. Quicker Maths: Sum

Difference in Interests 100 24 100 Times Difference in rate 2 1

= 1200 Note: The above formula is simply based on

Sum

Int. 100 Time Rate

20. Quicker Method:

8 3 1 100 5 5 300 12% Rate 100 Time 5 25 21. Total Interest = 8.13% of 10000 = 813 Remainder money = 10000 – (2000 + 4000 + 1400) = 2600 Then, 8% of 2000 + 7.5% of 4000 + 8.5% of 1400 + x% of 2600 = 813 or, 160 + 300 + 119 + 26x = 813

234 x 26 9% S S86 2 100 We can’t find the value of S. We also see that the above relationship is not correct. Thus, we conclude that the question is wrong. 23. Quicker Maths: Use the formula given in Q 19. 22.

Sum

Difference in Interests 100 Times Difference in rates

275 or, 500

2.5 100 2x

2.5 100 0.25% 2 500 24. Quicker Maths:

x

Sum

Difference in Interests 144 100 Rate Difference in times 15 1.5

= 640 25. Quicker Maths:

Sum

Interest 100 r1 t1 r2 t 2 r3 t 3 ......

1120 100 1120 100 = 2000 426 483 56 26. Let the sum lent at 6% rate of interest be x. Then, (1680 – x) is lent at 8% rate of interest. Then, SI = 19000 – 16800 = 2200

x 6 2 (16800 x ) 2 8 2200 100 100 or, 12x + 268800 – 16x = 2200 × 100 or, 268800 – 220000 = 4x 48800 = 12200 4 27. Let the amount invested in scheme A be x and that in B be 3x.

or, x =

Then,

x 4 8 3x 2 13 = 1320 100 100

or,

32x 78x 1320 100 100

or,

110x = 1320 100

1320 100 = 1200 110 28. Let the sum invested in scheme A be x. Then the amount invested in scheme B = (16000 – x)

x=

(16000 x ) 3 8 3480 Now, x 5 3 100 100 or, 15x + 384000 – 24x = 3480 × 100 or, 9x = 384000 – 348000 = 36000

x=

36000 = 4000 9

Quicker Maths

276 Quicker Method (Alligation Method): 3480 100 7.25% Average ratio of interest = 16000 3 Now,

32. Let the sum be P P 8 2 P 10 3 P 6 3 = 12800 100 100 100 or, 16P + 30P + 18P = 12800 × 100 or, 64P = 12800 × 100

Then,

12800 100 = 20000 64 33. Reqd difference

P=

Amount invested in A = 29. Let the sum be x. SI =

1600 1 = 4000 1 3

prt (formula) 100

x 12 5 x 8 5 = 800 100 100 or, 60x – 40x = 800 × 100 or, 20x = 800 × 100

Then,

800 100 = 4000 20 Quicker Method: 5 × (12 –8)% 800 or, 20% 800 100% 4000 30. According to the question,

x=

P 12 5 – P 8 2 1100 100 100 or, 60P – 16P = 1100 × 100 or, 44P = 1100 × 100 1100 100 = 2500 44 Quicker Method: (12 × 5)% – (8 ×2)% 1100 or, 44% 1100

P=

100%

1100 100 = 2500 44

24200 4 6 = 5808 100 Now, P = 24200 + 5808 = 30008 According to the question,

31. SI =

30008 4 4 = 4801.28. 100 Hence the last four years’ simple interest is 4801.28 Direct Method: Interest in last 4 years = 24200(100 + 24)% × 16% = 242 × 124 × 0.16 = 4801.28

SI =

P 10 2 12 5 P 10 2 – = P 100 100 100 100P 20P 60 20P – or, = 1300 100 100 100

or,

120P 3 20P – 1300 100 5 100

or,

6P 3 20P – 1300 5 5 100

72P – 20P 1300 100 or, 52P = 1300 × 100

or,

1300 100 = 2500 52 Quicker Approach: Suppose Ravi invested 100 in scheme A. Then, A B Investment 100 120 (= 100+20) Interest 20 72 (12 × 5 = 60%) (10 × 2 = 20%) According to the question, 72 – 20 = 52 1300

P=

1300 100 = 2500 52 34. Ratio of Nikhilesh’s investments in different schemes A, B and C

P = 100

= 100 :

150 100 : 150 8 : 5 : 12 240

Now, according to the question,

8k 10 5k 12 12k 15 3200 100 100 100 or, 80k + 60k + 180k = 3200 × 100 or, 320k = 3200 × 100 or, k = 1000 amount invested in scheme B = 1000 × 5 = 5000

Chapter 25

Compound Interest Money is said to be lent at Compound Interest (CI) when at the end of a year or other fixed period the interest that has become due is not paid to the lender, but is added to the sum lent, and the amount thus obtained becomes the principal for the next period. The process is repeated until the amount for the last period has been found. The difference between the original principal and the final amount is called Compound Interest (CI).

Important Formulae Let Principal = P, Time = t yrs and Rate = r% per annum Case I: When interest is compounded annually: t

r Amount P 1 100 Case II: When interest is compounded half-yearly: 2t

r 2t r 2 Amount P 1 P 1 200 100 Case III: When interest is compounded quarterly: r Amount P 1 4 100

time

rate Amount Principal 1 100 If the principal is 1, the amount for first, second and third years will be 2

3

r r r 1 , 1 and 1 respectively.. 100 100 100 And, if the rate of interest is 10%, 5% and 4%, these values will be 2

11 11 11 , , 10 10 10 21 , 20

3

2

3

2

3

21 21 , 20 20

26 26 26 , , 25 25 25

The above information can be put in the tabular form as given below: Principal = 1, then CI:

4t

r P 1 400

We have the basic formula:

4t

Case IV: When rate of interest is r1%, r2% and r3% for 1st year, 2nd year and 3rd year respectively, then

r1 r r 1 2 1 3 Amount = P 1 100 100 100 The above-mentioned formulae are not new for you. We think that all of you know their uses. When dealing with the above formulae, some mathematical calculations become lengthy and take more time. To simplify the calculations and save the valuable time we are giving some extra informations. Study the following sections carefully and apply them during your calculations. The problems are generally asked upto the period of 3 years and the rates of interest are 10%, 5% and 4%.

Time r 10 5 4

1 Year

2 Years

r r 1 1 100 100 11 121 10 100 21 441 20 400 26 676 25 625

3 Years 2

r 1 100 1331 1000 9261 8000 17576 15625

3

The above table should be remembered. The use of the above table can be seen in the following examples. Ex.: 7500 is borrowed at CI at the rate of 4% per annum. What will be the amount to be paid after 2 yrs?

Quicker Maths

278 Soln:

As the rate of interest is 4% per annum and the time is 2 yrs, our concerned fraction would be

3 102 103 Diff 1002 % of 8000 100

676 . From the above table, you know that Re 1 625 becomes

676 at 4% per annum after 2 yrs. So, 625

after 2 yrs 7500 will produce 7500 ×

676 625

= 8112. Another Useful Method of Calculating CI. 1. For 2 years: We know that Compound Interest and Simple Interest remain the same for the first year. In the second year, they differ because we also count Simple Interest on the first year's Simple Interest to get the Compound Interest. So, mathematically if our rate of interest is r%, then for 2 years Simple Interest (SI) = 2 × r = 2r% of capital.

r and Compound Interest (CI) = 2r 100 % of capital 2

r2 % of capital. 100 Now take the above solved example. We can calculate the SI mentally. That is 2 × 4 = 8% of 7500 = 600 Therefore, the difference is

Difference is

16 7500 = 12 100 100

CI = 612 required value = (7500 + 612) = 8112 Note: For 2 years, SI = 600 SI for one year = 300 diff is 4% of 300 = 12 CI = 600 + 12 = 612 required value = 7500 + 612 = 8112 2. For 3 years: If the rate of interest is r% then for 3 years Total Simple Interest = 3 × r = 3r% of capital 3r 2 r3 3r % of capital. Compound Interest = 100 1002 3r 2 r3 Therefore difference = 100 1002 % of capital Now see the following examples: Ex 1: Find the compound interest on 8000 at 10% per annum for 3 years. Soln: SI = 30% of 8000 = 2400

Ex 2: Soln:

= (3 + 0.1)% of 8000 = 3.1% of 8000 = 248 CI = (2400 + 248) = 2648 Find the amount after 3 years when 1000 is deposited at 5% compound rate of interest. SI = 15% of 1000 = 150

3 52 53 = 15 + 0.75 + 0.0125 100 1002 = 15.76125% of 1000 = 157.625 Amount = 1,157.625 To find the % difference between CI & SI 1. For 2 years: If rate of interest is 4%, then SI = 2 × 4 = 8% of capital CI = 15

CI = 4 + 4 +

44 42 = 24 = 8.16% of 100 100

capital(*) (*) It means if CI rate of 4% for 2 years is converted to SI rate for 1 year, it is equivalent to 8.16%. If rate interest is 5%, then SI = 2 × 5 = 10% of capital

55 = 10.25% of capital 100 If rate of interest is 8%, then SI = 16% of capital CI = 5 + 5 +

64 = 16.64% of capital 100 If rate of interest is 10%, then SI = 20% of capital CI = 16

CI = 20

102 21% of capital 100

r2 . 100 It has been discussed in the chapter 'Percentage'. Find the theory behind the similarity. (2) You may arrange the above calculation in tabular form to remember it. 2. For 3 years. (a) If R = 4%, then SI = 4 × 3 = 12% of capital Note: (1) You must have recognised the form: 2r

8.16 4 = 12.16 + 0.3264 100 = 12.4864% of capital

CI = 8.16 (*) + 4 +

Compound Interest Note:

279

(*) is % CI for 2 years. It means for 3 years, you have to find % CI for 2 years and apply the formula

4

Soln:

r1r2 r1 r2 for 3rd year.. 100

4

r 1 100 2

(b) R = 5%; then SI = 5 × 3 = 15% of capital

10.25 5 = 15.25 + 0.5125 100 = 15.7625% of capital (c) R = 8% then SI = 8 × 3 = 24% of capital

Cubing both sides, we get

16.64 8 100 = 24.64 + 1.3312 = 25.9712% of capital (d) R = 10% then SI = 30% of capital

r or, P 1 8P 100

CI = 10.25 + 5 +

12

r 3 1 2 8 100 12

CI = 16.64 + 8 +

21 10 = 31 + 2.1 100 = 33.1% of capital Note: (1) You may use the formula for difference in % as CI = 21 10

3r 2 r3 given earlier as 2 . When you don't 100 100 recall this, you are suggested to go for the detail method discussed above. (2) Arrange the calculation in tabular form and remember it.

Hence, the required time is 12 yrs. Quicker Approach: x becomes 2x in 4 yrs. 2x becomes 4x in next 4 yrs. 4x becomes 8x in yet another 4 yrs. Thus, x becomes 8x in 4 + 4 + 4 = 12 yrs. Ex. 3: Find the least number of complete years in which a sum of money at 20% CI will be more than doubled. t

Soln:

t

Miscellaneous Examples

t

4 390625 1 456976 100 t

1 456976 1 25 390625 t

26 26 25 25

Ex. 2:

4

t=4 the required time is 4 years. A sum of money placed at compound interest doubles itself in 4 yrs. In how many years will it amount to eight times itself?

6 6 6 6 2 5 5 5 5 The required time is 4 yrs. A sum of money at compound interest amounts to thrice itself in three years. In how many years will it be 9 times itself? Detail Method: Suppose the sum = x Then, we have By trial,

Ex. 4:

t

r P 1 A 100

20 We have, P 1 2P 100 6 5 2

To find time Ex. 1: In what time will 390625 amount to 456976 at 4% compound interest? Soln:

r We have P 1 2P 100

Soln:

r 3x x 1 100

3

r or, 3 1 100

3

Squaring both sides 3 r (3) 1 100 2

2

Quicker Maths

280 6

r or, 9 1 100 Now, multiply both sides by x; then 9x

Ex. 6: At what rate percentage (compound interest) will a sum of money become eight times in three years? Soln: By Direct Formula: 1 Rate% (8) t 1 100

6

r = x 1 100 the sum x will be 9 times in 6 years. Quicker Method: Remember the following conclusion: If a sum becomes x times in y years at CI then it

1 (8) 3 1 100 (2 1) 100 100%

Ex. 7:

n

will be ( x ) times in ny years. Thus, if a sum becomes 3 times in 3 years, it will be

2

Soln:

(3) 2 times in 2 × 3 = 6 years. Sample Questions: 1. If a sum deposited at compound interest becomes double in 4 years, when will it be 4 times at the same rate of interest? Soln: Using the above conclusion, we say that the sum 2

2. Soln:

will be (2) times in 2 × 4 = 8 years. In the above question, when will the sum be 16 times? (2)4 = 16 times in 4 × 4 = 16 years.

To find rate Ex. 5: At what rate per cent compound interest does a sum of money become nine-fold in 2 years? Soln: Detail method: Let the sum be x and the rate of compound interest be r% per annum; then

r 9x x 1 100

2

r or, 9 1 100

2

or, 3 1

2

1 r 100 (9) 2 1 100(3 1) 200%

2

r 21 100 20 r = 5% Another Quicker Approach: 80,000 amounts to 88200 in 2 years. It means interest is 8200 in 2 years, which is about 10%. This implies that the rate of interest is about 10 ÷ 2 = 5%. Suppose this is true. Then or, 1

52 10.25% of 80,000 = 8200 (*) 100 Which is the same as we found from the question. It means that our assumption is correct. (*) The same thing may be confirmed from the difference between SI and CI. CI = 10

Note:

52 % of 80000 = 200 100 Which is true. Hence our assumption that r = 5% is true. OR The difference is 5% of (5% of 80000) = 5% of 4000 = 200 The difference is

or,

1 100 (m) t 1 . In this case,

r We have 80,000 1 88,200 100 r 88, 200 441 21 or, 1 100 80,000 400 20

r 100

r 2 100 r = 200% Direct Formula: The general formula of compound interest can be changed to the following form: If a certain sum becomes ‘m’ times in ‘t’ years, the rate of compound interest r is equal to

At what rate per cent compounded yearly will 80,000 amount to 88,200 in 2 yrs?

Given CI, to find SI and vice versa Ex. 8: If the CI on a certain sum for 2 yrs at 3% be 101.50, what would be the SI? 2

Soln:

2

3 103 CI on 1 rupee = 1 1 1 100 100 =

609 10000

Compound Interest 23 6 = 100 100

SI on 1 =

281

40 100 = 500 42 Another Quicker Approach: Difference = 0.80,

SI 6 10000 200 CI 100 609 203

And sum =

200 200 SI = 203 of CI 203 101.5 = 100 Another Quicker Approach: For 2 years at 3% SI = 2 × 3 = 6% of capital 32 = 6.09% of capital 100 Here, 6.09% of capital = 101.5

40 = 20 2 It means r% of 20 = 0.80 SI for one year =

CI = 6

Note:

101.5 6% of capital = 6.09 6 = 100 If you don’t want to go through the details of the above method, remember the following direct formula and get the answer quickly. Simple Interest

rt r 100 1 1 100 t

Compound Interest

Given CI and SI, to find sum and rate Ex. 9:The compound interest on a certain sum for 2 yrs is 40.80 and simple interest is 40. Find the rate of interest per annum and the sum. Soln: A little reflection will show that the difference between the simple and compound interests for 2 yrs is the interest on the first year’s interest.

40 = 20 2 CI – SI = 40.8 – 40 = 0.80 Interest on 20 for 1 year = 0.80 First year’s SI =

Interest on 100 for 1 yr =

2 0.8 100 4% 40

Thus, in this case, rate =

80 100 = 4 100 20

rate = 4% Now, principal P is given by 100 I 100 40 = 500 tr 24 Quicker Method (Direct Formula) [for 2 yrs only]: P

2 Difference in CI and SI Rate 100 SI

80

r 20 4%

Division of sum Ex. 10: Divide 3903 between A and B, so that A’s share at the end of 7 yrs may equal B’s share at the end of 9 yrs, compound interest being at 4%. Soln:

We have, A’s share at present = 1

4 and B’s share at present = 1 100

4 100

7

9

A 's share at present 4 B' s share at present 1 100

2

2

676 26 25 625 Dividing 3903 in the ratio 676 : 625;

676 3903 = 2028 676 625 B’s present share = 3903 – 2028 = 1875 A’s present share =

When Difference Between SI and CI is given Theorem: When difference between the compound interest and simple interest on a certain sum of money for 2 years at r% rate is x, then the sum is given by: Difference 100 100 Sum Rate Rate

x(100) 2 100 x r r2

2

And when sum is given and difference between SI and CI is asked, then

r Difference = Sum 100

2

Quicker Maths

282 Proof: Let the sum be A. Then, SI =

3

r S CI = S 1 100

A 2 r 2Ar 100 100

3 r S 1 – 1 100

2

r r2 2r CI A 1 A A 1 – A 2 100 100 100

r3 3r 3r 2 S 1 1 3 2 (100) 100 100

Ar 2 2Ar Ar 2 2Ar A A 2 100 100 1002 100 Now, CI – SI =

Ar 2 2Ar 2Ar r A 2 100 100 100 100

100 A = Difference r

r3 3r 2 3r S 3 2 100 100 100

2

r3 3r 2 3r 3Sr Now, CI – SI S – 3 2 100 100 100 100

2

Ex. 11: The difference between the compound interest and the simple interest on a certain sum of money at 5% per annum for 2 years is 1.50. Find the sum. Soln: Using the above theorem:

r3 3r 2 S 3 2 100 100 or, Difference

2

100 Sum 1.5 1.5 400 = 600 5

Ex. 12: Find the difference between the compound interest and the simple interest for the sum 1500 at 10% per annum for 2 years. 2

Soln:

S

2

r 10 Sum 1500 100 100 = 15

Ex 13: The difference between the simple and the compound interests on a certain sum of money for 2yrs at 4% per annum is Re 1. Find the sum. 2

Soln:

Theorem: If the difference between CI and SI on a certain sum for 3 years at r% is x, the sum will be Difference (100)3 and if the sum is given and r 2 ( 300 r)

Sum

Sr 2 ( 300 r) (100)3

Proof: Let the sum be S. Then, SI =

S 3 r 3Sr 100 100

Difference (100)3 r 2 (300 r)

122 100 100 100 = 16,000 52 (300 5)

Ex 15: Find the difference between CI and SI on 8000 for 3 yrs at 2.5% p.a. Soln:

the difference is asked, then Difference

Sr 2 (300 r) (100)3

Ex 14: If the difference between CI and SI on a certain sum of money for 3yrs at 5% p.a. is 122, find the sum. Soln: By the above theorem:

2

100 100 Sum difference 1 r 4 = 625

Sr 2 r 3 1002 100

Difference

Sum r 2 (300 r) (100)3

8000 2.5 2.5(300 2.5) 100 100 100

8 25 25 3025 121 = 15.125 100 100 100 8 Ex 16: Find the difference between CI and SI on 2000 for 3 yrs at 5% p.a.

Soln:

Difference

2000 5 5(305) = 15.25 100 100 100

Compound Interest

283

Ex 17: The simple interest on a sum at 4% per annum for 2 yrs is 80. Find the compound interest on the same sum for the same period. Soln: Recall the formula used in Ex 6.

Rate

2 Diff in CI & SI 100 SI

Rate SI or, Diff in CI and SI = 2 100

Difference in CI and SI =

4 80 1.6 2 100

CI = 80 + 1.6 = 81.6 Other approach: If you don’t remember the formula, then understand the following conclusion: “Compound interest differs from simple interest for 2 yrs only because under compound interest, simple interest over first year’s simple interest is also included.” Simple interest for 2 yrs = 80 Simple interest for 1 yr = 80 ÷ 2 = 40 CI for 2yrs = SI for 2 yrs + SI over SI of first 40 4 1 80 1.6 = 81.6 100 If you have understood the above conclusion, you may write the direct solution as year = 80

80 4 CI 80 80 1.6 = 81.6 2 100 Ex 18: The compound interest on a certain sum of money for 2 years at 10% per annum is 420. Find the simple interest at the same rate and for the same time. Soln: By the formula given in Ex 8. we have SI

rt t r 100 1 – 1 100

CI

When t = 2, SI

SI

2r 2r CI 100 CI 2 2 r 200r r 2r 100 1 – 1 2 100 100

200r CI r(r 200)

Now, it is easy to use this form of the above formula. Therefore, 200 10 420 SI = 400 10 210

Another Quicker Approach: SI for 2 yrs at 10% = 20% of capital CI for 2 yrs at 10% = 2 10

102 = 21% of capital 100

Now, 21% of capital = 420 420 20% of capital = 21 20 = 400 Ex 19: If the compound interest on a certain sum of money for 2 years at 5% is 246, find the simple interest at the same rate for the same time. Soln: Using the above formula: SI

200 5 246 = 240 5(205)

Note: (1) When CI is given and SI is asked then we apply the above used formulae in Ex 18 and Ex 19. (2) But when SI is given and CI is asked we use the method as used in Ex 17. Another Quicker Approach: SI for 2 yrs at 5% = 10% of capital

25 = 10.25% of capital 100 Now, 10.25% of capital = 246 CI for 2 yrs at 5% = 10

246 10% of capital = 10.25 10 = 240. Ex 20: If the simple interest on a certain sum of money for 3 yrs at 5% is 150, find the corresponding CI. Soln: Whenever the relationship between CI and SI is asked for 3 yrs of time, we use the formula: SI

rt t r 100 1 – 1 100

150

CI

53 3 r 100 1 – 1 100

CI

9261 – 8000 150 100 8000 CI 5 3 150 100 1261 1261 = 157.625 5 3 8000 8 Another Quicker Approach: SI = 5 × 3 = 15% of capital

=

Quicker Maths

284 3 52 53 15 % of capital CI = 100 1002 = (15 + 0.75 + 0.0125)% of capital = 15.7625% of capital Now, 15% of capital = 150 15.7625% of capital = 157.625 Ex 21: The CI on a certain sum is 104 for 2 yrs and SI is 100. What is the rate per cent? Soln: Difference in CI and SI = 104 - 100 = 4 Therefore, by using the formula (used in Ex 5 & Ex 14)

2 Diff 100 2 4 100 8% SI 100 Ex 22:An amount of money grows upto 4840 in 2 yrs and upto 5324 in 3 yrs on compound interest. Find the rate per cent. Soln: We have, P + CI of 3 yrs = 5324 -------- (1) P + CI of 2 yrs = 4840 -------- (2) Subtracting (2) from (1), we get CI of 3rd year = 5324 – 4840 = 484 Thus, the CI calculated in the third year which is 484 is basically the amount of interest on the amount generated after 2 years which is 4840. Rate

Ex 24: Find the compound interest on 18,750 in 2 yrs, the rate of interest being 4% for the first year and 8% for the second year. Soln: After first year the amount

4 104 18750 = 18750 1 100 100 104 108 After 2nd year the amount = 18750 100 100 26 27 = 18750 = 21060 25 25 Ex. 25:

Soln:

4

r 4800 1 6000 100 4

r 6000 5 or, 1 100 4800 4 r Now, 1 100

484 100 10% 4840 1 Quicker Method (Direct Formula): r

Rate

(5324 4840) 484 100 100 10% 4840 4840 Note: The above generalised formula can be used for any positive value of n. See in the following example. Ex 23: A certain amount of money at compound interest grows upto 51168 in 15 yrs and upto 51701 in 16 yrs. Find the rate per cent per annum. Soln: Using the above formula:

3

5 125 4 64

12

12

r or, 4800 1 9375 100

In this case, n = 2 Difference of amount after 2yrs and 3yrs 100 Amount after 2 yrs

4 3

r 125 75 9375 or, 1 100 64 75 4800

Difference of amount after n yrs and(n 1)yrs 100 Amount after n yrs

rate

CI = 21060 – 18750 = 2310 4800 becomes 6000 in 4 years at a certain rate of compound interest. What will be the sum after 12 years? Detail method: We have:

The above equation shows that 4800 becomes 9375 after 12 years. Direct formula:

(51701 – 51168) 100 533 100 Rate 51168 51168

100 25 1 1 % 96 24 24

12

Required amount = Note:

(6000)

4

12 –1 4

(4800)

(6000)3 (4800) 2

= 9375 Thus, we can say that: “If a sum `A’ becomes `B’ in t1 years at compound rate of interest, then after t2 years the sum becomes

(B) (C)

t2

t2

t1

t1 1

rupees.”

Ex. 26: Find the compound interest on 10000 for 3 years if the rate of interest is 4% for the first year, 5% for the second year and 6% for the third year.

Compound Interest Soln:

285

The Compound Interest on x in ‘t’ years if the rate of interest is r1% for the first year, r2% for the second year ... and rt% for the tth year is given by r r r x 1 1 1 2 .... 1 t x 100 100 100 In this case; Compound interest 4 5 6 10000 1 1 1 10000 100 100 100

Soln:

n

r P 1 100 n 1 n2 l r r r A 1 1 .... 1 100 100 100 In the above case:

26 21 53 10000 10000 25 20 50 Note:

= 11,575.20 – 10,000 = 1,575.2 The more general formula for this type of question can be given as: If the compound rate of interest for the first t1 years is r1%, for the next t2 years is r2%, for the next t3 years is r3%, ... and the last tn years is rn%, then compound interest on x for (t1 + t2 + t3 + .... tn) years is

r1 r r 1 2 ...... 1 n – x x 1 100 100 100 In the above case, t1 = t2 = t3 = 1 year. Ex. 27: What sum of money at compound interest will amount to 2249.52 in 3 years if the rate of interest is 3% for the first year, 4% for the second year, and 5% for the third year? Soln: The general formula for such question is: t1

t2

tn

r r r A P 1 1 1 2 1 3 ... 100 100 100 where A = Amount, P = Principal and r1 , r2 , r3 are the rates of interest for different years. In the above case:

3 4 5 2249.52 P 1 1 1 100 100 100 or, 2249.52 = P (1.03) (1.04) (1.05) 2249.52 P 1.03 1.04 1.05 = 2000

Direct formula: By the rule of fraction:

100 100 100 = 2000 Principal = 2249.52 103 104 105 Ex. 28: A man borrows 3000 at 10% compound rate of interest. At the end of each year he pays back 1000. How much amount should he pay at the end of the third year to clear all his dues?

The general formula for the above question may be written as: If a man borrows P at r% compound interest and pays back A at the end of each year, then at the end of the nth year he should pay

3 2 1 10 10 10 3000 1 1000 1 1 100 100 100

11 2 11 11 11 11 3000 – 1000 10 10 10 10 10

Note:

121 11 3993 – 1000 1000 100 10 = 3993 – 1210 – 1100 = 1683 The above question may be solved like: Principal at the beginning of the 2nd year 10 3000 1 1000 3300 – 1000 = 2300 100 Principal at the beginning of the third year

10 – 1000 = 1530 = 2300 1 100

at the end of the third year, he should pay 10 = 1683 1530 1 100 Ex. 29: Geeta deposits 20,000 in a private company at the rate of 16% compounded yearly; whereas Meera deposits an equal sum in PNB Housing Finance Ltd at the rate of 15% compounded half1 yearly. If both deposit their money for 1 years 2 only, calculate which deposit earns better interest. Soln: If you are not asked to find the absolute values, the question becomes easier. 1 Amount after 1 years in both the cases will be 2 3 3 16 2 7.5 20000 1 20000 1 and 100 100 Now move for inequality:

16 20000 1 100

3

2

7.5 20000 1 100

3

Quicker Maths

286 16 or, 1 100

3

2

7.5 1 100

Ex. 30: A sum of money is lent out at compound interest rate of 20% per annum for 2 years. It would fetch 482 more if interest is compounded half-yearly. Find the sum. Soln: Suppose the sum is P. CI when interest is compounded

3

On raising both sides to the power 2 3 ,

16 7.5 or, 1 1 100 100

2

2

20 P yearly P 1 100 CI when interest is compounded half-yearly

2

or, 1 or,

16 7.5 7.5 1 2 100 100 100

4

10 P 1 P 100

16 15 7.5 7.5 100 100 100 100

4

Note:

4 2 P 1.1 – 1.2 482

56.25 or, 1 100 Clearly LHS is greater than RHS. Thus, Geeta gets better interest. If you are asked to find the value by which Geeta earns more than Meera, you will have to calculate 3

2

10 20 P 1 482 Now, we have, P 1 100 100

7.5 7.5 or, 1 100

P (1.1) 2 – (1.2) (1.1) 2 (1.2) 482 P[{1.21 1.2}{1.21 1.2}] 482 P[(0.01)(2.41)] 482

3

16 2 7.5 20000 1 and 20000 1 . 100 100

P

482 = 20,000 2.41 0.01

EXERCISES 1. Find the amount of 6400 in 1 year 6 months at 5 p.c. compound interest, interest being calculated halfyearly. 2. Find the compound interest on 10000 in 9 months at 4 p.c., interest payable quarterly. 3. Find the difference between the simple and the compound interests on 1250 for 2 years at 4 p.c. per annum. 4. I give a certain person 8000 at simple interest for 3

1 p.c. How much more should I have gained 2 had I given it at compound interest? 5. A merchant commences with a certain capital and gains annually at the rate of 25 p.c. At the end of 3 years he has 10,000. What was his original capital? 6. In what time will 1200 amount to 1323 at 5 p.c. compound interest? 7. In what time will 2000 amount to 2431.0125 at 5 p.c.per annum compound interest? years at 7

8. In what time will 6250 amount to 6632.55 at 4 p.c. compound interest payable half-yearly? 9. At what rate per cent compound interest will 400 amount to 441 in 2 years? 10. At what rate per cent compound interest will 625 amount to 676 in 2 years? 11. At what rate per cent compound interest does a sum of

9 times itself in 2 years? 4 12. At what rate per cent compound interest does a sum of money become fourfold in 2 years? 13. If the difference between the simple interest and the compound interest on a certain sum of money for 3 years at 5 per cent per annum is 122, find the sum. 14. The simple interest on a certain sum of money for 4 years at 4 per cent per annum exceeds the compound interest on the same sum for 3 years at 5 per cent per annum by 57. Find the sum. money become

Compound Interest 15. A sum of money at compound interest amounts in two years to 2809, and in three years to 2977.54. Find the rate of interest and the original sum. 16. A sum is invested at compound interest payable annually. The interest in two successive years was 225 and 236.25. Find the rate of interest and the principal. 17. Raghu invested a certain sum in Scheme X for 4 years. Scheme X offers simple interest @ 12 pcpa for the first two years and compound interest (compounded annually) @ 20 pcpa for the next two years. The total interest earned by him after 4 years is 11016. What was the sum invested by Raghu in Scheme X? 18. A person invested equal amounts in two schemes A and B at the same rate of interest. Scheme A offers simple interest while scheme B offers compound interest. After two years he got 1920 from scheme A as interest and 2112 from scheme B. If the rate of interest is increased by 4%, what will be the total interest after two years from both schemes? 19. Raman took a loan of 15,000 from Laxman. It was agreed that for the first three years the rate of interest charged would be at 8% simple interest per annum and at 10% compound interest (compounded annually) from the fourth year onwards. Ram did not pay anything until the end of the fifth year. How much would he have to repay if he clears the entire amount only at the end of the fifth year? (in rupees)

2 3 of the total money in scheme A for 6 years and rest of the money he invested in scheme B for 2 years. Scheme A offers simple interest at a rate of 12% p.a. and scheme B offers compound interest (compounded annually) at a rate of 10% p.a. If the total interest obtained from both the schemes is 2,750, what was the total amount invested by him in scheme A and scheme B together? (Approximate value) 21. Javed invested equal sums in schemes A and B. Both the schemes offer same rate of interest ie, 8 p.c.p.a. The only difference is scheme A offers compound interest (compounded annually) and scheme B offers simple interest. If the difference between interests accrued by Javed from both the schemes after two years is 53.76, what sum was invested by him in each of the schemes ? 22. A sum of money was invested for 14 years in Scheme A, which offered simple interest at a rate of 8% pa. The 20. Shashi had a certain amount of money. He invested

287

23.

24.

25.

26.

amount received from Scheme A after 14 years was then invested for two years in Scheme B, which offers compound interest (compounded annually) at a rate of 10% pa. If the interest received from Scheme B was 6678, what was the sum invested in Scheme A? 6100 was partly invested in Scheme A at 10% pa compound interest (compounded annually) for 2 years and partly in Scheme B at 10% pa simple interest for 4 years. Both the schemes earn equal interests. How much was invested in Scheme A? The respective ratio of the sums invested for 2 years each, in Scheme A offering 20% per annum compound interest (compounded annually) and in Scheme B offering 9% pa simple interest is 1 : 3. Difference between the interests earned from both the schemes is 1200. How much was invested in Scheme A? A certain sum is invested for 2 years in scheme A at 20% pa compound interest compounded annually. The same sum is also invested for the same period in scheme B at x% pa simple interest. The interest earned from scheme A is twice that earned from scheme B. What is the value of x? Shyama invested P for 2 years in scheme A, which offered 11% pa simple interest. She also invested 600 + P in scheme B, which offered 20% compound interest (compounded annually) for 2 years. If the amount received from scheme A was less than that received from scheme B by 1216 then what is the value of P?

1 27. A invests 800 at simple rate of interest 12 % pa for 2 2 years. B invests a certain sum at compound rate of interest 10% pa compounded annually for 3 years. The amounts that A and B receive individually are in the ratio of (10)3 : (11)3 respectively. What is the sum invested by B? 28. The compound interest (compounded annually) on 9300 for 2 years @ R% pa is 4092. Had the rate of interest been (R-10)%, what would have been the interest on the same sum of money for the same time? (2 years) 29. Poona invests 4200 in Scheme A, which offers 12% pa simple interest. She also invests (4200 – P) in scheme B offering 10% pa compound interest (compounded annually). The difference between the interests Poona earned from both the schemes at the end of 2 years is 294. What is the value of P?

Quicker Maths

288

ANSWERS 1.

2.5 Amount 6400 1 100

7. Same as Q 6.

3

3

6400 41 41 41 41 6400 = 6892.1 40 40 40 40 2.

3 1 CI 10000 1 1 100

30301 10000 = 303.01 100 100 100 3. Quicker Maths: Use the formula (Diff for 2 yrs) r Difference Sum 100

2

2

4. Quicker Maths: Use the formula (Diff for 3 yrs) Sum r 2 (300 r) (100)3 8000 (7.5) (300 7.5) (100)3 2

= 138.375 = 138.38 Therefore, I will get 138.38 more. 5.

25 10000 x 1 100

x

5 6. 1323 = 1200 1 100

or,

2

663255 51 625000 50

t

3

132651 51 51 or, 125000 50 50

t

t=3 t 3 2 2 Note: As the interest was compounded half-yearly; we r and t to 2t. 2 9. From the table we find the answer directly as 5%. 10. From the table we find the answer directly as 4%. changed r to

9 r 11. S S 1 4 100 2

2

r 3 or, 1 2 100 or, 1

t

t

or,

2

r 3 100 2

r 1 100 2

r 50% 12. Same as Q. 11 Difference (100)3 r 2 (300 r)

122 100 100 100 = 16000 25 305 14. Let the sum be x.

t

21 21 or, 20 20

t = 2 years

or,

13. Sum

t

441 21 400 20

6632.55 51 625000 50

3

10000 4 4 4 = 5120 555

1323 21 1200 20

or,

t

Hence, the time is

4 1250 1250 = 2 100 625

Difference

2 8. 6632.55 = 6250 1 100

t

3 x44 5 57 x 1 1 Then, 100 100

Compound Interest

or,

4x 1261 57 x 25 8000

4 1261 57 or, x 25 8000 1280 1261 57 or, x 8000

57 8000 = 24000 19 Note: As the time is different for simple and compound interests, we didn’t find the quicker method (direct formula). 15. Difference in amounts = 2977.54 - 2809 = 168.54 Now, we see that 168.54 is the interest on 2809 in one year (it is either simple or compound interest because both are the same for a year).

x

168.54 100 6% 2809 Now, for the original sum, Hence, rate of interest

6 2809 x 1 100 53 or, 2809 x 50

2

2

2809 50 50 = 2500 53 53 16. Method is the same as in Q. 15. Difference in interest = 236.25 – 225 = 11.25 This difference is the simple interest over 225 for

x

11.25 100 5% one year. Hence, rate of interest 225 1 Now, since any particular number of years is not mentioned, we can’t find the sum. 17. Let the sum of money invested by Raghu be P. Then, 2 P 12 2 P1 20 – 1 11016 100 100

6 2 24 P or, 100 P 5 – 1 11016

or,

24 P 11P = 11016 100 25

289 24 P 44 P = 11016 100 or, 68P = 11016 × 100

or,

P=

11016 100 = 16200 68

Quicker Method: Rate of 20% pa CI for 2 yrs is equivalent to 20 + 20 + 20 20 44% 100 Therefore, total interest = (2 × 12)% of x + 44% of x = 11016 68% fo x = 11016 11016 100 = 16200 68 18. CI – SI = 2112 – 1920 = 192

x =

1920 = 960 2 Interest on 960 for 1 years = 192

SI for 1 year =

Rate =

192 100 = 20% per annum 960 1

960 100 = 4800 20 1 New rate = 24% per annum

Principal =

SI =

4800 24 2 = 2304 100

T R 1 CI = P 100 1 2 24 1 = 4800 100 1

= 4800 [(1.24)2 – 1] = 4800 (1.5376 – 1) = 4800 × 0.5376 = 2580.48 Total interest = (2304 + 2580.48) = 4884.48 19. Simple interest for first three years 15000 8 3 = = 3600 100 Amount after three years = 15000 + 3600 = 18600 Now, on this amount he pays 10% compound interest for 2 years. Amount = 186001 10 100

2

121 = 18600 = 22506 100 Hence, to clear loans after five years Raman requires to pay 22506.

Quicker Maths

290 20. Let the total sum of money possessed by Shashi be P. Then, 2 P 12 6 1 21% of P 3 100 3 48 P 21P 2750 100 300 165P or, = 2750 300

or,

2750 300 = 5000 165 21. Quicker Method : Let the investment in each scheme be P.

Now, 21% of 212% of A = 6678 6678 100 100 = 15000 21 212 23. Let the amount invested in Scheme B be x. Then the amount invested in Scheme A = (6100 – x) Now, according to the question,

A=

x 10 4 = (6100 – x) 100

P=

PR ² Difference between CI and SI for two years= (100)²

P 8 8 53.76 10000

53.76 10000 = 8400 88 22. Let the Principal invested in scheme A be x.

P=

Then SI =

p r t x 8 14 112x 100 100 100

112x 212x 100 100 Now, this amount is invested in Scheme B on compound interest for 2 years at the rate of 10% pa.

Amount = x

212x 10 Then, A = 1 100 100 212x 110 = 100 100

2

2

212 121x 25652x 10000 10000 Now, CI = A – P

=

6678 = or, 6678 =

25652x 212x – 10000 100 4452x 10000

6678 10000 x= = 15000 4452 Quicker Method: In scheme A, total interest is (14 × 8)% of A = 112% of A Amount after 14 yrs = 100% of A + 112% of A = 212% of A

or,

2 10 1 – – 1 100

40 x 121 – 100 (6100 – x ) 100 100

2x (21) (6100 – x ) 5 100 or, 200x = 6100 × 5 × 21 – 21 × 5 × x or, 200x + 105x = 6100 × 5 × 21 or, 305x = 6100 × 5 × 21

or,

6100 5 21 = 2100 305 The amount invested in Scheme A = 6100 – 2100 = 4000 Quicker Method: 10% rate of compound interest for 2 yrs is equivalent

x=

10 10 to 10 10 = 21%. 100 So, according to the question, 21% of A = (4 × 10)% of B

A 40 A : B = 40 : 21 B 21

6100 40 = 4000 40 21 24. Let the amount invested in scheme A be x. Amount invested in scheme B = 3x

Amount invested in A =

T R 1 CI obtained from scheme A = P 100 1 2 1 2 20 1 1 = x 100 = x 1 5 1

=

6 2 36 x 5 1 = x 1 = 25

SI from scheme B= =

11x 25

Pr incipal Time Rate 100

3x 2 9 = 100

54 x 100

Compound Interest

54 x 11x 1200 100 25

54 x 44 x 1200 100

291 22 P 1216 (144 6) = 1216 – 864 = 352 100

10x = 1200 100 x = 1200 × 10 = 12000 Quicker Method: 20% rate of CI for 2 yrs is equivalent to 44% Scheme A gives interest of 44% of x and scheme B gives interest of (2 × 9)% of 3x = 54% of x. According to the question, 54% of x – 44% of x = 1200 10% of x = 1200 x = 12000 25. Let the sum be p.

2 6 2 20 Then, CI = p1 100 – 1 = p 5 – 1

11p 36 – 25 = p = 25 25

25 2 = 1000 2 100 The amounts received by A and B are in the ratio of (10)3 : (11)3. Now, (10)3 = 1000 (11)3 = 1331 Thus, amount recieved by B after three years = 1331 Now, let the amount invested by B be x.

= 800 + 800 ×

Then, A = P 1 r 100

p x 2 2px px 100 100 50 Again, according to the question, 11p 2px 25 50 11p px or, 25 25 x = 11% Quicker Method: We know that equivalent rate of 20% of compound 20 20 interest for 2 yrs = 20 20 = 44% 100 Scheme A gives 44% of P as interest. Also scheme B gives 2r% of P as interest. According to the question, 44% of P = 2(2r% of P) 44 = 4r r = 11 26. Amount received from scheme A P 2 11 100P 22P 122P = P 100 100 100 Amount recieved from scheme B 2

144 = (P + 600) 100

Now, according to the question, 122P 1216 P 600 144 100 100

n

or, 1331 = x 1 10 100 x=

Now, SI =

20 = (P + 600) 1 100

352 100 = 1600 22 27. After 2 years the amount received by A

P=

3

1331 10 10 10 = 1000 1331

n R 28. CI = P 1 100 – 1

or, 1 CI 1 R P 100

n

Now, 1 4092 1 R 9300 100 or, 1 + 0.44 = 1 R 100

2

2

R or, 1.44 = 1 + 100

or, 1.2 = 1 +

R 100

R = 0.2 100 R = 20% Again, new rate = (20 – 10)% = 10% CI = 2 years @ 10% per annum

or,

10 10 = 10 10 = 21% 100 21% of 9300 = 21 × 93 = 1953 Quicker Approach: 4092 100 44% 9300

Quicker Maths

292 CI of 4092 is 44% of sum 9300 in 2 years. Rate of interest = 20% 20 20 20 20 44% 100

New rate of interest = 20 – 10 = 10% Equivalent CI rate = 21% 10 10 10 10 21% 100

21% of 9300 = 1953 29. SI for two years @ 12% = 2 × 12 = 24% CI for two years @ 10% per annum = 10 + 10 +

10 10 21% 100

Now, SI – CI = 294

4200 P 21 294 Now, 4200 24 100 100 or, 42 × 24 – 42 × 21 + or, 1008 – 882 + or,

21P 294 100

21P 294 100

21P 294 126 168 100

P=

168 100 = 800 21

Chapter 26

Alligation Alligation is the rule that enables us (i) to find the mean or average value of mixture when the prices of two or more ingredients which may be mixed together and the proportion in which they are mixed are given (this is Alligation Medial); and (ii) to find the proportion in which the ingredients at given prices must be mixed to produce a mixture at a given price. This is Alligation Alternate. Note: (1) The word Alligation literally means linking. The rule takes its name from the lines or links used in working out questions on mixture. (2) Alligation method is applied for percentage value, ratio, rate, prices, speed etc and not for absolute values. That is, whenever per cent, per hour, per kg, per km etc are being compared, we can use Alligation. Rule of Alligation: If the gradients are mixed in a ratio, then

By the alligation rule: (Quanitity of cheaper rice) 35 7 . (Quantity of dearer rice) 15 3 They must be mixed in the ratio 7 : 3. Ex. 2: How many kg of salt at 42 P per kg must a man mix with 25 kg of salt at 24 P per kg so that he may, on selling the mixture at 40 P per kg, gain 25% on the outlay? Soln:

42

CP of unit quantity of cheaper (c) CP of unit quantity of dearer (d) Mean price (m)

(d–m)

(m–c)

Then, (cheaper quantity) : (dearer quantity) = (d - m) : (m - c)

Solved Problems Ex. 1:

Soln:

8 10 Ratio = 4 : 5 Thus, for every 5 kg of salt at 24 P, 4 kg of salt at 42 P is used.

the required no. of kg = 25×

108

(360 paise)

15

0 90

Mean price (325 paise) 35

4 =20. 5

Milk and Water Ex. 3: A mixture of certain quantity of milk with 16 litres of water is worth 90 P per litre. If pure milk be worth 1.08 per litre, how much milk is there in the mixture? Sol: The mean value is 90 P and the price of water is 0 P. milk water

In what proportion must rice at 3.10 per kg be mixed with rice at 3.60 per kg, so that the mixture be worth 3.25 a kg? C.P. of 1 kg cheaper rice C.P. of 1 kg dearer rice (310 paise)

24 32

Quantity of cheaper CP of dearer – Mean price Quantity of dearer Mean price CP of cheaper We represent it as under:

100 P 32P per kg 125 ------- (By the rule of fraction)

Cost price of mixture = 40

Ex. 4:

90 – 0 108 –90 By the Alligation Rule, milk and water are in the ratio of 5 : 1. quantity of milk in the mixture = 5 × 16 = 80 litres. In what proportion must water be mixed with spirit 2 to gain 16 % by selling it at cost price? 3

Quicker Maths

294 Soln:

Let C.P. of spirit be 1 per litre. Then, S.P. of 1 litre of mixture = 1.

We first find the proportion in which wheat at 127 P and 132 P must be mixed to produce a mixture at 130 P. (i) 1st wheat 3rd wheat

2 Gain = 16 %. 3 100 3 1 6 = C.P of 1 litre of mixture = 350 7 C.P. of 1 litre water

C.P. of 1 litre pure spirit

127

132 130

2 3 The proportion is 2 : 3. We next find the proportion in which wheat at 129 P and 132 P must be mixed to produce a mixture at 130 P. 2nd wheat 3rd wheat

129 1 (Quantity of water) 7 1 . (Quantity of spirit) 6 6 7 Ex. 5:

Soln:

130

or, Ratio of water and spirit = 1 : 6 A butler stole wine from a butt of sherry which contained 40% of spirit. He replaced what he had stolen by wine containing only 16% spirit. The butt was then of 24% strength only. How much of the butt did he steal? Wine containing Wine containing 40% Spirit

16% Spirit Wine containing 24% Spirit

8

16

By alligation rule: wine with 40% spirit 8 1 wine with 16% spirit 16 2 i.e., they must be mixed in the ratio (1 : 2). Thus,

1 rd of the butt of sherry was left and hence the 3 2 rd of the butt. 3 Three ingredients—Number of proportions unlimited Ex. 6: In what proportion may three kinds of wheat at 1.27, 1.29 and 1.32 per kg be mixed to produce mixture worth 1.30 per kg? Soln: 1st wheat 2nd wheat 3rd wheat Mean Price 127 P 129 P 132 P 130 P Here, the first two prices are less and the third price is greater than the mean price. butler drew out

132

2 1 The proportion is 2 : 1. Now, in whatever proportion these two mixtures are mixed, the price of the resulting mixture will always be 130 P per kg because both mixtures cost 130 P/kg. Now, 5 kg of the first mixture is composed of 2 kg of wheat at 127 P and 3 kg of wheat at 132 P, and 3 kg of second mixture is composed of 2 kg of wheat at 129 P and 1 kg of wheat at 132 P; hence 5 + 3 or 8 kg of the resulting mixture is composed of 2 kg at 127 P, 2 kg at 129 P and (3+1) or 4 kg at 132 P. Hence, the required proportion is 2:2:4 or 1:1:2. Take another case: If we use (say) 4 kg of the first wheat, we must use 6 kg of the third wheat. Again, if we use (say) 10 kg of the second wheat, we must use 5 kg of the third wheat. There is, thus, another proportion. 1st 2nd 3rd 4 kg : 10 kg : 6 + 5 = 11 kg or, 4 : 10 : 11 The student can verify this result also. In fact, we can use any number of kg of the 1st or 2nd wheat as long as we use the necessary corresponding number of kg of the 3rd and hence the number of proportions is unlimited. Note: The above calculations can be simplified further. For this, follow the following rule: Rule Reduce the several prices to one denomination (like, 1.24, 1.31, 1.20 can be written as 124, 131 and 120) and place them under one another in order of magnitude, the least being uppermost. Set down the mean price to the left of the prices. Link the prices in pairs so that the prices greater and

Alligation

295

lesser than the average price go together. Then find the difference between each price and the mean price and place it opposite to the price with which it is linked. These differences will give the required answer. For example, the above example can be solved as: 130

Ex. 7:

Soln:

2

(= 132 – 130)

129

2

(= 132 – 130)

132

3+1=4

[= (130 – 127) + (130 – 129)]

the required proportion is 2 : 2 : 4 or 1 : 1 : 2. In what ratio must a person mix three kinds of wheat costing him 1.20, 1.44 and 1.74 per kg, so that the mixture may be worth 1.41 per kg? 1st wheat 2nd wheat 3rd wheat 120 144 174 following the above rule, we have, 141

Note:

127

120

3 + 33

[(144 – 141) + (174 – 141)]

144

21

[= 141 – 120]

174

21

[= 141 – 120]

8 of the weight of mixture. 13 Now, we want to form a mixture in which milk will In vessel B, milk =

9 of the weight of this mixture. 13 By alligation rule: 5 8 7 13 be

9 13

Therefore, the required ratio = 36 : 21 : 21 = 12 : 7 : 7 Try to get the other ratios which satisfy the conditions.

Four ingredients Ex. 8: How must a grocer mix 4 types of rice worth 54 P, 72 P, 1.20 and 1.44 per kg so as to obtain a mixture at 96 P per kg? Soln: First solution:

54 96

48

[=144 – 96]

72

24

[= 120 – 96]

120

24

[= 96 – 72]

144

42

[=96 – 54]

required proportion is 48 : 24 : 24 : 42 =8:4:4:7 Second solution: 54 96

24

[=120 – 96]

2 91

1 13

required proportion is

30%

12% 18%

6%

72

48

[= 144 – 96]

120

42

[= 96 – 54]

This means that

42

[=96 – 72]

required proportion is 24 : 48 : 42 : 24 =4:8:7:4 Different ways of linking will give different solutions.

1 2 : 7 :2 13 91

A butler stealing wine Ex. 10: A butler stole wine from a butt of sherry which contained 30% of spirit and he replaced what he had stolen by wine containing only 12% of spirit. The butt was then 18% strong only. How much of the butt did he steal? Soln: By the alligation rule, we find that wine containg 30% of spirit and wine containing 12% of spirit should be mixed in the ratio 1 : 2 to produce a mixture containing 18% of spirit.

Ratio = 6 : 12 = 1 : 2

144

Note:

Mixture from two vessels Ex. 9: Milk and water are mixed in a vessel A in the proportion 5 : 2, and in vessel B in the proportion 8 : 5. In what proportion should quantities be taken from the two vessels so as to form a mixture in which milk and water will be in the proportion of 9 : 4? 5 Soln: In vessel A, milk = of the weight of mixture 7

12%

1 rd of the butt of sherry was left, 3 2 i.e. to say, the butler drew out rd of the butt . 3 2 rd of the butt was stolen. 3

Quicker Maths

296 Ex. 11: A goldsmith has two qualities of gold — one of 12 carats and another of 16 carats purity. In what proportion should he mix both to make an ornament of 15 carats purity? Soln: I II 12 16

15 1

3

he should mix both the qualities in the ratio 1 : 3. Ex. 12: 300 gm of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution? Soln: The existing solution has 40% sugar. And sugar is to be mixed; so the other solution has 100% sugar. So, by alligation method: 40%

average price = 20 Applying the alligation rule: Chemical 25

10%

50%

The two mixtures should be added in the ratio 5 : 1. Therefore, required sugar =

300 1 60 gm 5

Direct formula: Quantity of sugar added Solution (required% value – present% value) (100 required% value)

Average speed 285 6

300(50 – 40) 60 gms 100 – 50 Ex. 13: There are 65 students in a class. 39 rupees are distributed among them so that each boy gets 80 P and each girl gets 30 P. Find the number of boys and girls in that class. Soln: Here, alligation is applicable for “money per boy or girl”.

Boys 80

3900 60 P 65 Girls 30

45 45 6 6 time spent in bus : time spent in train

20

45 45 : 1 :1 6 6

285 = 142.5 km 2 Ex. 16: In what ratio should milk and water be mixed so that after selling the mixture at the cost price a

distance travelled by train =

60 30 Boys : Girls = 3 : 2

Water 0

16 9 C : W = 16 : 9 Ex. 15: A person travels 285 km in 6 hrs in two stages. In the first part of the journey, he travels by bus at the speed of 40 km per hr. In the second part of the journey, he travels by train at the speed of 55 km per hr. How much distance did he travel by train? Soln: In this question, the alligation method is applicable for the speed. Speed of Bus Speed of train 40 50

In this case, Ans

Mean value of money per student =

100 = 16 /litre 125

16

100% 50%

65 3 39 3 2 and number of girls = 65 – 39 = 26 Ex. 14: A person has a chemical of 25 per litre. In what ratio should water be mixed in that chemical so that after selling the mixture at 20/litre he may get a profit of 25%? Soln: In this question, the alligation method is applicable on prices, so we should get the average price of mixture. SP of mixture = 20/ litre; profit = 25%

Number of boys

Soln:

2 profit of 16 % is made? 3 See soln 4.

Alligation

297

Short-cut Method: In such questions the ratio is

2 : 100 = 1 : 6 3 Ex. 17: In what ratio should water and wine be mixed so that after selling the mixture at the cost price a profit of 20% is made? Soln: Water : Wine = 20 : 100 = 1 : 5 Ex. 18: A trader has 50 kg of pulses, part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. What is the quantity sold at 18% profit? Soln: Detail Method Let the quantity sold at 18% profit be x kg. Then, the quantity sold at 8% profit will be (50 – x) kg. For a matter of convenience suppose that the price of pulse is 1 rupee per kg. Then, price of x kg pulse = x and price of (50-x) kg pulse = (50-x) Now, we get an equation. 18% of x + 8 % of (50 – x) = 14% of 50 18x + 8 (50 – x) = 14 × 50 10x = 300 x = 30 By Alligation Method: I part II part water : milk = 16

8% profit

18% profit 14% (mean profit)

4%

6%

=4:6=2:3 Therefore, the quantity sold at 18% profit

50 3 30 kg 23 Note: For the above example, both the detailed and alligation methods are given so that you can compare them and understand the importance of alligation method in Quicker Maths. Ex. 19: A trader has 50 kg of rice, a part of which he sells at 10% profit and the rest at 5% loss. He gains 7 % on the whole. What is the quantity sold at 10% gain and 5% loss? Soln: I part II part =

10

(–) 5 7

12

3

Ratio of quantities sold at 10% profit and 5% loss = 12 : 3 = 4 : 1. Therefore, the quantity sold at 10% profit = 50 4 40 kg and the quantity sold at 5% loss 4 1 = 50 – 40 = 10 kg. Note: Whenever there is a loss, take the negative value. Here, difference between 7 and (–5) = 7–(–5) = 7 + 5 =12. Never take the difference that counts negative value. Ex. 20: A trader has 50 kg of rice, a part of which he sells at 14% profit and the rest at 6% loss. On the whole his loss is 4%. What is the quantity sold at 14% profit and that at 6% loss? Soln: I part II part 14

(–) 6 (–) 4

2

(as there is a loss on the whole) 18

ratio of quantities sold at 14% profit and 6% loss = 2 : 18 = 1 : 9. quantity sold at 14% profit 50 1 5kg and sold at 6% loss 1 9 = 50 – 5 = 45 kg. Note: Numbers in the third line should always be +ve. That is why (–) 6 – (–)4 = –2 is not taken under consideration. Ex. 21: Mira’s expenditure and savings are in the ratio 3 : 2. Her income increases by 10%. Her expenditure also increases by 12%. By how many % does her saving increase? Soln: Expenditure Saving 12 x =

(% increase in exp)

(% increase in saving)

10 (% increase in income) 3

2 (given)

We get two values of x, 7 and 13. But to get a viable answer, we must keep in mind that the central value (10) must lie between x and 12. Thus, the value of x should be 7 and not 13. required % increase = 7% Ex. 22: A vessel of 80 litre is filled with milk and water. 70% of milk and 30% of water is taken out of the vessel. It is found that the vessel is vacated by 55%. Find the initial quantity of milk and water.

Quicker Maths

298 Soln:

Here, the % values of milk and water that is taken from the vessel should be taken into consideration. milk water 70% 30%

55%

is invested at 5%, the annual income is 5% of 1,500 = 75 But, real income = 85 Applying the alligation rule, we have 6% 5%

25% 15% 5:3 Ratio of milk to water = 5 : 3

quantity of milk =

80 5 = 50 litres 53

2

80 3 = 30 litres and quantity of water = 53 Ex. 23: A container contained 80 kg of milk. From this container, 8 kg of milk was taken out and replaced by water. This process was further repeated two times. How much milk is now contained by the container? Soln: Amount of liquid left after n operations, when the container originally contains x units of liquid from x y which y units is taken out each time is x x units. Thus, in the above case, amount of milk left

n

3

Ex. 24: Nine litres are drawn from a cask full of water and it is then filled with milk. Nine litres of mixture are drawn and the cask is again filled with milk. The quantity of water now left in the cask is to that of the milk in it as 16 : 9. How much does the cask hold? Soln: Let there be x litres in the cask. From the above formula we have, after n operations: n

2

x 9 16 16 Thus, in this case, x 16 9 25 Ex. 25:

Soln:

Money invested at 5% =

1

1 1,500 = 500 3

Note: Solve with the help of average rate of interest. Ex 26: A mixture of 40 litres of milk and water contains 10% water. How much water must be added to make 20% water in the new mixture? Soln: This question is the same as Ex 12. The existing mixture has 10% water. Water is to be added, so the other solution has 100% water. So, by alligation method 10% 100%

20%

80 8 80 kg 58.32 kg 80

Water left in vessel after n operations x y x Whole quantity of milk in vessal

:

x = 45 litres 1500 is invested in two such parts that if one part be invested at 6%, and the other at 5% , the total interest in one year from both investments is 85. How much is invested at 5%? If the whole money is invested at 6%, the annual income is 6% of 1,500 = 90. If the whole money

80% 10% The two mixture should be added in the ratio 8 : 1. i.e. for every 8 litres of first mixture, 1 litre of water should be added. Therefore, for 40 litres of first mixture

40 1 5 litres of water should be 8

added. By Direct formula: By the formula given in Ex 12. Required quantity of water

40(20 10) 5 litres 100 20 Note: (1) Both of the above methods are fast-working. But, usually it seems to be difficult to recall the formula in the examination hall. So, we suggest you to solve this type of questions by the Rule of Alligation. (2) In Ex 12, sugar was added in a solution of sugar, but in the above example water is added in a mixture of milk and water. In both the cases this method works. The only thing you should keep in mind is that the % value should be given for the same component in both the mixtures. For example, see the following case:

Alligation

299

“A mixture of 40 litres of milk and water contains 90% milk. How much water must be added to make 20% water in the new mixture?” In the above example, percentage value of milk (90%) is given in the first mixture and percentage value of water (20%) is given in the resulting mixture. So, the above example can be solved by both the ways. (1) By finding the % value of water in first mixture, or, (2) by finding the % value of milk in second mixture. For case (1): Water % in the mixture = 100 – 90 = 10% Now, the rest is the same as given in Soln (26). For case (2) : Percentage value of milk in the resulting mixture = 100 – 20 = 80%. Now, we can apply the alligation rule. The second mixture is water which has 0% milk; so 1st mixture 2nd mixture (water) 90% 0%

80% 80% 10% Ratio in which the two mixtures should be added is 8 : 1. Thus, we get the same result by this method also. Ex 27: In a zoo, there are rabbits and pigeons. If heads are counted, there are 200 and if legs are counted, there are 580. How many pigeons are there? Soln: This question can be solved by many ways. If you suppose the quantities to be x and y, then you get two equations and by solving them you get the required answer. We give you a direct formula for the questions when 2-legged and 4-legged creatures are counted together. No. of 4-legged creatures Total legs 2 Total heads = 2 No. of 2-legged creatures 4 Total heads – Total legs 110 2 number of pigeons (2-legged) 4 200 – 580 110 2 By Alligation Rule: Rule of Alligation is applicable on number of legs per head.

Average number of legs per head = Rabbit 4

580 29 200 10 Pigeons 2

29 10 11 10

9 10

Rabbit : Pigeons = 9 : 111 200 11 110 9 11 Ex 28: A jar contains a mixture of two liquids A and B in the ratio 4 : 1. When 10 litres of the mixture is taken out and 10 litres of liquid B is poured into the jar, the ratio becomes 2 : 3. How many litres of liquid A was contained in the jar? Soln: This question should have been discussed under the chapter “Ratio and Proportion”. But, as it is easy to solve it by the method of alligation, it is being discussed here. First, we see the method of alligation. Method I: In original mixture, % of liquid B

number of pigeons

1 100 20% 4 1 In the resultant mixture, % of liquid B 3 100 60% = 23 Replacement is made by the liquid B, so the % of B in second mixture = 100% Then, by the method of Alligation: 20% 100% =

60% 40% 40% Ratio in which first and second mixtures should be added is 1:1. What does it imply? It simply implies that the reduced quantity of the first mixture and the quantity of mixture B which is to be added are the same. Total mixture = 10 + 10 = 20 litres and liquid A =

20 4 = 16 litres 5

Method II: The above method is explained through percentage. Now, method II will be explained through fraction.

Quicker Maths

300 1 Fraction of B in original mixture = 5 Fraction of B in second mixture (liquid B) = 1 Fraction of B in resulting mixture =

3 5

So, 1 5

1 3 5

2 2 5 5 Thus, we see that the original mixture and liquid B are mixed in the same ratio. That is, if 10 litres of liquid B is added then after taking out 10 litres of mixture from the jar, there should have been 10 litres of mixture left. So, the quantity of mixture in the jar = 10 + 10 = 20 litres and quantity of A in the jar =

20 4 = 16 litres. 5

Method III: This method is different from the Method of Alligation. Let the quantity of mixture in the jar be 5x litre. Then

4 1 4x 10 :x – 10 10 2 : 3 ---(*) 4 1 4 1 or, 4x – 8 : x – 2 + 10 = 2 : 3

4x 8 2 x8 3 x=4 Then, quantity of A in the mixture = 4x = 4 × 4 = 16 litre Note(*): Liquid A in original mixture = 4x Liquid A taken out with 10 litres of mixture or,

4 litres 4 1 Remaining quantity of A in the mixture = 10

4 = 4x –10 5 Liquid B in original mixture = x Liquid B taken out with 10 litres of mixture

1 = 10 litres 5

Liquid B added = 10 litres

1 Total quantity of liquid B = x – 10 + 10 5 And the ratio of the two should be 2 : 3. Ex 29: 729 litres of a mixture contains milk and water in the ratio 7 : 2. How much water is to be added to get a new mixture containing milk and water in the ratio 7 : 3? Soln: Similar questions were discussed in examples 12 and 26. Previously, the percentage of components of mixture were given, but in this example components are given in ratio. Some methods to solve this question are being discussed below. Method I: Change the ratio in percentage and use the formula given in Ex. 12. % of water in the original mixture

2 200 100 % 72 9 % of water in the resulting mixture

3 100 30% 10 Quantity of water to be added =

200 729 30 9 729 70 = 81 litres 100 30 9 70 Method II: It is a little easier than the above method. You don’t need to find the percentage value of water. You can use the fractional value of water in the mixture. Use the formula given below: Required quantity of water to be added

Solution(Required fractional value –

Present fractional value) 1 (Required fractional value)

2 3 3 2 729 – 729 3 7 2 7 10 9 3 3 1 1 3 7 10

7 729 90 729 81 litres 7 9 10

Alligation

301

Method III: To solve this question by the method of alligation, we can use either of the two, percentage or fractional value. 200 % 9

100% 30% 70 % 9

70%

Therefore, the ratio in which the mixture and water

1 are to be added is 1 : or 9 : 1. 9 Then, quantity of water to be added

729 1 81 litres = 9 Note: Solve this question by this method. You can use the fractional value also. Try it. Theorem: If x glasses of equal size are filled with a mixture of spirit and water. The ratio of spirit and water in each glass are as follows: a 1 : b1 , a 2 : b 2 .....a x : b x. If the contents of all the x glasses are emptied into a single vessel, then proportion of spirit and water in it is given by a1 a2 ax a b a b .... a b : 1 1 2 2 x x b1 b2 bx a b a b ... a b 1 1 2 2 x x Ex 30: In three vessels each of 10 litres capacity, mixture of milk and water is filled. The ratios of milk and water are 2 : 1, 3 : 1 and 3 : 2 in the three respective vessels. If all the three vessels are emptied into a single large vessel, find the proportion of milk and water in the mixture. Soln: By the above theorem the required ratio is

3 3 2 2 1 3 1 3 2 2 3 3 3 4 5

Note:

1 2 1 : 2 1 3 1 3 2

1 1 2 : 3 4 5

40 45 36 20 15 24 : 121 : 59 3 4 5 3 45 This question can also be solved without using this theorem. For convenience in calculation, you will

have to suppose the capacity of the vessels to be the LCM of (2 + 1), (3 + 1) and (3 + 2), i.e. 60 litres. Because it hardly matters whether the capacity of each vessel is 10 litres or 60 litres or 1000 litres. The only thing is that they should have equal quantity of mixture.

1 Ex 31: If 2 kg of metal, of which rd is zinc and the rest 3 is copper, be mixed with 3 kg of metal, of which

Soln:

1 th is zinc and the rest is copper, what is the ratio 4 of zinc to copper in the mixture? Quantity of zinc in the mixture

1 1 2 3 8 9 17 2 3 3 4 3 4 12 12 Quantity of copper in the metal 3 2

17 17 43 5 12 12 12

17 43 ratio = 12 : 12 17 : 43 Ex 32: A man mixes 5 kilolitres of milk at 600 per kilolitre with 6 kilolitres at 540 per kilolitre. How many kilolitres of water should be added to make the average value of the mixture 480 per kilolitre? Soln: This question should be solved by the method of alligation. Cost of milk when two qualities are 5 600 6 540 6240 = per kilolitre 56 11 Cost of water = 0/ kilolitre So, First mixture Second mixture (milk) (water) 6240 0 11 mixed =

480 480

Ratio of milk and water

960 11

960 2 1: 11: 2 11 11 Which implies that 11 kilolitres of milk should be mixed with 2 kilolitres of water. Thus, 2 kilolitres of water should be added. = 480 :

Quicker Maths

302 Ex 33: If goods be purchased for 450 and one-third be sold at a loss of 10%, what per cent of profit should be taken on the remainder so as to gain 20% on the whole transaction? Soln: Ist Part 2nd Part -10%

x%

Ex 35: A man buys two horses for 1350 and sells one so as to lose 6% and the other so as to gain 7.5% and on the whole he neither gains nor loses. What does each horse cost? Soln: Ist horse 2nd horse

–6%

7.5%

20%

0 2 3

1 3

or 1 : 2 We see that 20 – (–10) = 20 + 10 = 30. As 2 is written in place of 30, there should be 15 in place of 1. Therefore, x = 20 + 15 = 35%.

1 th of the 4 goods be sold at a loss of 20%, at what gain per cent should the remainder be sold so as to gain 20% on the whole? Ist Part 2nd Part

Ex 34: If goods be purchased for 840 and

Soln:

-20%

x% 20% 3 4

1 4

or 1 : 3 We see that, 20 – (–20) = 40 is replaced by 3, so there should be

40 in place of 1. Then, 3

40 100 1 33 . 3 3 3 To find the value of x, you may use 20 (–20) 3 x 20 1 40 or, x – 20 = 3 x 20

Note:

x

40 100 1 20 33 % 3 3 3

20 (–10) 2 x 20 1 or, 2x – 40 = 30; And in Ex 33 :

x

70 35% 2

7.5 6 or, 5 : 4 Thus, we see that the ratio of the costs of the two horses is 5 : 4.

Cost of 1st horse =

1350 5 = 750 54

1350 4 = 600 5 4 Ex 36: A merchant borrowed 2500 from two money lenders. For one loan he paid 12% p.a. and for the other 14% p.a. The total interest paid for one year was 326. How much did he borrow at each rate? Soln: This example is similar to example 25. But we will solve it differently. Previously, the amount was used, but in this we will use the rate of interest. The merchant paid 326 as interest for his total borrowed amount. Then, average per cent of interest paid and cost of 2nd horse =

=

326 326 100 % 2500 25

12%

14% 326 % 25

24 26 25 25 ratio in which the amount should be divided is

24 26 : = 12 : 13 25 25 Thus, the amount lent at 12% 2500 12 = 1200 = 12 13

2500 13 = 1300 12 13 Ex 37: How many kg of tea at 42 per kg must a man mix with 25 kg of tea at 24 per kg so that he may, on selling the mixture at 40 per kg, gain 25% on the outlay? Soln: Solve yourself (see Ex 2). and amount lent at 14% =

Alligation

303

Ex 38:

1050 was divided among 1400 men and women so that each man gets 1 and each woman 50 paise. Find the number of women. (Ans. 700) Soln: Solve yourself (see Ex 13). Ex 39: A man travelled a distance of 80 km in 7 hours partly on foot at the rate of 8 km per hour and partly on bicycle at 16 km per hour. Find the distance travelled on foot. Soln:

80 km/hr 7 By alligation method: Foot Bicycle 8 16 80 7 32 24 7 7 Ratio of time travelled on foot and by bicycle Average speed =

32 24 : 4:3 7 7

7 4 = 4 hrs 43 Distance travelled on foot = 8 × 4 = 32 km Ex 40: Some amount out of 7000 was lent at 6% per annum and the remaining was lent at 4% per annum. The total simple interest from both the parts in 5 yrs was 1600. Find the sum lent at 6% p.a. (Ans: 2000) Soln: Solve it yourself by both the methods discussed in Ex 25 and Ex 36. Ex 41: Milk and water are mixed in a vessel A as 4 : 1 and in vessel B as 3 : 2. For vessel C, if one takes equal quantities from A and B, find the ratio of milk to water in C. (Ans: 7 : 3) Soln: Try yourself (see Ex 30 and the theorem used in it). Ex. 42: An army of 12,000 consists of Europeans and Indians. The average height of Europeans is 5ft 10 inches and that of an Indian is 5ft 9 inches. The 3 average height of the whole army is 5ft 9 inches. 4 Find the number of Indians in the army. Soln: Detail method: Let the number of Indians be x; then x(5 ft 9 in) (12000 x) (5 ft 10 in) 12000

or, x (69 in) + (12000 – x) (70 in) = 69.75 in × 12000 or, x = 12000 (70 – 69.75) = 12000 × 0.25 = 3000 By Method of Alligation (Quicker Method): Europeans Indians 70 69

69.75 0.75 0.25 ratio = 0.75 : 0.25 = 3:1 12000 1 3000 no. of Indians = 31 Ex. 43: In an alloy, zinc and copper are in the ratio 1:2. In the second alloy the same elements are in the ratio 2:3. In what ratio should these two alloys be mixed to form a new alloy in which the two elements are in ratio 5:8? Soln: Detail Method: Let them be mixed in the ratio x : y. Then, in 1st alloy, Zinc

Time travelled on foot =

5 ft 9

3 in 4

2nd alloy : Zinc = Now, we have or,

x 2x and Copper 3 3

2y 3y and Copper 5 5

x 2y 2x 3y : 5:8 3 5 3 5

5x 6y 5 10x 9y 8

or, 40x + 48y = 50x + 45y or, 10x = 3y

x 3 y 10

Thus, the required ratio = 3 : 10 By Method of Alligation (Quicker Method): You must know that we can apply this rule over the fractional value of either zinc or copper. Let us consider the fractional value of zinc. 1st alloy 2nd alloy 1 2 3 5 5 (mixture) 13 1 2 65 39 Therefore, they should be mixed in the ratio

Note:

1 2 1 39 3 : or, or 3 : 10 65 39 65 2 10 Try to solve it by taking fractional value of Copper.

Quicker Maths

304 Ex. 44: Jayashree purchased 150 kg of wheat at the rate of 7 per kg. She sold 50 kg at a profit of 10%. At what rate per kg should she sell the remaining to get a profit of 20% on the total deal? Soln: Selling price of 150 kg wheat at 20% profit

x 8.4

120 150 7 100 = 1260

0.7 0.35; 2

x 8.75 Selling price per kg of remaining 100 kg = 8.75

Selling price of 50 kg wheat at 10% profit

110 = 385 = 50 7 100

Selling price per kg of remaining 100 kg wheat 1260 – 385 = 8.75 100 By Method of Alligation: Selling price per kg at 10% profit = 7.70 Selling price per kg at 20% profit = 8.40 Now, the two lots are in ratio = 1 : 2

7.7

Ex. 45: How much water must be added to a cask which contains 40 litres of milk at cost price 3.5/litre so that the cost of milk reduces to 2/litre? Soln: This question can be solved in so many different ways. But the method of alligation method is the simplest of all the methods. We will apply the alligation on price of milk, water and mixture. Milk Water 3.5 0

Mean 2 2 1.5 ratio of milk and water should be 2 : 15 =4:3

x 8.4

1

8.4 7.7 2 x 8.4 1

added water

2

40 3 30 litres 4

EXERCISES 1. Gold is 19 times as heavy as water and copper 9 times. In what ratio should these metals be mixed so that the mixture may be 15 times as heavy as water? 2. How much chicory at 4 a kg should be added to 15 kg of tea at 10 a kg so that the mixture be worth 6.50 a kg? 3. A mixture of 40 litres of milk and water contains 10% water. How much water must be added to make water 20% in the new mixture? 4. A sum of 6.40 is made up of 80 coins which are either 10-paise or five-paise coins. How many are of 5 P? 5. In a zoo, there are rabbits and pigeons. If heads are counted, there are 200 and if legs are counted, there are 580. How many pigeons are there? 6. A man has 30,000 rupees to lend on loan. He lent some of his capital to Mohan at an interest rate of 20% per annum and the rest to Suresh at an interest rate of 12% per annum. At the end of one year, he got 17% of his capital as interest. How much did he lend to Mohan? 7. A vessel of 80 litre is filled with milk and water. 70% of milk and 30% of water is taken out of the vessel.

It is found that the vessel is vacated by 55%. Find the initial quantity of milk and water. 8. Mohan’s expenditures and savings are in the ratio of 4 : 1. His income increases by 20%. If his savings increase by 12%, by how much % should his expenditure increase? 9. Mukesh earned 4000 per month. From the last month, his income increased by 8%. Due to rise in prices, his expenditures also increased by 12% and his savings decreased by 4%. Find his increased expenditure and initial saving. 10. A man has 60 pens. He sells some of these at a profit of 12% and the rest at 8% loss. On the whole, he gets a profit of 11%. How many pens were sold at 12% profit and how many at 8% loss? 11. A man has 40 kg of tea, a part of which he sells at 5% loss and the rest at the cost price. In this business he gets a loss of 3%. Find the quantity which he sells at the cost price. 12. The ratio of milk to water in 66 kg of adulterated milk is 5 : 1. Water is added to it to make the ratio 5 : 3. The quantity of water added is ___________.

Alligation 13. Some amount out of 7000 was lent at 6% p.a. and the remaining at 4% p.a. If the total simple interest from both the fractions in 5 years was 1600, the sum lent at 6% p.a. was ______. 14. 729 ml of a mixture contains milk and water in the ratio 7:2. How much more water is to be added to get a new mixture containing milk and water in the ratio 7:3? 15. A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is ____________. 16. A sum of 41 was divided among 50 boys and girls. Each boy gets 90 paise and each girl 65 paise. The number of boys is ________. 17. A can contains a mixture of two liquids A and B in proportion 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the proportion of A and B becomes 7:9. How many litres of liquid A was contained by the can initially? 18. In a mixture of 60 litres, the ratio of milk to water is 2:1. If the ratio of milk to water is to be 1:2, then the amount of water to be further added is ___________. 19. A vessel contains 56 litres of a mixture of milk and water in the ratio 5:2. How much water should be mixed with it so that the ratio of milk to water be 4:5? 20. A sum of 39 was divided among 45 boys and girls. Each girl gets 50 P. whereas each boy gets one rupee. How many girls are there? 21. I mixed some water in pure milk and sold the mixture

2 at the cost price of the milk. If I gained 16 % , in what 3 ratio did I mix water in the milk? 22. Milk and water are mixed in vessel A in the ratio of 5:2 and in vessel B in the ratio of 8:5. In what ratio should quantities be taken from the two vessels so as to form a mixture in which milk and water will be in the ratio of 9:4? 23. There are two vessels A and B. Vessel A is containing 40 litres of pure milk and vessel B is containing 22 litres of pure water. From vessel A, 8 litres of milk is taken out and poured into vessel B. Then 6 litres of mixture (milk and water) is taken out and from vessel B poured into vessel A. What is the ratio of the quantity of pure milk in vessel A to the quantity of pure water in vessel B? 24. A vessel contains 64 litres of mixture of milk and water in the ratio 7 : 3 respectively. 8 litres of mixture is replaced by 12 litres of milk. What is the ratio of milk and water in the resulting mixture ?

305 25. From a container of milk, 5 litres of milk is replaced with 5 litres of water. This process is repeated again.Thus in two attempts the ratio of milk and water became 81 : 19. The initial amount of milk in the container was ______. 26. There was 120 litres of pure milk in a vessel. Some quantity of milk was taken out and replaced with 23 litres of water in such a way that the resultant ratio of the quantity of milk to that of water in the mixture was 4 : 1. Again 23 litres of the mixture was taken out and replaced with 28 litres of water. What is the ratio of milk to water in the resultant mixture? 27. In 120 litres of mixture of milk and water, water is only 25%. The milkman sold 20 litres of this mixture and then he added 16.2 litres of pure milk and 3.8 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture? 28. There are two jars - A and B containing a mixture of milk and water only. Jar A has 8 litres of mixture which has 25% water. Jar B has 14 litres of mixture. Milk and water in jar B are in the respective ratio of 6 : 1 . The content of both the jars are mixed. What is the percentage of water in the new mixture ? 29. 18 litres of pure water was added to a vessel containing 80 litres of pure milk. 49 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the ratio of 2 :1. If the resultant ratio of milk to water in the vessel was 4 : 1, what was the quantity of pure milk added to the vessel? (in litres) 30. A vessel contains a mixture of Grape, Pineapple and Banana juices in the respective ratio of 4 : 6 : 5. 15 litres of this mixture is taken out and 8 litres of grape juice and 2 litres of pineapple juice is added to the vessel. If the resultant quantity of grape juice is 10 litres less than the resultant quantity of pineapple juice, what was the initial quantity of mixture in the vessel ? (in litres) 31. In a 140-litre mixture of milk and water, the percentage of water is only 30%. The milkman gave 20 litres of this mixture to a customer. Then he added equal quantities of pure milk and water to the remaining mixture. As a result the ratio of milk to water in the mixture became 2 : 1. What was the quantity of milk added? (in litres) 32. A wholesaler blends two varieties of tea, one costing 60 per kilo and another costing 105 per kilo. The ratio of quantities they were mixed in was 7 : 2. If he sold the mixed variety at 100 per kilo, what was his profit percentage?

Quicker Maths

306 33. In a 90-litre mixture of milk and water, the percentage of water is only 30%. The milkman gave 18 litres of this mixture to a customer and then added 18 litres of water to the remaining mixture. What is the percentage of milk in the final mixture? 34. A jar has 40 litres milk. From the jar, 8 litres of milk was taken out and replaced with an equal quantity of water. If 8 litres of the newly formed mixture is taken out of the jar, what is the final quantity of milk left in the jar? 35. A mixture contains wine and water in the ratio of 3 : 2 and another mixture contains them in the ratio of 4 : 5. How many litres of the latter must be mixed with 3litres of the former so that the resultant mixture may contain equal quantities of wine and water ? 36. A vessel contains 180 litres of a mixture of milk and water in the ratio of 13 : 5. Fifty-four litres of this mixture was taken out and replaced with 6 litres of water. What is the approximate percentage of water in the resultant mixture? 37. In Jar A, 180 litres milk was with 36 litres water. Some of the mixture was taken out from Jar A and

put in jar B. If after adding 6 litres of water in the mixture, the ratio of milk to water in Jar B was 5 : 2, then what was the amount of mixture that was taken out from Jar A (in litres)? 38. A jar contains a mixture of milk and water in the ratio

1 of the mixture is taken out and 24 25 litres water is added to it. If the resultant ratio of milk to water in the jar was 2 : 1, what was the initial quantity of mixture in the jar? (in litres) of 3 : 1. Now,

39. If 2 kg of metal, of which

1 is zinc and the rest is 3

1 is 4 zinc and the rest is copper, then what will be the ratio of zinc to copper in the mixture? 40. A mixture of milk and water in a jar contains 28 L milk and 8 L water. X L milk and X L water are added to form a new mixture. If 40% of the new mixture is 20 L, then find the value of X? (in litres) copper, be mixed with 3 kg of metal, of which

SOLUTIONS 1.

Gold

Copper

19

4. Average value per coin =

9 15

10

6 4 Gold : Copper = 6 : 4 = 3 : 2 2. Chicory Tea 4 10 6.5

100%

3 2 10-paise coin : 5-paise coin = 3 : 2

Rabbits

4

10%

40 1 = 5 litres. 8

580 2.9 200 Pigeons

2 2.9

Impure Milk : Water = 8:1 Therefore, if there is 40 litres of impure milk, quantity of water is

80 2 = 32 5

5. Average legs per head =

20% 80%

5 8

5-paise coins =

3.5 2.5 Chicory : Tea = 3.5 : 2.5 = 7 : 5 Therefore, when tea is 15 kg, chicory should be 21 kg. 3. Impure Milk Water 10%

640 = 8P 80

0.9

1.1

Rabbits : Pigeons = 0.9 : 1.1 = 9 : 111 No. of pigeons =

200 11 110 20

Alligation 6.

307

Mohan

Suresh

20%

10.

12%

5% Ratio = 5 : 3

19% 1% Therefore, ratio of pens sold at profit to those sold at loss = 19 : 1

3%

30000 5 = 18750 amount given to Mohan = 8 7. Milk Water

70%

number of pens sold at 12% profit = 11.

At loss 5%

30%

Therefore, ratio of milk and water in the vessel = 5:3

80 5 = 50 litres 8

80 3 = 30 litres 8

8. Expenditure x

Savings 12

(% increase in exp)

At cost 0%

3 2 Ratio of quantity of tea sold at loss to those sold at cost price = 3:2

15%

Thus, milk =

60 19 57 20

3%

55% 25%

Loss 8%

11%

17%

Water =

Profit 12%

40 2 = 16 kg 5 Note: In Q. 10, we took loss as -ve because there was overall profit and thus each was presented in terms, of profit. [Profit = – (loss)] But in Q. 11 there is overall loss, and each is presented in terms of loss. Therefore, loss is taken as positive. 5 12. F1 = Fraction of milk in the adulterated milk = 6 F2 = Fraction of milk in water = 0

quantity sold at cost price =

(% increase in savings)

F3 = Fraction of milk in the new mixture =

20 (% increase in income) 4

1

or, 8 We see that x = 22% 9. % Inc. in Exp.

2

F1 5 6

(given)

5 8

% Inc in Sav.

12

–4 8 % Inc in income

12 4 Therefore, expenditure : savings = 12 : 4 = 3 : 1

4000 3 = 3000 4 and saving = 4000 – 3000 = 1000

Expenditure =

112 = 3360 Now, increased expenditure = 3000 100

5 8

F2

0 5 8

5 24

5 5 : 3:1 8 24 If we have 66 kg of adulterated milk, water 66 1 22 litres. = 3 1600 100 32 % 13. Overall rate of interest 5 7000 7 Rate for I amount Rate for II amount 4% 6% 32 % 7 (average rate)

4% 7

F1 : F2

10 % 7

Quicker Maths

308 ratio of two amounts = 2:5 amount lent at 6% =

7000 2 = 2000 7

14. Same as Q. 12 15. We will apply alligation on % profit. If he sells the milk at CP, he gains 0%. But if he sells water at CP, he gains 100%. Milk Water 0% 100%

25% 75% 25% Ratio of milk to water in the mixture should be 3:1 1 100 25% % of water in mixture = 31 16. Apply the alligation method on paise per head.

4100 82 Paise per student = 50 Boys Girls 90 65

82 17 Boys : Girls = 17 : 8

8

50 17 34 25 17. Apply alligation on fraction of A in each mixture. Original mixture B 7 0 12 New mixture 7 16 7 7 48 16 7 7 : 3:1 Ratio of original mixture to B 16 48 When 9 litres of B is mixed, original mixture should 9 be 3 27 litres. 1 Therefore, initial quantity in can = 27 + 9 = 36 litres. 18. Apply the alligation on fraction of milk in each mixture. Mixture Water 2 0 3

no. of boys =

1 3 1 3

19. 20. 21. 22.

Ratio of mixture to water = 1 : 1 Therefore, if there is 60 litres of solution, 60 litres of water should be added. Same as Q. 18. Apply the alligation on paise per head. Same as Q. 16. Same as Q. 15. Apply the alligation on fraction of milk in each vessel. Vessel A Vessel B 8 5 13 7 9 13 2 1 91 13 Ratio of quantity taken from vessel A and vessel B

1 2 : 7:2 13 91 23. Initially Milk in Vessel A = 40 lit Water in Vessel B = 22 lit After first operation: Milk in Vessel A = 40 – 8 = 32 lit Water in Vessel B = 22 lit Milk in Vessel B = 8 lit Mixture in Vessel B = 22 + 8 = 30 lit

After second operation (when 6 lit or

6 1 of the 30 5

mixture is taken out from B, it means

22 lit of water 5

and

8 lit of milk is taken out): 5

8 168 Milk in Vessel A = 32 lit 5 5 22 88 Water in Vessel B = 22 lit 5 5 168 : 88 = 21 : 111 5 5 24. In 64 litres of mixture, Reqd ratio =

7 × 64 = 44.8 litres 10 Water = 64 – 44.8 = 19.2 litres In 8 litres of mixture,

Milk =

7 × 8 = 5.6 litres 10 Water = 2.4 litres

Milk =

1 3

Alligation

309

In resulting mixture, Milk = 44.8 – 5.6 + 12 = 51.2 litres Water = 19.2 – 2.4 = 16.8 litres Required ratio = 51.2 : 16.8 = 64 : 21 25.

92 4 : 23 4 28 5 5 = (92 × 4) : (23 × 4) + (28 × 5) = 92 : (23 + 35) = 46 : 29 27. Quantity of water in 120 litres of mixture reqd ratio =

Milk ( remaining ) 81 Water 19

1 = 30 litres 4 Now, quantity of water in 20 litres mixture = 120 ×

Milk (remaining) 81 81 Milk (initial) 81 19 100

Remaining milk Quantity taken out = Initial concentration 1 Total Amount

5 81x = 100x 1 k 81 5 1 100 k

n

2

28.8 100 = 24% 120 28. In vessel A,

2

2

5 9 1 10 K

=

2

5 9 5 9 1 1 1 k = 50 litres k 10 k 10 10 Quicker Method : Initially the milk was 100% and after two operations it

81 100 81% 81 19 Now, if we suppose x% of milk is taken out in each operation, then after 2 opertaions (100 –x)% (100 – x)% = 81% = 90% × 90% = (100 10)% (100 10)% x =10% We are given that 10% 5 litres litres 26. Initially milk =120 litres. Aftar change, ratio of m : w = 4 : 1 1 23 litres; 4 92 litres milk is 92 litres and water is 23 litres out of total (92 + 23=) 115 litres reduces to

1 of 115 litres) is taken out and 5 28 litres water is added. Now 23 litres (ie

4 4 th of 92 milk and th of (23) + 28 litres 5 5 water is in the final mixture .

1 = 5 litres 4 Reamaining water in the remaining mixture = 30 – 5 = 25 litres Now, 3.8 litres of water is added to the mixture. Then the quantity of water = 25 + 3.8 = 28.8 litres Required percentage of water in the new mixture = 20 ×

Milk = 8 ×

3 = 6 litres 4

Water = 8 ×

1 = 2 litres 4

In vessel B, Milk =

6 × 14 = 12 litres 7

1 × 14 = 2 litres 7 In 22 litres of mixture, Water = 4 litres Water =

4 100 200 18 2 % 22 11 11 29. Initial amount of water = 18 l Amount of milk = 80 l Amount of mixture = 18 + 80 = 98 l 49l or half of the mixture was sold. In the remaining Required percent =

18 80 49l of mixture, there is 9 l of water and 2 2 40 l of milk. Now, suppose 2x litres of milk and x litres of water are added.

40 2x 4 9 x 1 or, 40 + 2x = 36 + 4x or, x = 2 litres 2x = 4 litres

Quicker Maths

310 Quicker Method : When 49 liters of mixture is taken out from (80 + 18 =) 98 litres of the mixture, 49 litres is left in the ratio M : W = 40 : 9 (the same ratio as it was in the beginning as 80 : 18 = 40 : 9)

In ratio terms addition of water or milk = 1 litres 32. The ratio of quantities is 7 : 2. Then, reqd % profit =

9 100 (7 60 2 105) 100 7 60 2 105

900 – 630 270 300 6 100 100 % 42 % 630 630 7 7 90 30 33. Water in the mixture = = 27 litres 100 Milk in the mixture = 90 – 27 = 63 litres Now, in 18 litres mixture the quantity of milk

=

Which is equal to the final ratio given in the question. 4 litres of milk and 2 litres of water was added. 30. Total initial quantity of juice in the vessel = 4x + 6x + 5x = 15x litres In 15 litres of juice, Grape juice = 4 litres Pineapple juice = 6 litres Banana juice = 5 litres Now, according to the question, (6x – 6 + 2) – (4x – 4 + 8) = 10 6x – 4 – 4x – 4 = 10 2x – 8 = 10 2x = 10 + 8 = 18 x=9 Intitial quantity of mixture = 15x = 15 × 9 = 135 litres 31. According to the question, milk in 20 litres mixture 20 70 = = 14 litres 100 20 30 Water in 20 litres mixture = = 6 litres 100 Suppose x litres of milk and water each was added to the mixture. 140 70 – 14 x 2 100 Now, 140 30 1 – 6 x 100

or,

98 – 14 x 2 (42 – 6) x 1

84 x 2 36 x 1 or, 2x + 72 = 84 + x or, x = 84 – 72 = 12 litres Quicker Method: Ratio of milk to water remains the same M : W = 7 : 3 in (140 – 20 =) 120 litres

or,

18 63 = 12.6 litres 90 Remaining quantity of milk in the mixture = 63 – 12.6 = 50.4 litres 50.4 100 Reqd % of milk in the mixture = = 56% 90 Quicker Method : Initially M = 70% 18 litres out of 90 litres is taken out. is taken out of milk remains final % of milk = 80% of 70% = 56% 34. Milk in a jar = 40 litres After first operation: 40 – 8 = 32 litres milk. Now, 8 litres of water is added to the milk. Then, new mixture = 32 + 8 = 40 litres Again, in new mixture the ratio of milk to water = 32 : 8 = 4 : 1 Now, 8 litres mixture is drawn out. Then, quantity of milk in 8 litres mixture

=

4 32 = 8 = 6.4 litres mixture 5 5 Remaining milk in jar = 32 – 6.4 = 25.6 litres Quicker Method : 8 litres out of 40 litres is taken out 20% of milk is taken out of milk remains So after two operations, quantity of remaining milk = 80% 80% of 40 litres = 64% of 40 litres = 25.6 litres 35. Wine Water First mixture 3x 2x Second mixture 4y 5y In 3 litres of first mixture: Wine = 1.8 l

Alligation

311

Water = 1.2 l When 9y of second mixture is added 1.8 l + 4y = 1.2 l + 5y or, y = 0.6 l Resultant mixture = 9y = 9 × 0.6 = 5.4 litres 36. Total mixture = 180 litres Now, 54 litres mixture is taken out Then the remaining mixture = 180 – 54 = 126 litres Quantity of milk in the mixture = 126

Clearly, 2 – 1= 1 6 (1 + 5) 6 × 6 = 36 (As ratio of milk to water in Jar ‘A’ is also 5 : 1) 38. Let the initial quantity of the mixture be 100x. Then, milk = 75x Water = 25x

1 the mixture is taken out. 25 Then quantity of milk = 72x Quantity of water = 24x So, Now,

13 = 91 litres 18

Quantity of water in the mixture = 126

5 = 35 18

litres When 6 litres of water is replaced, the new mixture = 126 + 6 = 132 litres In the new mixture quantity of water = 35 + 6 = 41 litres 41 100 31% 132 37. The ratio of milk to water in Jar A = 180 : 36 = 5 : 1 Now, let 6x litres of mixture be taken out from Jar A and put in Jar B. Then, milk in Jar B = 5x Water in Jar B = x

Reqd % of water =

5x 5 x6 2 or, 10x = 5x + 30 or, 5x = 30 x=6 Hence the mixture that was taken out from Jar A = 6x = 6 × 6 = 36 litres Quicker Approach: Initially, the ratio of milk to water remains the same in both the jars A and B.

72x 2 24x 24 1 or, 72 x = 48 x + 48 or, 24x = 48 x= 2 Initial quantity of mixture = 100 × 2 = 200 litres

1 1 39. In new metal, zinc = 2 3 3 4 2 3 And copper = 2 3 3 4 23 3 4 Now, new ratio = 4 9 3 4

So,

40.

8 9 12 17 12 16 27 43 = 17 : 43 40% 20L

20 100 = 50 litres 40 Now, 28 + x + 8 + x = 50 2x = 50 – 36 = 14 100%

x=

14 = 7 litres 2

312

Quicker Maths

Chapter 27

Time and Work If ‘M1’ persons can do ‘W1’ work in ‘D1’ days and ‘M2’ persons can do W2 work in ‘D 2, days then we have a very general formula in the relationship of

M 1 D1 W2 M 2 D 2 W1 . The above relationship can be taken as a very basic and all-in-one formula. We also derive 1) More men less days and conversely more days less men. 2) More men more work and conversely more work more men. 3) More days more work and conversely more work more days. If we include the working hours (say T1 and T2) for the two groups then the relationship is M1D1T1W2 = M2D2T2W1 Again, if the efficiency (say E1 and E2) of the persons in two groups is different then the relationship is M1D1T1E1W2 = M2D2T2E2W1 Now, we should go ahead starting with simpler to difficult and more difficult questions. Ex. 1: ‘A’ can do a piece of work in 5 days. How many days will he take to complete 3 works of the same type? Soln: We recall the statement: “More work more days” It simply means that we will get the answer by multiplication. Thus, our answer = 5 × 3 = 15 days. This way of solving the question is very simple, but you should know how the “basic formula” could be used in this question. Recall the basic formula: M1D1W2 = M2 D2 W1 As ‘A’ is the only person to do the work in both the cases, so M1 = M2 = 1 (Useless to carry it) D1 = 5 days, W1 = 1, D2 = ? and W2 = 3 Putting the values in the formula we have, 5 × 3 = D2 × 1 or, D2 = 15 days. Ex. 2: 16 men can do a piece of work in 10 days. How many men are needed to complete the work in 40 days? Soln: To do a work in 10 days, 16 men are needed. Or, to do the work in 1 day, 16 ×10 men are needed.

So, to do the work in 40 days,

16 10 4 men 40

are needed. This was the method used for non-objective exams. We should see how the “basic formula” works here. M1 = 16, D1 = 10, W1 = 1 and M2 = ?, D2 = 40, W2 = 1 Thus, from M1D1W2 = M2D2W1 16 × 10 = M2 × 40

16 10 = 4 men 40 By rule of fractions: To do the work in 40 days we need less number of men than 10. So, we should multiply 10 with a fraction which is less than 1. or, M2 =

And that fraction is

10 . Therefore, required 40

10 4 40 40 men can cut 60 trees in 8 hrs. If 8 men leave the job how many trees will be cut in 12 hours? 40 men – working 8 hrs – cut 60 trees number of men 16

Ex. 3: Soln:

or, 1 man – working 1 hr – cuts

60 trees 40 8

Thus, 32 men – working 12 hrs – cut

60 32 12 40 8

= 72 trees By our “basic - formula” M1 = 40, D1 = 8 (As days and hrs both denote time) W1 = 60 (cutting of trees is taken as work) M2 = 40 – 8 = 32, D2 = 12, W2 = ? Putting the values in the formula M1 D1 W2 = M2 D2 W1 We have, 40 × 8 × W2 = 32 × 12 × 60

32 12 60 = 72 trees 40 8 By rule of fractions: First, there were 40 men, but when 8 men leave the job we are left with 32 men. As the number or, W2 =

Quicker Maths

314 of men is reduced, less number of trees will be cut by them. So, 60 should be multiplied with less-than-one

32 . Furthermore, as the number of hours 40 increases, more number of trees will be cut. So, the previous product will be multiplied by more-than-one

Note:

5 16 6 6 = 3 days. 12 10 8

fraction,

fraction,

12 . Therefore, the required number of trees 8

32 12 = 60 72 trees. 40 8 Note: Ex. 4:

Soln:

Theorem: If A can do a piece of work in x days and B can do it in y days then A and B working together will do the

xy same work in x y days. Proof: A’s work in 1 day =

Try to solve this question without writing the initial steps. 5 men can prepare 10 toys in 6 days working 6 hrs a day. Then, in how many days can 12 men prepare 16 toys working 8 hrs a day? This example has an extra variable ‘time’ (hrs a day), so the ‘basic-formula’ can’t work in this case. An extended formula is being given: M1 D1 T1 W2 = M2 D2 T2 W1 Here, 5 × 6 × 6 × 16 = 12 × D2 × 8 × 10

5 6 6 16 D2 12 8 10 = 3 days Note: Number of toys is considered as work in the above example. By rule of fractions: See the steps: 1. We have to find number of days, so write the given number of days first. 2. Number of men increases work will be done in less days multiplying fraction should be 5 . 12 3. Number of toys increases it will take more days multiplying fraction should be more less than 1, which is

16 than 1, which is . 10 4. Number of working hours increases it will take less days multiplying fraction should be less than 1, which is

If you understand the method of fraction, your writing work reduces and you need to write only

6 . 8

1 x

1 B’s work in 1 day = y 1 1 xy (A+B)’s work in 1 day = x y xy xy (A+B) do the whole work in x y days Ex. 5:

A can do a piece of work in 5 days, and B can do it in 6 days. How long will they take if both work together?

Soln:

‘A’ can do

1 work in 1 day.. 5

‘B’ can do

1 work in 1 day.. 6

1 1 Thus, ‘A’ and ‘B’ can do work in 1 day.. 5 6

‘A’ and ‘B’ can do the work in

1 days 1 1 5 6

30 8 2 days 11 11 By the theorem: A+B can do the work in

56 30 8 days 2 days 56 11 11 Theorem: If A, B and C can do a work in x, y and z days respectively then all of them working together can finish

5 16 6 Thus, required number of days = 6 12 10 8

the work in

= 3 days

Proof: Try yourself.

xyz xy yz xz days.

Time and Work

315

Ex. 6: In the above question, if C, who can do the work in 12 days, joins them, how long will they take to complete the work? Soln: By the theorem: ‘A’, ‘B’ and ‘C’ can do the work in 5 6 12 360 2 2 days 5 6 6 12 5 12 162 9 Note: Do you find it easier to remember the direct formula in examples 5 and 6? Try to solve some more examples by this method. Ex. 7: Mohan can do a piece of work in 10 days and Ramesh can do the same work in 15 days. How long will they take if both work together?

10 15 150 6 days 10 15 25 Ex. 8: In the above question if Suresh, who can finish the same work in 30 days, joins them, how long will they take to complete the work? Sol: Ans Sol:

A + C can do in 20 days. By the theorem: We see that 2 (A + B + C) can do the work

12 15 20 5 days 12 15 12 20 15 20 A + B + C can do the work in 5 × 2 = 10 days (Less men more days) in

Now, A can do the work in

in Ex.7) [As A =( A + B + C) – (B + C)]

10 20 20 days 20 10 [As, B = (A + B + C) – (A + C)] B can do the work in

Ans =

10 15 30 10 15 30 5 days 10 15 10 30 15 30 900 Theorem: If A and B together can do a piece of work in x days and A alone can do it in y days, then B alone can do

xy the work in x y days. Proof: Try yourself. Ex. 9: A and B together can do a piece of work in 6 days and A alone can do it in 9 days. In how many days can B alone do it? 1 Soln: A and B can do th of the work in 1 day.. 6 A alone can do

1 th of the work in 1 day.. 9

1 1 1 B alone can do th of the work in 6 9 18 1 day. B alone can do the whole work in 18 days. By the theorem: B alone can do the whole work in 6 9 54 18 days 96 3 More uses of the above formula Ex. 10: A and B can do a piece of work in 12 days, B and C in 15 days, C and A in 20 days. How long would each take separately to do the same work? Soln: A + B can do in 12 days. B + C can do in 15 days.

10 15 30 days (As 15 10

10 12 60days. 12 10 [As, C = (A + B + C) – (A + B)] C can do the work in

Working alternately Ex. 11: Two women, Ganga and Saraswati, working separately can mow a field in 8 and 12 hrs respectively. If they work in stretches of one hour alternately, Ganga beginning at 9 a.m., when will the mowing be finished? Soln:

1 th of the field. 8 1 In the second hour, Saraswati mows th of the 12 field. In the first hour, Ganga mows

5 1 1 in the first 2 hrs of the field is 8 12 24 mown.

5 5 in 8 hrs 24 4 6 of the field is mown. ------- (*) 5 1 Now, 1 th of the field remains to be 6 6 mown. In the 9th hour Ganga mows

1 th of the 8

field. Saraswati will finish the mowing of

1 1 1 1 1 1 of the field in or of 6 8 24 24 12 2 an hour.

Quicker Maths

316 the total time required is 1 1 8 1 or 9 hrs. 2 2 Thus, the work will be finished at

1 1 1 9 9 18 or 6 p.m. 2 2 2 Note (*): We calculated the work for 4 pairs of hours only because if we calculate for 5 pairs of hours, the work done is more than 1. And it leads to absurd result. Ex. 12: A and B together can do a piece of work in 12 days which B and C together can do in 16 days. After A has been working at it for 5 days, and B for 7 days, C takes up and finishes it alone in 13 days. In how many days could each do the work by himself? Soln:

A and B in 1 day do

1 th work. 12

1 th work. 16 Now, from the question, A’s 5 days’ + B’s 7 days’ + C’s 13 days’ work = 1 or, A’s 5 days’ + B’s 5 days’ + B’s 2 days’ + C’s 2 days’ + C’s 11 days’ work = 1 (A +B)’s 5 days’ + (B+C)’s 2 days’ + C’s 11 days’ work = 1 B and C in 1 day do

5 2 C’s 11 days’ work = 1 12 16

5 2 11 C’s 11 day’s work = 1 12 16 24

Add B’s daily work to both sides. 4 times B’s daily work = (A + B + C)’s daily 1 work = 10 1 B’s daily work = 40 Also, 2 times C’s daily work = (A + B)’s daily work. Add C’s daily work to both sides. 3 times C’s daily work = (A + B + C)’s daily work =

1 10

C’s daily work =

1 30

Now, A’s daily work

1 1 1 1 10 40 30 24

A, B and C can do the work in 24, 40 and 30 days respectively. Quicker Method: Number of days taken by B = (Number of days taken by A+B+C) × (3+1) = 10 (3+1) = 40 days Similarly, Number of days taken by C = 10 (2+1) = 30 days Number of days taken by A 1 24 days 1 1 1 10 40 30 Ex. 14: If 3 men or 4 women can reap a field in 43 days, how long will 7 men and 5 women take to reap it? Soln: First Method 1 rd of the field in 1 day.. 43

11 1 C’s 1 day’s work = 24 11 24

3 men reap

1 1 1 B’s 1 day’s work = 16 24 48

1 man reaps

1 rd of the field in 1 day.. 43 3

4 women reap

1 rd of the field in 1 day.. 43

1 1 1 A’s 1 day’s work = 12 48 16 A, B and C can do the work in 16, 48 and 24 days respectively. Ex. 13: To do a certain work B would take three times as long as A and C together and C twice as long as A and B together. The three men together complete the work in 10 days. How long would each take separately? Soln: By the question 3 times B’s daily work = (A + C)’s daily work.

1 woman reaps

7

men

1 th of the field in 1 day.. 43 4 and 5 women reap

5 1 7 th of the field in 1 day.. 43 3 43 4 12

7 men and 5 women will reap the whole field in 12 days.

Time and Work

317

Second Method 3 men = 4 women

1 man =

5 (12 men + 16 boys) can do the work in

4 women 3

= 1 day 4 (13 men + 24 boys) can do the work in

28 7 men = women 3 28 43 5 women 3 3 Now, the question becomes: If 4 women can reap a field in 43 days, how long

7 men + 5 women

43 women take to reap it? 3 The “basic-formula” gives will

4 43

43 D2 3

4 43 3 = 12 days 43 Quicker Method: or, D2 =

Required number of days =

1 5 7 43 3 43 4

43 3 4 = 12 days 7 45 3 The above formula is very easy to remember. If we divide the question in two parts and call the first part as OR-part and the second part as AND

Note:

part then

5 5

7 43 3

Number of men in AND part Number of days Number of men in OR part

Similarly, you can look for the second part in denominator. Ex. 15: If 12 men and 16 boys can do a piece of work in 5 days and 13 men and 24 boys can do it in 4 days, how long will 7 men and 10 boys take to do it? Soln: 12 men and 16 boys can do the work in 5 days. ------- (1) 13 men and 24 boys can do the work in 4 days. ------- (2) Now, it is easy to see that if the no. of workers be multipled by any number, the time must be divided by the same number (derived from: more workers less time). Hence, multiplying the no. of workers in (1) and (2) by 5 and 4 respectively, we get

4 4

= 1 day or, 5(12 m + 16 b) = 4(13 m + 24 b) or, 60 m + 80 b = 52 m + 96 b ---------- (*) or, 60 m – 52 m = 96 b – 80 b or, 8 m = 16 b 1 man = 2 boys Thus, 12 men + 16 boys = 24 boys + 16 boys = 40 boys and 7 men + 10 boys = 14 boys + 10 boys = 24 boys The question now becomes: “If 40 boys can do a piece of work in 5 days, how long will 24 boys take to do it?” Now, by “basic formula”, we have 40 × 5 = 24 × D2 ----------- (*) (*)

40 5 1 8 days 24 3 Note: During practice session (*) should be your first step to be written down. Further calculations should be done mentally. Once you get that 1 man = 2 boys, your next step should be (*) (*). This way you can get the result within seconds. Ex. 16: A certain number of men can do a work in 60 days. If there were 8 men more it could be finished in 10 days less. How many men are there? Soln: Let there be x men originally. (x + 8) men can finish the work in (60 – 10) = 50 days Now, 8 men can do in 50 days what x men can do in 10 days, then by “basic formula” we have 8 × 50 = x × 10 or, D2 =

8 50 40 men 10 Another Approach: We have: x men do the work in 60 days and (x + 8) men do the same work in (60 – 10=) 50 days. Then, by “basic formula”, 60x = 50 (x + 8) x

50 8 40 men. 10 Quicker Method (Direct Formula): There exists a relationship:

x=

Quicker Maths

318 Original number of workers No. of more workers

Number of days taken by the second group No. of less days

8 (60 – 10) 8 50 40 man 10 10 Ex. 17: A is thrice as fast as B, and is therefore able to finish a work in 60 days less than B. Find the time in which they can do it working together. Soln: A is thrice as fast as B, means that if A does a work in 1 day then B does it in 3 days. Hence, if the difference be 2 days, then A does the work in 1 day and B in 3 days. But the difference is 60 days. Therefore, A does the work in 30 days and B in 90 days. Now, A and B together will do the work in

30 90 45 days 22.5 days 30 90 2 Ex. 18: I can finish a work in 15 days at 8 hrs a day. You can 2 days at 9 hrs a day. Find in how 3 many days can we finish it working together 10 hrs a day. Soln: First, suppose each of us works for only one hr a day. Then, I can finish the work in 15 × 8 = 120 20 9 = 60 days and you can finish the work in 3 days Now, together we can finish the work in 120 60 40 days 120 60 But, here, we are given that we do the work 10 hrs a day. Then, clearly we can finish the work in 4 days. Ex. 19: A can do a work in 6 days. B takes 8 days to complete it. C takes as long as A and B would take working together. How long will it take B and C to complete the work together? 6 8 24 days. Soln: (A+B) can do the work in 68 7 24 C takes days to complete the work. 7 24 8 24 8 2 7 2 days. (B+C) takes 24 5 8 24 56 7 finish it in 6

Ex. 20: A is twice as good a workman as B. Together, they finish the work in 14 days. In how many days can it be done by each separately? Soln: Let B finish the work in 2x days. Since A is twice as active as B therefore, A finishes the work in x days. (A+B) finish the work in

2x 2 14 3x

or x = 21 A finishes the work in 21 days and B finishes the work in 21 × 2 = 42 days. Quicker Approach: Twice + One time = Thrice active person does the work in 14 days. Then, one-time active person (B) will do it in 14 × 3 = 42 days and twice

42 = 21 days. 2 Efficient person takes less time. In other words, we may say that “Efficiency (E) is indirectly proportional to number of days (D) taken to complete a work”. Then, mathematically, active person (A) will do it in

Note:

1 K or , E , where K is a constant. D D or, ED = Constant or, E1D1 = E2D2 = E3D3 = E4D4 = .....EnDn And, we see in the above case: E1D1 = E2D2 = E3D3 or, 3 ×14 = 2 × 21 = 1 × 42 Thus, our answer verifies the above statement. Ex. 21: 5 men and 2 boys working together can do four times as much work per hour as a man and a boy together. Compare the work of a man with that of a boy. Soln: The first group is 4 times as much efficient as the second group. What does it mean? It simply means that the second group will take 4 times as many days as the first group (See the Note given under Ex. 20). Therefore, (5m + 2b)’s 1 day’s work = (1m + 1b)’s 4 days’ work or, (5m + 2b)’s 1 day’s work = (4m + 4b)’s 1 day’s work or, 5m + 2b = 4m + 4b or, m = 2b m 2 b 1 That is, a man is twice as efficient as a boy. Ex. 22: 12 men or 15 women can reap a field in 14 days. Find the number of days that 7 men and 5 women will take to reap it. Eα

Time and Work

319

Soln:

This example is the same as Ex. 14. Three methods have been discussed for Ex. 14. If you remember the direct formula, you get the required number of days 1 1 7 5 1 1 14 12 14 15 24 42 24 42 168 3 15 days 24 42 11 11 Ex. 23: 10 men can finish a piece of work in 10 days, whereas it takes 12 women to finish it in 10 days. If 15 men and 6 women undertake to complete the work, how many days will they take to complete it? Soln: 10 men do the work in 10 days. 10 20 15 men do the work in days (by rule 15 3 of fraction) 12 Similarly, 6 women do the work in 10 6 = 20 days (by rule of fraction) 15 men + 6 women do the work in

20 20 20 20 3 5 days 20 80 20 3

Quicker Approach: We see that the above question is: “10 men or 12 women do a work in 10 days. In how many days can 15 men and 6 women complete the work?” Thus, this question is the same as Ex 14 or Ex 22.

required number of days =

1 15 6 10 10 12 10

1 20 5 days. 3 1 4 20 20 Ex 24: A and B can do a work in 45 and 40 days respectively. They began the work together, but A left after some time and B finished the remaining work in 23 days. After how many days did A leave? Soln: B works alone for 23 days. 23 Work done by B in 23 days = work 40

23 17 A + B do together 1 work 40 40

Now, A + B do 1 work in

40 45 40 45 days 40 45 85

17 40 45 17 9days work in 40 85 40 Quicker Maths (Direct formula): If we ignore the intermediate steps, we can write a direct formula

A + B do

as:

40 45 40 23 9 days. 40 45 40

Ex. 25: A certain number of men complete a work in 160 days. If there were 18 men more, the work could be finished in 20 days less. How many men were originally there? Soln: This question is the same as in Ex 16. See the Quicker Method (direct formula) and apply here.

18 (160 20) 126 20 Ex. 26: 4 men and 6 women finish a job in 8 days, while 3 men and 7 women finish in 10 days. In how many days will 10 women finish it? Soln: Method I. Considering one day’s work: 1 4m + 6w = --------- (1) 8 1 3m + 7w = --------- (2) 10 (1) × 3 – (2) × 4 gives Original number of men =

3 4 1 – or, 10w = 8 10 40 10 women can do the work in 40 days. Method II: See the theory used in Ex 15. We find that 8(4m + 6w) = 10 (3m + 7w) or, 2m = 22w 4m = 44w 4men + 6women = 50 women do in 8 days 8 50 = 40 days 10 women do in 10 Ex. 27: 1 man or 2 women or 3 boys can do a work in 44 days. Then, in how many days will 1 man, 1 woman and 1 boy do the work? Soln: This is an extended form of Ex 14. Thus, by our extended formula, number of required days 18w – 28w =

1 44 1 2 3 24 days 1 1 1 63 2 44 1 44 2 44 3

Quicker Maths

320

Note:

1 Number of men in AND part 44 1 No. of days No. of men in OR part 1 Number of women in AND- part = 44 × 2 No. of days × No. of women in OR- part

1 . 44 3 Ex. 28: 3 men and 4 boys do a work in 8 days, while 4 men and 4 boys do the same work in 6 days. In how many days will 2 men and 4 boys finish the work? Soln: This question is the same as Ex 15. Try yourself. Ex. 29: A is thrice as good a workman as B. Together they can do a job in 15 days. In how many days will B finish the work? Soln: This question is the same as Ex 20. Thrice + One time = 4 times efficient person does in 15 days One-time efficient (B) will do in 15 × 4 = 60 days Ex. 30: A group of men decided to do a work in 10 days, but five of them became absent. If the rest of the group did the work in 12 days, find the original number of men. Soln: Suppose there were x men originally. Then, by “basic formula”, M1D1 = M2D2, we have 10x = 12 (x – 5) Similarly, you can define

(12 5) 30 men. 2 10 Ex. 31: A builder decided to build a farmhouse in 40 days. He employed 100 men in the beginning and 100 more after 35 days and completed the construction in stipulated time. If he had not employed the additional men, how many days behind schedule would it have been finished? Soln: Let 100 men only complete the work in x days. Work done by 100 men in 35 days + work done by 200 men in (40 – 35=) 5 days = 1. x

or, or,

35 200 5 1 x 100x 45 1 x

x 45 days. Therefore, if additional men were not employed, the work would have lasted 45 – 40 = 5 days behind schedule time.

Quicker Approach: 200 men do the rest of the work in 40 – 35 = 5 days 100 men can do the rest of the work in 5 200 10 days 100 required number of days = 10 – 5 = 5 days Ex. 32: A team of 30 men is supposed to do a work in 38 days. After 25 days, 5 more men were employed and the work was finished one day earlier. How many days would it have been delayed if 5 more men were not employed? Soln: This question is similar to Ex 31. To solve it, our quicker approach should be the same. 35 men do the rest of the job in 12 days. [12 = 38 – 25 – 1] 30 men can do the rest of the job in

12 35 14 days 30 Thus, the work would have been finished in 25 + 14 = 39 days, that is, (39 – 38 = ) 1 day after the scheduled time. Ex. 33: A can do a work in 25 days and B can do the same work in 20 days. They work together for 5 days and then A goes away. In how many days will B finish the work? Soln: A + B can do the work in 5 days 1 5 45 9 1 = 5 25 20 25 20 20 Rest of the work = 1 –

9 11 20 20

11 11 days. 20 Ex. 34: A and B working separately can do a work in 9 and 12 days respectively. A starts the work and they work on alternate days. In how many days will the work be completed? Soln: This question is the same as in Ex 11. B will do the rest of the work in 20

(A + B)’s 2 day’s work =

1 1 7 9 12 36

7 35 (just less than 1), ie, (A 36 36 + B) work for 5 pairs of days, ie, for 10 days. We see that 5

35 1 Now, rest of the work 1 is to be 36 36 done by A.

Time and Work 1 1 1 day.. A can do work in 9 36 36 4 1 1 Total days = 10 4 10 4 days. Ex. 35: 8 children and 12 men complete a work in 9 days. Each child takes twice the time taken by a man. In how many days will 12 men finish the same work? Soln: 2 children = 1 man 8 children + 12 men = 4 + 12 = 16 men 16 12 men finish the work in 9 = 12 days 12 Note:

---------- (*) (*) To find the result either use M1D1 = M2D2 (basic formula) or the “rule of fraction”. We suggest you to use “rule of fraction”. Since less men will do the work in more days. Therefore, 9 should be

16 (a more-than-one fraction). 12 Ex. 36: 30 men working 7 hrs a day can do a work in 18 days. In how many days will 21 men working 8 hrs a day do the same work? Soln: Using the formula: M1D1T1W2 = M2D2T2W1 Since work is the same for the two cases, M1D1T1 = M2D2T2 --------- (*)

321 Soln:

400 Thus, the rest of the food will last for 3 120 days for the 120 men left.

400 Ans 3 120 10 days Note:

Note(*): Man-day-hour is constant for a work. Ex 37: A, B and C can do a work in 8, 16 and 24 days respectively. They all begin together. A continues to work till it is finished, C leaving off 2 days and B one day before its completion. In what time is the work finished? Soln: Let the work be finished in x days. Then, A’s x day’s work + B’s (x – 1) day’s work + C’s (x – 2) day’s work = 1 x x 1 x 2 1 or, 8 16 24

6x 3x 3 2x 4 1 48 or, 11x = 55 x = 5 days Ex. 38: There is sufficient food for 400 men for 31 days. After 28 days, 280 men leave the place. For how many days will the rest of the food last for the rest of the men? or,

For less persons the food will last longer, therefore

140 , a more-than-one fraction. 120 Ex. 39: A takes as much time as B and C together take to finish a job. A and B working together finish the job in 10 days. C alone can do the same job in 15 days. In how many days can B alone do the same work? Soln: Quicker Method: 3 is multiplied by

multiplied by

M1 D1 T1 30 18 7 1 22 days D2 M T 21 8 2 2 2

The rest of the food will last for (31 – 28=) 3 days if nobody leaves the place.

(A + B) + (C) can do in

15 10 = 6 days 15 10

Since, A’s days = (B+C)’s days B + C can do in 6 × 2 = 12 days

B [B = {B + C} – C] can do in

15 12 15 12

= 60 days

4 days 5 and 32 days respectively. They started the work together but after 4 days A left. B left the work 3 days before the completion of the work. In how many days was the work completed? Suppose the work is completed in x days. A’s 4 days’ work + B’s (x – 3) days’ work + C’s x days’ work = 1

Ex. 40: A, B and C can do a work in 16 days, 12

Soln:

or,

4 (x – 3) 5 x 1 16 64 32

16 5x 15 2x 1 64 or, 7x + 1= 64 x = 9 days Ex. 41: Raju can do a piece of work in 16 days. Ramu can or,

4 do the same work in 12 days while Gita can do 5 in 32 days. All of them started to work together but Raju leaves after 4 days. Ramu leaves the job

Quicker Maths

322

Soln:

3 days before the completion of the work. How long would the work last? Suppose the work lasted for x days. Then, Raju’s 4 days’ work + Ramesh’s (x – 3) days’ work + Gita’s x days’ work = 1 or,

or,

4 x3 x 1 16 12 4 32 5

1 5(x 3) x 1 4 64 32

5(x 3) 2x 3 64 4 or, 7x – 15 = 48 or,

48 15 63 = 9 days 7 7 Ex. 42: A and B undertake to do a work for 56. A can do it alone in 7 days and B in 8 days. If with the assistance of a boy they finish the work in 3 days then the boy gets ______. Soln: A’s 3 days’ work + B’s 3 days’ work + Boy’s 3 days’ work = 1

x

3 3 or, Boy 's 3 day's work 1 7 8 3 3 11 or, Boy’s 3 days’ work = 1 7 8 56 Ratio of shares 3 3 11 3 56 3 56 11 56 : : : : 7 8 56 7 8 56 = 24 : 21 : 11 =

56 Boy’s share = 24 21 11 11 = 111 Ex. 43: A started a work and left after working for 2 days. Then B was called and he finished the work in 9 days. Had A left the work after working for 3 days, B would have finished the remaining work in 6 days. In how many days can each of them, working alone, finish the whole work? Soln: Detailed Method: Suppose A and B do the work in x and y days respectively. Now, work done by A in 2 days + work done by B in 9 days = 1 2 9 or, x y 1 3 6 Similarly, x y 1

To solve the above equation put

1 1 = a and y = x

b. Thus 2a + 9b = 1 --------(1) and 3a + 6b = 1 --------(2) Performing (2) × 3 – (1) × 3 we have 5a = 1 1 a= 5 or, x =

1 = 5 days a

1 = 15 days b Quicker Method: Direct formula: In such case: A will finish the work in and y =

3 9 2 6 15 5 days. 96 3 For B, we should use the above result. B does 1

2 3 work in 9 days. 5 5

5 = 15 days. 3 Ex. 44: If 5 men and 3 boys can reap 23 acres in 4 days, and 3 men and 2 boys can reap 7 acres in 2 days, how many boys must assist 7 men in order that they may reap 45 acres in 6 days? Soln: Firstly, we should find the relationship between man and boys. We may find two equations from the two given statements. The first statement implies that

B does 1 work in 9

5 men and 3 boys can reap

23 acres in 1 day.. 4

We write this as:

23 ---------- (1) 4 Similarly, from the second statement 5m + 3b =

7 ----------- (2) 2 Now, to find the relationship, 3m + 2b =

5m 3b 23 7 23 3m 2b 4 2 14 or, 70m + 42b = 69m + 46b m = 4b (or, one man is equal to 4 boys) 5m + 3b = 5 × 4 + 3 = 23 boys (for the first statement)

Time and Work

323

Now, use M1D1 W2 M 2 D 2 W1

M2

M1 D1 W2 23 4 45 30 boys. D 2 W1 6 23

30 – 7 × 4 = 2 boys should assist them. Ex. 45: A contractor undertakes to dig a canal 12 km long in 350 days and employs 45 men. After 200 days he finds that only 4.5 km of the canal has been completed. Find the number of extra men he must employ to finish the work in time. Soln: To apply the rule of fraction or our direct formula, we rewrite the above question as: 45 men prepare 4.5 km canal in 200 days. Then, how many more persons are needed to prepare 12 – 4.5 = 7.5 km in 350 – 200 = 150 days? 200 7.5 = 100 men By rule of fraction: 45 150 4.5

required no. of persons to be added = 100 – 45 = 55 men By Direct Formula: M1 D1 W2 M 2 D 2 W1 or, 45 × 200 × 7.5 = M 2 150 4.5

45 200 7.5 100 150 4.5 required number of persons to be added = 100 – 45 = 55 men Ex. 46: 8 men and 16 women can do a work in 8 days. 40 men and 48 women can do the same work in 2 days. How many days are required for 6 men and 12 women to do the same work? Soln: Method I: The man-power (ie, no. of persons doing the job) is indirectly proportional to number of days (i.e., more man-power, less days or, less manpower, more days). So, we can’t write the equation like; 8 m + 16 w = 8 or 40 m + 48 w = 2 Now, we have to find the two things which are directly proportional to each other. Clearly these two things in this respect are man-power and work. So, we change the relationship and find the work done by each group in one day. Then, we have the equations M2

8 m + 16 w =

1 8

and 40 m + 48 w =

---------- (i)

1 2

----------- (ii)

and we have to find: 6m + 12 w = ? Now, (2) – 3 × (1), gives

16m

1 3 1 2 8 8

6 16 8 Again, 5 × (1) – (2), gives

6m

5 1 1 32w 8 2 8 12w

12 32 8

Now, 6 12 3 3 6 3 16 8 32 8 64 64 64 32 Therefore, 6 men and 12 women will do the job in 6m 12w

32 2 10 days. 3 3 Method II: We will compare the capacity of a man and woman. To do so, we apply: 8 (8m + 16 w) = 2 (40 m + 48 w) or, 64m + 128 w = 80 m + 96 w or, 16 m = 32 w 1m=2w Now, 8 m + 16 w = 16 w + 16 w = 32 w (from first information) 6 m + 12 w = 12 w + 12 w = 24 w (from required information) Now, apply the formula: M1 D1 M 2 D2 Then; 32 × 8 = 24 × D2 32 8 32 2 D2 10 days 24 3 3 Note: This method works very fast, so we suggest you to follow only this method. One more method for special cases (which is applicable in this case) is being discussed below: Method III: (Very Quicker, but for special cases only): First, you should know the type of question where this method can be applied. See the number of men and women in the question part. Find the ratio of these two numbers, like in this case: men : women = 6 : 12 = 1 : 2. Now, look at the question-parts for the same ratio. In this case, the first question-part has the same ratio, i.e., 8 : 16 = 1 : 2. Now, we can use this method. If there is no such ratio in question part, we can’t use this method.

Quicker Maths

324

1 man does 1 work working 1 hr/day in (12 × 19 × 4.5 × 6) days 57 men do 2 work working 8 hrs/day in

8 m + 16 w do the work in 8 days or, 8 (m + 2 w) '' '' 8 days or, (m + 2 w) '' '' 8 × 8 = 64 days

6 (m + 2 w) ''

''

64 32 days 6 3

2 days 3 Ex. 47: 38 men, working 6 hours a day can do a piece of work in 12 days. Find the number of days in which 57 men working 8 hrs a day can do twice the work. Assume that 2 men of the first group do as much work in 1 hour as 3 men of the second group do 1 in 1 hrs. 2 Soln: Detailed Method: 2 × 1 men of first group = 3 × 1.5 men of second group or, 2 men of first group = 4.5 men of second group

6m + 12w ''

'' 10

12 19 4.5 6 2 27 days 57 8 Quicker Method: Ratio of efficiency of persons in first group to the second group = E1 : E2 = (3 × 1.5) : 2 × 1 = 4.5 : 2 -------- (*) Now, use the formula: M1D1T1E1 W2 = M2D2T2E2W1 --------- (*)(*)

38 12 6 4.5 2 27 days 57 8 2 1 (*) Less number of persons from the first group do the same work in less number of days, so they are more efficient. (*)(*) M represents the number of men. D represents the number of days. T represents the number of working hours. E represents the efficiency. W represents the work. and the suffix represents the respective groups. D2

Note:

4.5 38 19 4.5 2 (19 × 4.5) men do 1 work, working 6 hrs/day in 12 days.

38 men of first group =

EXERCISES 1. Mohan can do a job in 20 days and Sohan can do the same job in 30 days. How long would they take to do it working together? 2. Raju, Rinku and Ram can do a work in 6, 12 and 24 days respectively. In what time will they altogether do it? 3. A and B working together can do a piece of work in 6 days. B alone can do it in 8 days. In how many days A alone could finish? 4. A and B can finish a field work in 30 days, B and C in 40 days while C and A in 60 days. How long will they take to finish it together? 5. A can copy 75 pages in 25 hrs. A and B together can copy 135 pages in 27 hrs. In what time can B copy 42 pages? 6. A, B and C can finish a work in 10, 12 and 15 days respectively. If B stops after 2 days, how long would it take A and C to finish the remaining work? 7. B can do a job in 6 hrs, B and C can do it in 4 hrs and A, 2 B and C in 2 hrs. In how many hrs can A and B do it? 3 8. I can finish a work in 15 days at 8 hrs a day. You can

2 days at 9 hrs a day. Find in how many 3 days we can finish it together, if we work 10 hrs a day? 9. A can do a work in 7 days. If A does twice as much work as B in a given time, find how long A and B would take to do the work. 10. A can do a work in 6 days. B takes 12 days. C takes as long as A and B would take working together. How long will it take B and C to complete the work together? 11. A is twice as good a workman as B; and together they finish a work in 16 days. In how many days can it be done by each separately? 12. If 3 men or 5 women can reap a field in 43 days, how long will 5 men and 6 women take to reap it? 13. If 5 men and 2 boys working together can do four times as much work per hour as a man and a boy together, compare the work of a man with that of a boy. 14. One man, 3 women and 4 boys can do a work in 96 hrs; 2 men and 8 boys can do it in 80 hrs; and 2 men finish it in 6

Time and Work and 3 women can do it in 120 hrs. In how many hours can it be done by 5 men and 12 boys? 15. A and B working separately can do a piece of work in 9 and 12 days respectively. If they work for a day alternately, A beginning, in how many days will the work be completed? 16. A sum of money is sufficient to pay A’s wages for 21 days or B’s wages for 28 days. The money is sufficient to pay the wages of both for _____days. 17. A does half as much work as B in three-fourth of the time. If together they take 18 days to complete a work, how much time shall B take to do it? 18. 10 men can finish a piece of work in 10 days, whereas it takes 12 women to finish it in 10 days. If 15 men and 6 women undertake to complete the work, how many days will they take to complete it? 19. A and B can do a piece of work in 30 and 40 days respectively. They began the work together, but A left after some days and B finished the remaining work in 12 days. After how many days did A leave? 20. A certain number of men complete a piece of work in 60 days. If there were 8 men more, the work could be finished in 10 days less. How many men were originally there? 21. 8 children and 12 men complete a certain piece of work in 9 days. If each child takes twice the time taken by a man to finish the work, in how many days will 12 men finish the same work? 22. 2 men and 3 women can finish a piece of work in 10 days, while 4 men can do it in 10 days. In how many days will 3 men and 3 women finish it? 23. 3 men and 4 boys do a piece of work in 8 days, while 4 men and 4 boys finish it in 6 days. 2 men and 4 boys will finish it in ____ days. 24. If 1 man or 2 women or 3 boys can do a piece of work in 44 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in _______ days. 25. A and B together can complete a task in 20 days. B and C together can complete the same task in 30 days. A and C together can complete the same task in 40 days. What is the ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone ? 26. 24 men can complete a piece of work in 14 days. 2 days after they started working, 4 more men joined them and after 2 more days 6 men left. How many more days will they now take to complete the remaining work? 27. A, B and C have to type 506 pages to finish an assignment. A can type a page in 12 minutes, B in 15

325 minutes and C in 24 minutes. If they divide the task into three parts so that all three of them spend equal amount of time in typing, what is the number of pages that B should type? 28. If 36 persons are engaged on a piece of work, the work

3 th 4 of the work was completed. How many more persons are required to complete the work on time ? Three typists P, Q and R have to type 368 pages. P types one page in 8 minutes, Q in 18 minutes and R in 24 minutes. In what time will these pages be typed if they work together? 28 men can complete a piece of work in 15 days and 15 women can complete the same piece of work in 24 days. What is the ratio of the amount of work done by 30 men in 1 day to the amount of work done by 18 women in 1 day? 18 men can complete a project in 30 days and 16 women can complete the same project in 36 days. 15 men start working and after 9 days they are replaced by 18 women. In how many days will 18 women complete the remaining work ? 10 men can finish a piece of work in 15 days. 8 women can finish the same piece of work in 25 days. Only 10 women started working and in a few days completed a certain amount of work. After that 3 men joined them. The remaining work was completed by 10 women and 3 men together in 5 days. After how many days did 3 men join 10 women? A project manager hired 16 men to complete a project in 38 days. However, after 30 days, he realised that only can be completed in 40 days. After 32 days, only

29.

30.

31.

32.

33.

5 9 of the work is complete. How many more men does he need to hire to complete the project on time? 34. A can do a piece of work in 8 days which B can estroy in 3 days. A has worked for 6 days, during the last 2 days of which B has been destroying. How many days must A now work alone to complete the work? 35. C is 40% less efficient than A. A and B together can finish a piece of work in 10 days. B and C together can do it in 15 days. In how many days can A alone finish the same piece of work? 36. A can complete a given task in 24 days. B is twice as efficient as A. A started on the work initially, and was joined by B after a few days. If the whole work was completed in 10 days, after how many days, from the time A started working, did B join A?

Quicker Maths

326 37. To complete a project, 18 women take 4 more days than the number of days taken by 12 men. If eight men complete the project in 9 days, how much work will be left when 15 women and 12 men together work for 3 days?

4 5 times as efficient as A. If A can complete 3 8 of a given task in 15 days, what fraction of the same

38. B is

task would remain incomplete if B works on it independently for 10 days only? 39. A alone can do a piece of work in 24 days. The time taken by A in completing

1 of the work is equal to 3

1 of the work. In 2 what time will A and B together complete the work? the time taken by B in completing

ANSWERS 1. 2.

20 30 12 days 20 30 6 12 24 6 12 24 6 12 12 24 6 24 72 288 144

6 12 24 3 3 days 504 7

68 24 days 86 4. 2 (A + B + C) will do the work in 30 40 60 40 30 40 30 60 40 60 3 80 2 26 days 3 3

75 = 3 pages in 1 hr.. 25 135 A + B can copy = 5 pages in 1 hr.. 27 B can copy 5 – 3 = 2 pages in 1 hr..

5. A can copy

B can copy 42 pages in

46 12 hrs 64 From (1) & (4); C can do it in

3.

A + B + C will do in

8 hrs --------(1) 3 B + C can do the work in 4 hrs --------(2) B can do the work in 6 hrs --------(3) From (2) and (3);

7. A + B + C can do the work in

42 = 21 hrs. 2

1 1 1 6. A + B + C in 2 days, do 2 work. 10 12 15 1 1 2 work. 4 2 Now, B withdraws. A + B will do the whole work in

10 12 60 days 12 10 11 1 30 8 2 days A + B will do work in 2 11 11

--------(4)

8 12 32 3 24 3 3 3 hrs A + B can do it in 8 28 7 7 12 3 8. Change the time into hours. I finish in 15 × 8 = 120 hrs

20 9 = 60 hrs 3 both of us working together finish the work in

You finish in

120 60 = 40 hrs 120 60 40 = 4 days 10 Neglecting the intermediate steps, the direct formula can be written as:

number of days =

20 (15 8) 9 3 1 1 120 60 4 days 10 20 10 180 (15 8) 9 3 Note: See Ex 18 (solved). The method is different there. You are suggested to adopt the earlier method. 9. A is twice as efficient as B, hence B will do the work in 14 days.

Time and Work A + B will do in

327 7 14 14 2 4 days. 7 14 3 3

5m do the work in 240

6 12 = 4 days 6 12 C takes 4 days

From (2) and (6); we have,

10. A + B take

8b do the work in

12 4 3 days 12 4 11. Suppose B does in 2x days. A does in x days. Now, working together, they can do in

12b do the work in

2x 2 = 16 days 3x or, x = 24 days A does in 24 days and B does in 48 days. Note: For quicker approach see solved Ex 20. 12. 3 men = 5 women 5 25 women 5 men = 5 3 3 25 43 women 5 men + 6 women = 6 3 3 Now, we are given that 5 women do in 43 days.

43 3 women do in (43) 5 15 days. 43 3 Quicker Method: (See Ex 14)

1 6 5 3 43 5 43

15 days

13. 5m + 2b = 4 (1m + 1b) or, m = 2b m 2 b 1 Therefore, a man does twice as much work as a boy does. 14. 1m + 3w + 4b in 96 hrs ---------(1) 2m + 8b in 80 hrs ---------(2) or, 1m + 4b in 160 hrs ---------(3) 2m + 3w in 120 hrs ---------(4) From (1) and (3); we have,

160 96 = 240 hrs 160 96 From (4) and (5); we have, 3w do the work in

80 240 = 120 hrs 240 80

120 8 = 80 hrs ---------(8) 12 Now, from (7) and (8) we have,

B + C take

Required no. of days

2 = 96 hrs---------(7) 5

---------(5)

240 120 2m do the work in = 240 hrs---------(6) 240 120

5m + 12 b do the work in 15. (A + B)’s work in 2 days

96 80 480 7 43 hrs 96 80 11 11 1 1 43 7 9 12 36 36

In 5 pairs of days they will complete That is, after 5 × 2 = 10 days, 1

7 5 35 36 36

35 1 work is left 36 36

which will be done by A alone. A does 1 work in 9 days.

A does

1 1 1 days work in 9 36 36 4

1 1 10 days. 4 4 16. Let the sum be equal to LCM of 21 and 28, ie. 84.

Total number of days = 10 +

84 84 = 4/day and B gets = 3/day 21 28 A + B get 4 + 3 = 7/day Then, A gets

84 is sufficient for

84 = 12 days to pay both of 7

them. Quicker Method (Direct formula): Number of days =

Multiplication of no. of days Addition of no.of days

21 28 12 days 21 28 17. Suppose B does the work in x days.

Then, A does

1 3x work in days. 2 4

A does 1 work in

3x days. 2

Quicker Maths

328 3x 2 18 (given) A + B do the work in 3x x 2 x

3 2 x 2 18; or, 5 x 2

x

10 20 = 20 days 20 10

40 days 3 3 men + 3 women do in

and 3 men do in

40 3 20 40 8 days 40 100 20 3 20

18 5 30 days 3

18. 15 men do the work in

10 10 20 days 15 3

12 10 20 days 6 women do the work in 6 15 men + 6 women do in 20 20 20 20 3 5 days 20 80 20 3

23. 3 men + 4 boys do in 8 days -------(1) 4 men + 4 boys do in 6 days -------(2) Subtracting (1) from (2); we have, 1 man does in

86 = 24 days 86

-------(3)

3 men do in

24 = 8 days 3

-------(4)

From (1) and (4); we conclude that boys do no work. 2 men + 4 boys = 2 men will finish the work in

24 = 12 days. 2

19. See Ex 24. By direct formula, reqd no. of days

3 women do in

24. 1 man = 2 women = 3 boys

30 40 40 12 12 days 30 40 40

20. Let there be x men originally, then 1 man will do the work in 60x days. In the second case, 1 man does the work in (x + 8) 50 days. Now, 60x = 50 (x + 8)

400 x = = 40 men 10 Quicker Maths (Direct formula):

No. of more men (60 10) Number of men = 10 8 50 40 men 10 21. If each child takes twice the time taken by a man, 8 children = 4 men. 8 children + 12 men = 16 men do the work in 9 days.

12 men finish the work in 22. 4 men do in 10 days 2 men do in 20 days

9 16 = 12 days 12

1 man + 1 woman + 1 boy = 3 boys + 1 boy =

3 boys + 2

11 boys 2

Now, 3 boys do the work in 44 days.

11 44 3 boys do the work in 2 = 24 days 2 11

25. A and B can finish the work in 20 days. A and B’s one day’s work =

1 20

B and C and finish the work in 30 days. B and C’s one day’s work =

1 30

A and C can finish the work in 40 days A and C’s one day’s work =

1 40

Adding we get 2(A + B + C)’s one day’s work =

1 1 1 643 13 20 30 40 120 120

Time and Work

329

(A + B + C)’s one day’s work = A’s one day’s work =

13 13 120 2 240

13 1 13 – 8 5 1 – 240 30 240 240 48

A alone can finish the work in 48 days. C’s one day work =

13 1 13 – 12 1 – 240 20 240 240

C alone can finish the work in 240 days. Reqd ratio =

48 1: 5 240

36 × 40 = M2 × 32 36 40 45 32 Additional men = 45 – 36 = 9

M2 =

1 1 1 943 29. Page printed in one minute 8 18 24 72

Quicker Approch : Suppose total work is 240 units. Then

240 =12 units /day 12

... (1)

B + C do

240 = 8 units /day 30

... (2)

240 6 units/day ... (3) 40 C) do 26 units/day A + B + C do 13 untits /day ...(4) (4) — (1) C does 1 units/day (4) — (2) does 5 units/day ratio of days taken by A to that by C = A+ C do

240 : 240 =1:5 5 1 26. M D = M1 D1 + M2 D2 + M3 D3 24 × 14 = 24 × 2 + 28 × 2 + 22 × D3 336 = 48 + 56 + 22 × D3 22 × D3 = 336 – 104 = 232

9 = 1656 minutes = 27.6 hours 2 30. 28 men complete the work in 15 days. 1 man completes the work in 15 × 28 days.

232 116 6 10 days 22 11 11 27. All three spend equal amount of time on typing. Ratio of work done by them in 1 minute

15 28 = 14 days 30 Again, 15 women can complete the work in 24 days. 1 woman can complete the work in 24 × 15 days.

30 men complete the work in

18 women can complete the work in

1 : 1 : 1 = 10 : 8 : 5 12 15 24 In equal time they work in the ratio of 10 : 8 : 5 =A: B: C =

3 1 4 4 Remaining time = 8 days

M1D1 M 2 D2 W1 W2

8 506 = 176 23

24 15 18

= 20 days 1 1 : = 20 : 14 = 10 : 7 14 20 31. Let the work done by 15 men in 9 days be W2. M1D1 M 2 D2 W W 1 2

Reqd ratio =

D3

28. Remaining work = 1 –

16 2 72 9

Time taken = 368 ×

A + B do

So, the number of pages typed by B =

36 40 M 2 8 1 1 4

18 30 15 9 1 W2 18 × 30 × W2 = 15 × 9

W2 =

15 9 1 18 30 4

1 3 4 4 Again, 16 women complete the project in 36 days.

Remaining work = 1

M1D1 M 2 D2 W1 W2

16 36 18 D2 3 1 4

18 × D2 =

3 × 16 × 36 = 27 × 16 4

Quicker Maths

330 27 16 = 24 days 18 Quicker Approach : Total work = 18 30 man days or 16 36 woman days

D2 =

15 9 1 18 30 4 18x x Work done by 18 women in x days = 16 36 32 1 x 1 Now , 4 32 3 x 32 24 days 4 32. Total work =10 × 15 man-days or 8 × 25 woman-days =150 man-days or 200 woman-days Suppose 3 men joined after x days. 10 women worked for (5 + x) days and 3 men worked for 5 days Work done by 15 men in 9 days =

So, total work done =

And (B + C)’s one day’s work =

60 = 6 units 10 60 = 4 units 15

According to the question, C : A = 60 : 100 = 3 : 5 or,

C 3 A 5

5C 3 Again, A + B = 6 units ... (i) B + C = 4 units ... (ii) Putting the value of A in equation (i), we get

or, A =

C = 3 units

5C 5 3 = 5 units 3 3 Now, total work is 60 units Then, A =

9 20 18 5 x 10 x = 13 days 33. Men Days Work

?

(A + B)’s one day’s work =

10(5 x ) 3 5 1 200 150

5 x 1 1 20 10

16

30

5 9

8

4 9

Using M1D1W2 = M2D2W1 =

60 Then, A alone can do the work in 12 days 5

16 30 4 9 = 48 589

men No. of more men required to the complete the work on time = 48 – 16 = 32 men 34. A can do the work in 8 days. B can destroy it in 3 days. Suppose total work = 24 units (ie LCM of 8 and 3) So, A can do = 3 units/per day B can destroy = 8 units/per day Now, A has done the work in 6 days = 6 × 3 = 18 units But B has been destroying the work in 2 days = 2 × 8 = 16 units In 8 days the work completed = 18 – 16 = 2 units Remaining work = (24 – 2) = 22 units 22 1 22 units of work is done by A in 7 days 3 3 35. Suppose total work = 60 units (LCM of 10 and 15)

Quicker Approach : Ratio of efficiency of A : C = 100 : 60 = 5 : 3 Suppose total work = 60 units

...(i)

60 = 60 units/day 10

...(ii)

A + B do

60 = 4 units/day ...(iii) 15 From (i), (ii) and (iii), we may calculate that A’s efficiency is 5, so he does 5 units/day; C's efficiency is 3 so he does 3 units/day. So, B does 6 – 5 =1 unit/day (from (ii)) It also satisfies eqn. (iii). So, we conclude that efficiency ratio A : B : C =5:1:3 and work per day ratio A : B : C = 5 : 1 : 3 and B + C do

60 A can do 60 units in 5 12 days 36. A can complete the work in 24 days Efficiency of B is twice that of A. B can complete the work in 24 ×

1 = 12 days 2

Time and Work According to the question, the work is completed in 10 days. Let the total work be 24 units (LCM of 24 and 12). 24 A can do = 1 unit/day 24 24 And B can do = 2 units/day 12 Total work done by A in 10 days = 10 × 1 = 10 units Remaining work = 24 – 10 = 14 units 14 Now, 14 units of work is done by B in 2 7 days Hence B joined the work after (10 – 7=) 3 days. 37. 8 men complete the project in 9 days

89 = 6 days 12 18 women can complete the project= (6 + 4 =) 10 days Now, (12 × 6)M = (18 × 10)W or, 4M = 10W 12 men complete the project in

10 w = 2.5 women 4 12M = 2.5 × 12 = 30 women So, (15 + 30) women work together for 3 days. 18 women = 180 × 10 = 180 units 45 women work for 3 days = 45 × 3 = 135 units Remaining work = 180 – 135 = 45 units M=

45 1 Work left = part 180 4 Quicker Approach : ( in terms of mandays or womandays ) Total work = 8 mandays 72 = 6 days 12 women complete in 6 + 4 = 10 days 12 men complete in

331 Total work = 72 man-days or 180 woman-days Work done by 15 × 3 = 45 woman-days and 12 × 3 = 36 man-days =

45 36 1 1 3 180 72 4 2 4

work remaining = 1

31 4 4

38. A can complete the task in

15 8 = 24 days 5

3 B can complete the task in 24 = 18 days 4 4 times as efficient as A) 3 Now, work left incomplete by B after 10 days ( B is

10 8 = 1– 18 18 39. A can do the work in 24 days A can do

1 1 work in 24 = 8 days 3 3

1 work is completed by B in 8 days 2 1 work will be completed by B in 8 × 2 = 16 days Now,

(A + B) together complete the work in =

24 16 24 16

24 16 48 = days 40 5

Quicker Method: Ratio of efficiency (A : B) = 2 : 3 With 2 efficiency work is done in 24 days. With 1 efficiency work is done in 24 × 2 = 48 days. With 2 + 3 = 5 efficiency work is done in days.

48 5

332

Quicker Maths

Chapter 28

Work and Wages Theorem: Wages are distributed in proportion to the work done and in indirect (or inverse) proportion to the time taken by the individual. Ex.1: A can do a work in 6 days and B can do the same work in 5 days. The contract for the work is 220. How much shall B get if both of them work together? Soln: Method I: A’s 1 day’s work =

1 1 ; B’s 1 day’s work = 6 5

ratio of their wages =

1 1 : =5:6 6 5

220 6 = 120 56 Method II: As wages are distributed in inverse proportion of number of days, their share should be in the ratio 5 : 6. B’s share =

220 6 = 120 11 Ex.2: A man can do a work in 10 days. With the help of a boy he can do the same work in 6 days. If they get 50 for that work, what is the share of that boy?

B’s share =

Soln: The boy can do the work in

10 6 15 days. [Recall 10 6

the theorem] Man’s share : Boy’s share = 15 : 10 = 3 : 2 Man’s share =

50 3 = 30 5

Ex.3: A, B and C can do a work in 6, 8 and 12 days respectively. Doing that work together they get an amount of 1350. What is the share of B in that amount? Soln: A’s one day’s work = C’s one day’s work =

1 1 ; B’s one day’s work = ; 6 8 1 12

1 1 1 : : 6 8 12 Multiplying each ratio by the LCM of their denominators, the ratios become 4 : 3 : 2 A’s share : B’s share : C’s share =

B’s share =

1350 3 = 450. 9

Direct Method: A’s share : B’s share : C’s share = B’s time × C’s time : A’s time × C’s time : A’s time × B’s time = 96 : 72 : 48 = 4 : 3 : 2

1350 3 = 450 9 Ex.4: A, B and C contract a work for 550. Together, A

B’s share =

and B are supposed to do

7 of the work. How much 11

does C get? Soln: A + B did

7 work and C did 11

7 4 work. 1 11 11

(A + B)’s share : C’s share =

7 4 : =7:4 11 11

550 4 = 200. 11 Ex. 5: Two men undertake to do a piece of work for 200. One alone could do it in 6 days, the other in 8 days. With the assistance of a boy they finish it in 3 days. How should the money be divided?

C’s share =

Soln: 1st man’s 3 days’ work =

3 ; 6

2nd man’s 3 days’ work =

3 8

3 3 1 The boy’s 3 days’ work = 1 6 8 8 Their share will be in the ratio

3 3 1 : : =4:3:1 6 8 8

Quicker Maths

334 200 4 = 100 8 200 3 = 75 2nd man’s share = 8 200 1 = 25 The boy’s share = 8 Ex. 6: Wages for 45 women amount to 15525 in 48 days. How many men must work 16 days to receive 5750, the daily wages of a man being double those of a woman? 15525 115 Soln: Wage of a woman for a day = = 45 48 16 115 115 Thus, wage of a man for a day = 2 = 16 8 Now, number of men Total wage No. of days 1 man's 1 day's wage

1st man’s share =

5750 8 25 men 16 115 Note: We should remember the relationship: Total wage = One person’s one day’s wage × No. of persons × No. of days Ex.7: 3 men and 4 boys can earn 756 in 7 days. 11 men and 13 boys can earn 3008 in 8 days. In what time will 7 men with 9 boys earn 2480?

Soln: (3m + 4b) in 1 day earn

756 = 108 7

---- (1)

3008 = 376 ---- (2) 8 From (1), we see that to earn 1 in 1 day, there 3m 4b should be persons. Similarly, from (2), to 108 11m 13b earn 1 in 1 day there should be 376 persons. 3m 4b 11m 13b And also; -------- (*) 108 376 or, m (3 × 376 – 11 × 108) = b (108 × 13 – 4 × 376) ----------- (*) (*) (11m + 13b) in 1 day earn

m 100 5 b 60 3 Now, from (1); (3m + 4b) in 1 day earn 108

3 or, 3m + 4 m in 1 day earn 108 5

or,

27 m in 1 day earn 108 5

108 5 = 20 27 Thus, we get that a man earns 20 daily and a boy

1m in 1 day earns

3 earns 20 = 12 daily.. 5 7m + 9b earn (7 × 20 + 9 × 12) = 248 in 1 day.. 7m + 9b earn 2480 in 10 days. Note : (*) Since both the LHS and the RHS denote the same quantity: “Number of persons earning 1 in 1 day”. (*) (*) We can arrive at this step directly be using cross-multiplication-division rule. Arrange the given information as follows: Men Boys Earning Days 3 4 × 756 ÷ 7 11 13 × 3008 ÷ 8 3 3008 11 756 Now, Men 8 7

13 756 4 3008 Boys 7 8 or, m (3 × 376 – 11 × 108) = b (108 × 13 – 4 × 376)

m 5 b 3 Ex 8: 12 men with 13 boys can earn 326.25 in 3 days. 5 men with 6 boys can earn 237.5 in 5 days. In what time will 3 men with 4 boys earn 210? Soln: Solve yourself (same as Ex 7). Ex 9: A, B and C together earn 1350 in 9 days. A and C together earn 470 in 5 days. B and C together earn 760 in 10 days. Find the daily earning of C. or,

Soln: Daily earning of A + B + C =

1350 = 150 ---(1) 9

Daily earning of A+C =

470 = 94 5

---(2)

Daily earning of B+C =

760 = 76 10

---(3)

From (1) and (2); daily earning of B = 150 – 94 = 56 ---(4) From (3) and (4); daily earning of C = 76 – 56 = 20

Work and Wages

335

EXERCISES 1. Two men A and B working together complete a piece of work which it would have taken them respectively 12 and 18 days to complete if they worked separately. They received in payment 149.25. Find their shares. 2. A, B and C together do a piece of work for 53.50. A working alone could do it in 5 days, B working alone could do it in 6 days and C working alone could do it in 7 days. How should the money be divided among them? 3. It the wages of 45 women amount to 15525 in 48 days, how many men must work 16 days to receive

5750, the daily wages of a man being double those of a woman? 4. If 3 men with 4 boys can earn 210 in 7 days, and 11 men with 13 boys can earn 830 in 8 days, in what time will 7 men with 9 boys earn 1100? 5. If 12 men with 13 boys can earn 326.25 in 3 days, and 5 men with 6 boys can earn 237.50 in 5 days, in what time will 3 men with 4 boys earn 210?

ANSWERS 1. Wages are distributed in inverse proportion of number of days. Hence, the money will be divided in the ratio 18 : 12.

A gets

A’s share =

149.25 18 = 89.55 and 30

149.25 12 = 59.70 B gets 30 2. A’s share : B’s share : C’s share = 6 × 7 : 5 × 7 : 5 × 6 = 42 : 35 : 30

B’s share =

53.50 42 42 35 30 53.50 42 = 21 = 107

53.50 30 = 15 107

3. See Ex. 6. 4. See Ex. 7. 5. Same as Ex. 4.

336

Quicker Maths

Chapter 29

Pipes and Cisterns Introduction: Pipes and Cisterns problems are almost the same as those of Time and Work problems. Thus, if a

1 pipe fills a tank in 6 hrs, then the pipe fills th of the tank 6 in 1 hour. The only difference with Pipes and Cisterns problems is that there are outlets as well as inlets. Thus, there are agents (the outlets) which perform negative work too. The rest of the process is almost similar. Inlet: A pipe connected with a tank (or a cistern or a reservoir) is called an inlet, if it fills it. Outlet: A pipe connected with a tank is called an outlet, if it empties it.

hrs, and all of them are opened together, the net

1 1 1 part filled in 1 hr = x y z xyz time taken to fill the tank = yz xz xy hrs. (vi) A pipe can fill a tank in x hrs. Due to a leak in the bottom it is filled in y hrs. If the tank is full, the time

xy taken by the leak to empty the tank = y x hrs. Ex 1:

Two pipes A and B can fill a tank in 36 hours and 45 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

Soln:

Part filled by A alone in 1 hour =

1 36

Part filled by B alone in 1 hour =

1 45

FORMULAE (i)

If a pipe can fill a tank in x hours, then the part filled

1 . x (ii) If a pipe can empty a tank in y hours, then the part of in 1 hour =

1 the full tank emptied in 1 hour = y .

Part filled by (A + B) in 1 hour

(iii) If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours, then the net part filled

1 1 in 1 hour, when both the pipes are opened = x y .

time taken to fill the tank, when both the pipes are

1 9 1 1 = . 36 45 180 20 Hence, both the pipes together will fill the tank in 20 hours. Direct Method: [By formula (iv)]

xy opened = y x (iv) If a pipe can fill a tank in x hrs and another can fill the same tank in y hrs, then the net part filled in 1 hr, when

1 1 both the pipes are opened = x y . xy time taken to fill the tank = y x (v)

If a pipe fills a tank in x hrs and another fills the same tank in y hrs, but a third one empties the full tank in z

Time taken = Ex 2:

Sol:

36 45 = 20 hrs. 36 45

A pipe can fill a tank in 15 hours. Due to a leak in the bottom, it is filled in 20 hours. If the tank is full, how much time will the leak take to empty it? Work done by the leak in 1 hour

1 1 1 = . 15 20 60

the leak will empty the full tank in 60 hrs.

Quicker Maths

338 Direct Method: [By formula (vi)] Required time = Ex 3:

Sol:

Note:

15 20 = 60 hrs. 20 15

pipe. Since 2 cm diametre fills

Pipe A can fill a tank in 20 hours while pipe B alone can fill it in 30 hours and pipe C can empty the full tank in 40 hours. If all the pipes are opened together, how much time will be needed to make the tank full?

Ex 4:

Soln:

Ex 5:

1 cm, 2 cm, running 3 together, when the largest alone will fill it in 61 minutes, the amount of water flowing in by each pipe being proportional to the square of its diametre? 1 In 1 minute, the pipe of 2 cm diametre fills 61 of the cistern. 1 1 In 1 minute, the pipe of 1 cm diametre fills 61 4 of the cistern. ----- (*) whose diametres are 1 cm, 1

Soln:

In 1 minute, the pipe of 1

1 4 of the cistern. 61 9

1 cm diametre fills 3

----- (**)

1 4 1 1 In 1 minute of the 61 61 4 61 9 36 cistern is filled. the whole is filled in 36 minutes. Ans.

1 1 1 1 of the 61 2 61 4

cistern.

1 4 (**) 1 cm diametre fills 3 3 2

The tank will be full in

20 30 40 120 1 17 hrs. 30 40 20 40 20 30 7 7 Two pipes A and B can fill a cistern in 1 hour and 75 minutes respectively. There is also an outlet C. If all the three pipes are opened together, the tank is full in 50 minutes. How much time will be taken by C to empty the full tank? Work done by C in 1 min. 1 1 3 1 1 . = 60 75 50 300 100 C can empty the full tank in 100 minutes. In what time would a cistern be filled by three pipes

1 of the cistern, 61

2

1 cm diametre fills

1 1 7 1 Net part filled in 1 hour = . 20 30 40 120

120 1 i.e. 17 hours. 7 7 Direct Method: [(By formula (v)]

(*) We are given that amount of water flowing is proportional to the square of the diametre of the

1 1 4 1 4 of the cistern. 61 4 3 61 9 Ex 6:

There is a leak in the bottom of a cistern. When the cistern is thoroughly repaired, it would be filled

1 in 3 hrs . It now takes half an hour longer. If 2 the cistern is full, how long would the leak take to empty the cistern? Soln: Ex 7:

Soln:

3.5 4 = 28 hrs. 4 3.5 Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 18 minutes? Let B be closed after x minutes. Then, part filled by (A+B) in x min. + part filled by A in (18 – x) min. = 1. Required time =

1 1 1 1 x (18 x) 24 32 24

7x 18 x 1 96 24 or, 7x + 4(18 – x) = 96 or, 3x = 24 x = 8. So, B should be closed after 8 min. Direct Formula: or

18 Pipe B should be closed after 1 32 24 = 8 min. Ex. 8: Two pipes P and Q would fill a cistern in 24 hours and 32 hours respectively. If both pipes are opened together, find when the first pipe must be turned off so that the cistern may be just filled in 16 hours.

Pipes and Cisterns Soln:

339

Suppose the first pipe was closed after x hrs. Then, first’s x hrs’ supply + second’s 16 hrs’ supply =1 or,

x 16 1 24 32

16 The first pipe should work for 1 24 hrs. 32

Soln:

= 12 hrs. If two pipes function simultaneously, the reservoir is filled in 12 hrs. One pipe fills the reservoir 10 hours faster than the other. How many hours does the faster pipe take to fill the reservoir? Let the faster pipe fills the tank in x hrs. Then, the slower pipe fills the tank in x + 10 hrs. When both of them are opened, the reservoir will be filled in

x(x 10) 12 x (x 10)

or, x 2 14x 120 0 x = 20, -6 But x can’t be -ve, hence the faster pipe will fill the reservoir in 20 hrs. Ex. 10: Three pipes A, B and C can fill a cistern in 6 hrs. After working together for 2 hours, C is closed and A and B fill the cistern in 8 hrs. Then, find the time in which the cistern can be filled by pipe C. Soln:

A + B + C can fill in 1 hr =

1 th of the cistern. 6

A + B + C can fill in 2 hrs =

2 1 rd of the cistern. 6 3

1 2 Unfilled part = 1 rd is filled by A + B in 3 3 8 hrs.

83 = 12 hrs. 2 And, we have (A + B + C) can fill the cistern in 6 hrs. C = (A + B + C) – (A + B) can fill the cistern in

(A + B) can fill the cistern in

12 6 12 hrs. 12 6

12 8 = 24 hrs. 12 8 Capacity of tank = 24 × 60 × 6 = 8640 litres Ex. 12: A tank is normally filled in 8 hours but takes 2 hrs longer to fill because of a leak in its bottom. If the cistern is full, in how many hrs will the leak empty it? Soln: It is clear from the question that the filler pipe fills the tank in 8 hrs and if both the filler and the leak work together, the tank is filled in 8 hrs. Therefore, Soln:

x 1 1 24 1 2 2 x = 12 hrs. Direct Formula:

Ex. 9:

Ex. 11: A tank has a leak which would empty it in 8 hrs. A tap is turned on which admits 6 litres a minutes into the tank, and it is now emptied in 12 hrs. How many litres does the tank hold? The filler tap can fill the tank in

8 10 = 40 hrs. 10 8 Ex. 13: A pipe can fill a tank in 12 minutes and another pipe in 15 minutes, but a third pipe can empty it in 6 minutes. The first two pipes are kept open for 5 minutes in the beginning and then the third pipe is also opened. In what time is the cistern emptied? the leak will empty the tank in

Soln:

1 1 3 Cistern filled in 5 minutes = 5 12 15 4 Net work done by 3 pipes in 1 minute

1 1 1 1 12 15 6 60 -ve sign shows that

1 th part is emptied in 1 60

minutes.

3 3 th part is emptied in 60 = 45 minutes. 4 4 Ex. 14: If three taps are opened together, a tank is filled in 12 hrs. One of the taps can fill it in 10 hrs and another in 15 hrs. How does the third tap work? Soln: We have to find the nature of the third tap — whether it is a filler or a waste pipe. Let it be a filler pipe which fills in x hrs.

10 15 x 12 10 15 10x 15x or, 150x = 150 × 12 + 25x × 12 or, –150x = 1800 x = – 12 -ve sign shows that the third pipe is a waste pipe which vacates the tank in 12 hrs. Then,

Quicker Maths

340 Ex. 15: A, B and C are three pipes connected to a tank. A and B together fill the tank in 6 hrs. B and C together fill the tank in 10 hrs. A and C together fill the tank

Soln:

1 in 7 hrs. In how much time will A, B and C fill 2 the tank separately? A + B fill in 6 hrs. B + C fill in 10 hrs. 1 15 A + C fill in 7 hrs 2 2

6 10 6

15 2

15 15 10 2 2

6 5 15 5 hrs 180 2 A + B + C fill the tank in 5 hrs. Now, A [= (A + B + C) – (B + C)] fills in

10 5 10 hrs 10 5 15 5 Similarly, B fills in 2 15hrs. and C fills in 15 5 2

56 = 30 hrs. 65 Ex. 16: Two pipes can separately fill a tank in 20 hrs and 30 hrs respectively. Both the pipes are opened to 1 fill the tank but when the tank is rd full a leak 3 1 rd of the water 3 supplied by both the pipes leak out. What is the total time taken to fill the tank? Time taken by the two pipes to fill the tank develops in the tank through which

Soln:

2 as efficient 3

as earlier. the work of (12 – 4 =) 8 hrs will be completed

2 83 12 hrs 3 2 total time = 4 + 12 = 16 hrs OR

now in 8

1 rd of the supplied water leaks out, the 3 leakage empties the tank in 12 × 3 = 36 hrs. Now, time taken to fill the tank by the two pipes and the Since

6 10

2 (A + B + C) fill in

1 3

the filler pipes are only 1

20 30 hrs 12 hrs. 20 30

1 12 rd of the tank is filled in = 4 hrs. 3 3 1 Now, rd of the supplied water leaks out 3

36 12 = 18 hrs. 36 12 time taken by the two pipes and the leakage to leakage =

2 2 rd of the tank 18 = 12 hrs. 3 3 total time = 4 hrs + 12 hrs = 16 hrs. Ex. 17: A cistern is normally filled in 8 hrs but takes two hrs longer to fill because of a leak in its bottom. If the cistern is full, the leak will empty it in ______ hrs. Soln: (Detailed): Suppose the leak can empty the tank in x hrs. fill

Then, part of cistern filled in 1 hr = Cistern will be filled in

1 1 x 8 8 x 8x

8x hrs. x 8

8x 8 2 = 10 hrs. x 8 or, 8x = 10x - 80 x = 40 hrs. Quicker Approach: The filler takes 2 hrs more the leak empties in 10 hrs what the filler fills in 2 hrs. Now,

the leak empties in 10 hrs

2 1 th of the 8 4

cistern the leak empties the full cistern in 4 × 10 = 40 hrs. Direct formula: The leak will empty in = 40 hrs.

8 (8 2) 2

Pipes and Cisterns

341

EXERCISE 1. Pipes A and B can fill a tank in 10 hours and 15 hours respectively. Both together can fill it in ______ hrs. 2. A tap can fill the cistern in 8 hours and another can empty it in 16 hours. If both the taps are opened simultaneously, the time (in hours) to fill the tank is ______. 3. A pipe can fill a tank in x hours and another can empty it in y hours. They can together fill it in (y > x) ______. 4. One tap can fill a cistern in 2 hours and another can empty the cistern in 3 hrs. How long will they take to fill the cistern if both the taps are opened? 5. A cistern can be filled by two pipes A and B in 4 hours and 6 hours respectively. When full, the tank can be emptied by a third pipe C in 8 hours. If all the taps be turned on at the same time, the cistern will be full in ______. 6. A tank is filled by pipe A in 32 minutes and pipe B in 36 minutes. When full, it can be emptied by a pipe C in 20 minutes. If all the three pipes are opened simultaneously, half of the tank will be filled in ______ minutes. 7. If two pipes function simultaneously, the reservoir will be filled in 6 hours. One pipe fills the reservoir 5 hours faster than the other. How many hours does the faster pipe take to fill the reservoir? 8. Three pipes A, B and C can fill a cistern in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill it in 7 hours. The time taken by C alone to fill the cistern is ______ hrs. 9. A cistern has a leak which would empty it in 8 hours. A tap is turned on which admits 6 litres a minute into the cistern, and it is now emptied in 12 hours. How many litres does the cistern hold ?

10. Two taps can separately fill a cistern in 10 minutes and 15 minutes respectively and when the waste pipe is open, they can together fill it in 18 minutes. The waste pipe can empty the full cistern in ______ minutes