Magical Book on Quicker Maths PDF

 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 Enlar...

 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 Page No. 1. How to Prepare for Maths 1 2. Addition 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 time- saving 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 Maths- based 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) (i) Remember the TABLE upto 20 (at least): You should know that tables have been prepared to Quantitative Aptitude is a compulsory paper. You make calculations faster. You can see the use of table in can't neglect. So make sure you are ready to improve the following example: your mathematical skills. Each question values 1.2 Evaluate: 16 × 18 marks whereas each question of Reasoning values only If you don’t remember the table of either 16 or 18 1.066 marks in PO exam. So, if you devote relatively you will proceed like this: more time on this paper you get more marks. Also, 16 the answers of Maths questions are more confirmed 18 than answers of Reasoning questions, which are often 128 confusing. Most of you feel it is a more time-consuming 16 paper, but if you follow our guidelines, you can save 288 your valuable time in examination hall. Other exams: There are very few competitive exams without Maths paper. SSC exams have different types of But if you know the table of 16, your calculations Maths paper. The mains exam of SSC contains Subjective would be: Question paper. Keeping this in mind, I have also given 16 × 18 = 16(10 + 8) = 16 × 10 + 16 × 8 = 160 the detail method of each short-cut or Quicker Method + 128 = 288 given in this book. Each theorem, which gives you a direct Or, if you know the table of 18; your calculation formula also contains proof of the theorem, which is would be nothing but a general form (denoting numerical values 16 × 18 = 18(10 + 8) = 18 × 10 + 18 × 6 = 180 by letters say X, Y, Z etc) of detail method. + 108 = 288 If you can, you remember the table upto 30 or 40. 2. Preparation for this paper It will be precious for you. Note: (1) You should try the above two methods on some (A) How to start your preparation more examples to realise the beauty of tables. Maths is a very interesting subject. If you don't find Try to evaluate: it interest-ing, it simply means you havn't tried to 19 × 13; l7 × 24; 18 × 32 (18 × 30 + 18 × 2 = understand it. Let me assure you it is very simple and 540 + 36 = 576); 19 × 47; 27 × 38; 33 × 37 100% logical. There is nothing to be assumed and nothing etc. to be confused about. So, nothing to worry if you come (2) All the above calculations should be done forward with firm determination to learn maths. mentally. Try it. The most basic things in Maths are: (a) Addition - Subtraction (ii) LEARN the one-line Addition or Subtraction (b) Multiplication - Division method from this book All these four things are most useful. At least one of In the first chapter we have given some methods of these four things is certainly used in any type of faster addition and subtraction. Suppose you are given mathematical question. So, if we do our basic calculations to calculate: faster we save our valuable time in each question. To 789621 – 32169 + 4520 – 367910 =.... calculate faster, I suggest the following tips: If you don’t follow this book you will do like: 2 Quicker Maths 789621 32169 (iv) Learn the Rule of Fractions +4520 +367910 In the chapter Ratio and Proportion on Page No. 794141 400079 269, I have discussed this rule. It is the faster form of unitary method. It is nothing but simplified form of Rule 794141 of Three and Rule of Proportional Division. No doubt, it –400079 works faster and is used in almost all the mathematical 394062 questions where unitary method (Aikik Niyam) is used. The above method takes three steps, i.e. (i) add the See the following example: two +ve values; (ii) add the two -ve values; (iii) subtract Ex. If 8 men can reap 80 hectares in 24 days, how many the second addition from the first addition. hectares can 36 men reap in 30 days? But you can see the one-step method given in the Soln: I don't know how much time you will take to chapter. Have mastery over this method. It takes less answer the question but if we follow the rule of writing as well as calculating time. fraction our calculation would be: (iii) Learn the one-line Multiplication or Division 36 30 method from this book 80    450 hectares. 8 24 Method of faster multiplication is given in the second chapter. I think it is the most important chapter of this In this book this method is used very frequently. It is book. Multiplication is used in almost all the questions, so only when you go through the various chapters of this if your multiplication is faster you can save at least 35% book that you will find how wonderful the method is. of your usual time. You should learn to use the faster It saves at least half of your usual time. I think this one-line method. It needs some practice to use this method method should be adopted by all of you at any cost. First frequently. The following example will show you how learn it and then use it wherever you can. this method saves your valuable time. So, these four points are necessary for your strong Ex. Multiply: 549 × 36 and firm start. And only strong and firm start is the key If you don't follow the one-line method of to sure success. multiplication, you will calculate like: 549 (B) Clear the Fundamentals behind each chapter 36 There are 39 chapters in this book. Each chapter has 3294 some important basic fundas. Those fundas should be 1647 clear to you. Any doubt with the basics will hamper your 19764 further steps. Now the question arises - what are those If this method takes 30 seconds I assures: you that basic fundas? Naturally, for each chapter, there are one-line method given in this book will take at the most different fundas. I will discuss one chapter and its basic 15 seconds. Try it. fundas. You can understand and find the same with One-line method of calculation for Division is also different chapters. My chapter is PARTNERSHIP (on very much useful. You should learn and try it if you find page 309) it interesting. But, as division is less used, some of you may avoid this chapter. Partnership Simple (Different investments Compound [Different investments for the same period) (P & Q); Diff periods (t 1 & t 2) Ratio of profit = P : Q. 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 sources: Guides, Books, Magazines etc. The most standard before going through this chapter. Now, you can and reliable sets are available in the magazine Banking understand what I actually mean by the fundas. In a similar Services Chronicle. Also, with our Correspondence Course way, you can collect all the fundas and basic formulae at we give at least 60 sets of Maths papers separately and one place. 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 From the beginning to the end, the complete session as possible. Each question given in examples has some of practice should be divided into five parts. uniqueness. Mark it and keep it in mind. To collect more Part (i): Take your first test with previous paper without examples of different types you may consult different taking time into consideration. Try to solve all books available in the market. questions. Note down the total time and score in your performance diary. Also note down the (D) Use of Quicker Formulae questions which took more than one minute. Now Before going for quicker formula I suggest you to you have to find out the reason of your low know the detail method of the solution as well. So, see performance, if it is so. Naturally, you would find the proofs of all the formulae carefully. Once you get the following reasons: familiar with the detail solution, you find it easier to  Some questions were difficult and time understand the quicker method. Direct Formulae or consuming. Quicker Methods save your valuable time but they have  Some questions were unsolvable for you. very high potential of creating confusion in their usage.  You lost your concentration. So you should know where the particular formula should  You lost your patience. be used. A little change in the questions may lead you to  You did more writing job. wrong solution. So be careful before using them. In case  You could not use Quicker Methods. of any confusion, you are suggested to solve the questions Try to find out the solutions to all the above without using direct formula or quicker method. problems. If any of the questions was difficult for Only frequent use of the quicker methods can make you, it means your initial preparation was not good. you perfect in Quicker Maths. But don't worry. Go through that chapter again and After covering all the chapters and knowing all the clear your basic concepts. Because the standard of methods, you should be prepared for practice. a question is always within your reach. If you have passed your 10th exam with maths, you can solve 3. Practice of Maths Paper all the questions. You should take at least 10-12 tests (A) Pattern of paper in this part. You should know the pattern and style of the question Part (ii): This time the paper may be either previous or paper of the exam for which you are going to appear. model (sample). Fix your alloted period (say 50 Suppose you are preparing for Bank PO exam. You should minutes for PO). And solve as many questions as know that maths paper consists of 35 questions in prelims possible within that period. Once you have exam, 35 questions main exam and 50 questions in various completed your test, count the number of correct PGDBF courses. Out of which 15-20 questions are from questions. Note down the number of questions Data Analysis, 5 Questions from Data Sufficiency, 5-10 solved by you and the no. of correct solutions in are from Numericals (Calculation based), and 20-25 are your performance diary. Now, you can find the from Mathematical Chapters (like Profit & Loss, reasons for your low scoring. If the reasons are the Percentage, Partnership, Mensuration, Time & Work, same as in Part (i), you try to resolve the problem Train, Speed etc.). In other exams it may be different. again. Take at least 5-6 tests in this part. After The pattern can be known from previous papers. analysing your performance and problems you should be ready for your third part of test. (B) Collection of Previous Papers as well as Part (iii): This time you try to solve all the questions sample papers within a time period fixed by you in advance. This If you can, you should arrange as many as possible period should be less than that for numbers of previous and sample papers. There are many the tests in part (ii). If you couldn't do it, try it again on another test paper. Part (iii) should not be 4 Quicker Maths considered completed unless you have achieved possible time (say, 30-35 minutes for PO paper). your goal. This part may take 3 to 4 months. Keep patience Part (iv): After completion of part (iii), you need to and go on practicing. increase your speed. This part is penultimate stage Part (v): This is the last part of your practice. After part of your final achievement. You should try to solve (iv), you should take your test with complete full- the complete paper of maths within a minimum length paper for PO.  Chapter 2 Addition In the problem of addition we have two main factors 350 49.24 (speed and accuracy) under consideration. We will discuss 9989 a method of addition which is faster than the method The conventional method is to add the figures down used by most people and also has a higher degree of the right-hand column, 4 plus 8 plus 6, and so on. You accuracy. In the latter part of this chapter we will also can do this if you wish in the new method, but it is not discuss a method of checking and double-checking the compulsory; you can begin working on any column. But results. for the sake of convenience, we will start on the right- In using conventional method of addition, the average hand column. man cannot always add a fairly long column of figures We add as we go down, but we “never count higher without making a mistake. We shall learn how to check than 10”. That is, when the running total becomes greater the work by individual columns, without repeating the than 10, we reduce it by 10 and go ahead with the reduced addition. This has several advantages: figure. As we do so, we make a small tick or check- 1) We save the labour of repeating all the work; mark beside the number that made our total higher than 2) We locate the error, if any, in the column where it 10. occurs; and For example: 3) We are certain to find error, which is not 4 necessary in the conventional method. 8 4 plus 8,12: this is more than 10, so we subtract This last point is something that most people do not 10 from 12. Mark a tick and start adding again. realise. Each one of us has his own weaknesses and 6 6 plus 2, 8 own kind of proneness to commit error. One person may 1 1 plus 8, 9 have the tendency to say that 9 times 6 is 56. If you ask 0 0 plus 9, 9 him directly he will say “54”, but in the middle of a long 9 9 plus 9, 18: mark a tick and reduce 18 by 10, calculation it will slip out as “56”. If it is his favourite say 8. error, he would be likely to repeat it when he checks by The final figure, 8, will be written under the column repetition. as the “running total”. Next we count the ticks that we have just made as Totalling in columns we dropped 10’s. As we have 2 ticks, we write 2 under As in the conventional method of addition, we write the column as the “tick figure”. The example now looks the figures to be added in a column, and under the bottom like this: figure we draw a line, so that the total will be under the 4234 column. When writing them we remember that the 8238 mathematical rule for placing the numbers is to align the 646 right-hand-side digits (when there are whole numbers) and the decimal points (when there are decimals). For 5321 example: 350 Right-hand-side-digits Decimal 9989 alignment alignment running total: 8 4234 13.05 ticks: 2 8238 2.51 If we repeat the same process for each of the columns 646 539.652 we reach the result: 5321 2431.0003 6 Quicker Maths 4234 4 2 3 4 8238 8 2 3 8 646 6 4 6 5321 53 2 1 350 3 5 0 9989 9 9 8 9 running total: 6558 ticks: 2222 Total: 287 7 8 Now we arrive at the final result by adding together Note:We see that in the leftmost column we are left the running total and the ticks in the way shown in the with 2 ticks. Write down the number of ticks in a following diagram, column left to the leftmost column. Thus we get the running total: 0 6 5 5 8 0 answer a little earlier than the previous method. One more illustration : Q: 707.325 + 1923.82 + 58.009 + 564.943 + 65.6 = ? ticks: 0 2 2 2 2 0 Solution: Total: 2 8 7 7 8 707.325 Save more time: We observe that the running total 1923.82 is added to the ticks below in the immediate right column. 58.009 This addition of the ticks with immediate left column can 564.943 be done in single step. That is, the number of ticks in the 65.6 first column from right is added to the second column from right, the number of ticks in the 2nd column is Total: 3319.697 added to the third column, and so on. The whole method You may raise a question:is it necessary to write the can be understood in the following steps. numbers in column-form? The answer is ‘no’. You may 4 23 4 get the answer without doing so. Question written in a row-form causes a problem of alignment. If you get 8 2 3 8 command over it, there is nothing better than this. For 64 6 initial stage, we suggest you a method which would bring Step I. 5321 you out of the alignment problem. 350 Step I. “Put zeros to the right of the last digit after decimal 9 9 8 9 to make the no. of digits after decimal equal in each Total: 8 number.” [4 plus 8 is 12, mark a tick and add 2 to 6, which is For example, the above question may be written as 8; 8 plus 1 is 9; 9 plus 0 is 9; 9 plus 9 is 18, mark a tick 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 and write down 8 in the first column of total-row.] Step II. Start adding the last digit from right. Strike off 4234 the digit which as been dealt with. If you don’t cut, 8 2 3 8 duplication may occur. During inning total, don’t 6 4 6 exceed 10. That is, when we exceed 10, we mark a Step II. 5321 tick anywhere near about our calculation. Now, go 3 50 ahead with the number exceeding 10. 9 9 8 9 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = __ _ _. __7 Total: 78 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 [3 plus 2 (number of ticks in first column) is 5; 5 down 7. During this we strike off all the digits which are plus 3 is 8; 8 plus 4 is 12, mark a tick and carry 2; 2 plus used. It saves us from confusion and duplica-tion. 2 is 4; 4 plus 5 is 9; 9 plus 8 is 17, mark a tick and write Step III. Add the number of ticks (in rough) with the down 7 in 2nd column of total-row.] digits in 2nd places, and erase that tick from rough. In a similar way we proceed for 3rd and 4th columns. 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 7 is 5; 5 plus 4 is 9 and 9 plus 0 is 9; so write down 9 in its (–1) (10) (–1) (10) place. (4) (0) (–1) Step IV. (3) (9) (9) Note: The above explanation is easy to understand. And 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = _ _ _ _.697 the method is more easy to perform. If you practise 3 plus 8 is 11; mark a tick in rough and carry over 1; well, the two steps (I & II) can be performed 1 plus 0 is 1; 1 plus 9 is 10, mark another tick in rough simultaneously. The second step can be performed and carry over zero; 0 plus 6 is 6, so put down 6 in its in another way like: place. (4) (0) (–l) = 400 – 1= 399 Step V. Ex. 2: 5124 – 829 + 731– 435 Last Step: Following the same way get the result: Soln: According to step I, the temporary figure is: 707.325 + 1923.820 + 58.009 + 564.943 + 65.600 = 3319.697 (5) (– 4) (0) (– 9) Addition of numbers (without decimals) written in a Step II: Borrow 1 from 5. Thousands place becomes 5 row form – 1= 4.1 borrowed from thousands becomes 10 at Q. 53921 + 6308 + 86 + 7025 + 11132 = ? hundreds. Now, 10 – 4 = 6 at hundreds place, but 1 Soln: is borrowed for tens. So digit at hundreds becomes Step I: 53921 + 6308 + 86 + 7025 + 11132 = ___ _ 2 6 –1 = 5.1 borrowed from hundreds becomes 10 at Step II: 53921 + 6308 + 86 + 7025 + 11132 = _ _ _ 72 tens place. Step III: 53921 + 6308 + 86 + 7025 + 11132 = _ _ 472 Again we borrow 1 from tens for units place, after Step IV: 53921 + 6308 + 86 + 7025 + 11132 = _ 8472 which the digit at tens place is 9. Now, 1 borrowed Step V: 53921 + 6308 + 86 + 7025 + 11132 = 78472 from tens becomes 10 at units place. Thus the result Note: One should get good command over this method at units place is 10 – 9 = 1. Our required answer because it is very much useful and fast-calculating. If = 4591 you don’t understand it, try again and again. Note: After step I we can perform like: 5 (– 4) (0) (– 9) = 5000 – 409 = 4591 Addition and subtraction in a single row But this method can’t be combined with step I to Ex. 1: 412 – 83 + 70 = ? perform simultaneously. So, we should try to Step I: For units digit of our answer add and subtract understand steps I & II well so that in future we can the digits at units places according to the sign attached perform them simultaneously. with the respective numbers. For example, in the Ex. 3: 73216 – 8396 + 3510 – 999 = ? above case the unit place of our temporary result is Soln: Step I gives the result as: 2–3+0=–l (7) (–2) (–5) (–16) (–9) So, write as: Step II: Units digit = 10 – 9 = 1 [1 borrowed from (–16) 412 – 83 + 70 = _ _ (–1) results –16 –1 = –17] Similarly, the temporary value at tens place is 1 – 8 + Tens digit = 20 –17 = 3 [2 borrowed from (–5) results 7 = 0. So, write as: –5 – 2 = –7] 412 – 83 + 70 = _ (0) (–1) Hundreds digit = 10 – 7 = 3 [1 borrowed from –2 Similarly, the temporary value at hundreds place is results –2 –1 = – 3] 4. So, we write as: Thousands digit = 10 – 3 = 7 [1 borrowed from 7 412 – 83 + 70 = (4) (0) (–l) results 7–1=6] Step II: Now, the above temporary figures have to be So, the required value is 67331. changed into real value. To replace (–1) by a +ve The above calculations can also be started from the digit we borrow from digits at tens or hundreds. leftmost digit as done in last two examples. We have As the digit at tens is zero, we will have to borrow started from rightmost digit in this case. The result from hundreds. We borrow 1 from 4 (at hundreds) is the same in both cases. But for the combined which becomes 10 at tens leaving 3 at hundreds. operation of two steps you will have to start from Again we borrow 1 from tens which becomes 10 at rightmost digit (i.e. units digit). See Ex. 4. units place, leaving 9 at tens. Thus, at units place 10 Note: Other method for step II: (–2) (–5) (–16) (–9) = – 1 = 9. Thus our final result = 399. (–2) (–6) (–6) (–9) = (–2669) The above explanation can be represented as  Ans = 70000 – (2669) = 67331 8 Quicker Maths Ex. 4: 89978 - 12345 - 36218 = ? Ex. 7: Solve Ex. 3 in a single step without writing anything Soln: Step I: (4) (1) (4) (2) (–5) other than the answer. Try it yourself. Don’t move Step II: 4 1 4 1 5 to next example until you can confidently solve such questions within seconds. Single step solution: Ex. 8: 10789 + 3946 – 2310 – 1223 = ? Now, you must learn to perform the two steps Soln: Whenever we get a value more than 10 after simultaneously. This is the simplest example to understand addition of all the units digits, we will put the units the combined method. At units place: 8 - 5 - 8 = (-5). digit of the result and carry over the tens digit. To make it positive we have to borrow from tens. You We add the tens digit to +ve value, not to the should remember that we can’t borrow from -ve value –ve value. Similar method should be adopted for i.e., from 12345. We will have to borrow from positive all digits. value i.e. from 89978. So, we borrowed 1 from 7 (tens +1 +l +1 digit of 89978): 1 0 7 8 9 + 3946 – 2310 – 1223 = 11202 (-1) Note: 1. We put +1 over the digits of +ve value 10789. 8 9 9 7 8 - 12345 - 36218 = _ _ _ _ 5 It can also be put over the digits of 3946. But it Now digit at tens: (7 - 1 =) 6 - 4 - 1 = 1 can’t be put over 2310 and 1223. Digit at hundreds: 9 - 3 - 2 = 4 2. In the exam when you are free to use your pen Digit at thousands: 9 - 2 - 6 = 1 on question paper you can alter the digit with Digit at ten thousands: 8 - 1 - 3 = 4 your pen instead of writing +1, +2, –1, –2....  the required value = 41415 over the digits. Hence, instead of writing 8, Ex. 5: 28369 + 38962 - 9873 = ? you should write 9 over 8 with your pen. Soln: Single step solution: 1 Units digit = 9 + 2 – 3 = 8 Similarly, write 8 in place of 7. Tens digit = 6 + 6 – 7 = 5 Ex. 9: 765.819 – 89.003 + 12.038 – 86.89 = ? Hundreds digit = 3 + 9 – 8 = 4 Soln: First, equate the number of digits after decimals Thousands digit = 8 + 8 - 9 = 7 by putting zeros at the end. Ten thousands digit = 2 + 3 = 5 So, ? = 765.819 – 89.003 + 12.038 – 86.890  required value = 57458 Now, apply the same method as done in Ex. 4, 5, Ex. 6: Solve Ex. 2 by single-step method. 6, 7 & 8. Soln: 5124 – 829 + 731– 435 = -1 -1 -l -l +1 Units digit: 4 – 9 + 1 – 5 = (–9). Borrow 1 from 7 6 5. 8 1 9 - 89.003 + 12.038 – 86.890 tens digit of the positive value. Suppose we = 601.964 borrowed from 3 of 731. Then Ex 10: 5430 - 4321 + 3216 - 6210 = ? –1 Soln: The above case is different. The final answer comes 5124 – 829 + 731 – 435 = _ _ _ 1 negative. But as we don't know this in the beginning, Tens digit: 2 – 2 + 2 – 3 = (–1). Borrow 1 from we perform the same steps as done earlier. hundreds digit of +ve value. Suppose we borrowed Step I: (-2) (1) (1) (5) from 7 of 731. Then Step II: The leftmost digit is negative. It can't be made –1 –1 positive as there is no digit at the left which can 5124 – 829 + 731 – 435 = _ _ 9 1 lend. So, our answer is Hundreds digit: 1 – 8 + 6 – 4 = (–5). Borrow 1 from thousands  2000 digit of +ve value. We have only one such digit,  115 i.e. 5 of 5124. Then  1885 -1 -1 -1 Note: The second step should be done mentally keeping 5 1 2 4 – 829 + 731 - 435 = 4591 in mind that except the leftmost digit all the other (Thousands digit remains as 5 - 1 = 4) digits are positive. So, the final answer will be -ve Now you can perform the whole calculation in a but not (–)2115. It should be -2000 +115= –1885. single step without writing anything extra. Addition 9 Ex 11: 2695 - 4327 + 3214 - 7350 = ? from the digit-sum of the numbers must be equal to Soln: Step I: (-6) (2) (3) (2) the digit-sum of the answer. For example: So, required answer =  6000 The number: 23 + 49 + 15 + 30 = 117  232 The digit-sum: 5 + 4 + 6 + 3 = 0  5768 Which reduces to : 0 =0 This rule is also applicable to subtraction, Method of checking the calculation: Digit multiplication and upto some extent to division also. These will be discussed in the coming chapters. We should take sum Method another example of addition. This method is also called the nines-remainder 1.5 + 32.5 + 23.9 = 57.9 method. The concept of digit-sum consists of this : digit-sum: 6 + 1 + 5 =3 I. We get the digit-sum of a number by “adding or, 3 =3 across” the number. For instance, the digit-sum of Thus, if we get LHS = RHS we may conclude that 13022 is 1 plus 3 plus 0 plus 2 plus 2 is 8. our calculation is correct. II. We always reduce the digit-sum to a single figure if Sample Question: Check for all the calculations done it is not already a single figure. For instance, the in this chapter. digit-sum of 5264 is 5 plus 2 plus 6 plus 4 is 8 (17, Note: Suppose two students are given to solve the or 1 plus 7 is 8). following question: 1.5 + 32.5 + 23.9 = ? III. In “adding across” a number, we may drop out 9’s. One of them gets the solution as 57.9. Another student Thus, if we happen to notice two digits that add up gets the answer 48.9. If they check their calculation by to 9, such as 2 and 7, we ignore both of them; so the this method, both of them get it to be correct. Thus this digit-sum of 990919 is 1 at a glance. (If we add up method is not always fruitful. If our luck is against us, 9’s we get the same result.) we may approve our wrong answer also. 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 Addition of mixed numbers of zero. The digit-sum of 441, for example, is zero. 1 4 1 Quick Addition of Digit-sum: When we are “adding Q. 3  4  9  ? 2 5 3 across” a number, as soon as our running total reaches Solution:A conventional method for solving this question two digits we add these two together, and go ahead with is by converting each of the numbers into pure fractional a single digit as our new running total. numbers first and then taking the LCM of denominators. For example: To get the digit-sum of 886542932851 To save time, we should add the whole numbers and the we do like: 8 plus 8 is 16, a two – figure number. We fractional values separately. Like here, 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 1 4 1  1 4 1 3  4  9  (3  4  9 )      is 9, forget it. 4 plus 2 is 6. Forget 9.... Proceeding this 2 5 3  2 5 3 way we get the digit-sum equal to 7. For decimals we work exactly the same way. But we 5  24  10 19 = 16   16  1 don’t pay any attention to the decimal point. The digit- 30 30 sum of 6.256, for example, is 1. 19 19 19 Note: It is not necessary in a practical sense to understand = (16  1)   17   17 why the method works, but you will see how 30 30 30 interesting this is. The basic fact is that the reduced 2 1 3 1 digit-sum is the same as the remainder when the Q. 5  4  2  1 3 6 4 4 number is divided by 9. For example: Digit-sum of 523 is 1. And also when 523 2 1 3 1 Soln: (5  4  2  1)       is divided by 9, we get the remainder 1. 3 6 4 4 Checking of Calculation 8 2  9 3 12 Basic rule: Whatever we do to the numbers, we also = 2 2 = 2 + 1 =3  12  12 do to their digit-sum; then the result that we get 10 Quicker Maths  Chapter 3 Multiplication Special Cases The answer is 64812. We suggest you to remember the tables up to 30 As you see, each figure of the long number is used because it saves some valuable time during calculation. twice. It is first used as a “number”, and then, at the next Multiplication should be well commanded, because it is step, it is used as a neighbour. Looking carefully, we can needed in almost every question of our concern. use just one rule instead of three rules. And this only rule Let us look at the case of multiplication by a number can be called as “add the right neighbour” rule. more than 10. We must first write a zero in front of the given number, or at least imagine a zero there. Multiplication by 11 Then we apply the idea of adding the neighbour to every figure of the given number in turn: Step I: The last digit of the multiplicand (number multiplied) is put down as the right-hand figure 05892  11 As there is no neighbour on the right, so of the answer. 2 Step II: Each successive digit of the multiplicand is we add nothing. added to its neighbour at the right. 05892  11 Ex. 1. Solve 5892 × 11 = ? -------- As we did earlier 4812 Soln: Step I: Put down the last figure of 5892 as the right-hand figure of the answer: 05892  11 64812 -------- Zero plus 5 plus carried-over 1 5892  11 2 is 6 Step II: Each successive figure of 5892 is added to its This example shows why we need the zero in front right-hand neighbour. 9 plus 2 is 11, put 1 below the line of the multiplicand. It is to remind us not to stop too and carry over 1. 8 plus 9 plus 1 is 18, put 8 below the soon. Without the zero in front, we might have neglected line and carry over 1. 5 plus 8 plus 1 is 14, put 4 below the last 6, and we might then have thought that the answer the line and carry over 1. was only 4812. The answer is longer than the given number by one digit, and the zero in front takes care of 5892  11 that. (9 + 2 = 11, put 1 below the line and 12 Sample Problems: Solve the following: carry over 1) 1) 111111 × 11 2) 23145 × 11 3) 89067 × 11 4) 5776800 × 11 5892  11 5) 1122332608 × 11 (8 + 9 + 1 = 18, put 8 below the line and 812 Ans: 1) 1222221 2) 254595 carry over 1) 3) 979737 4) 63544800 5) 12345658688 5892  11 (5 + 8 + 1 = 14, put 4 below the line and 4812 Multiplication by 12 carry over 1) To multiply any number by 12, we Step III: The first figure of 5892, 5 plus 1, becomes the “Double each digit in turn and add its neighbour.” left-hand figure of the answer: This is the same as multiplying by 11 except that 5892  11 now we double the “number” before we add its 64812 “neighbour.” 12 Quicker Maths For example: 09483 13 Ex 1: Solve:5324 × 12 Step IV. (9×3 + 4 + 2 = 33, write down 3 3279 Soln: 05324  12 and carry over 3) Step I. (double the right-hand figure 09483 13 8 Last Step. 123279 (0 × 3 + 9 + 3 = 12, write it and add zero, as there is no neighbour) down) 05324  12 The answer is 1,23,279. Step II. (double the 2 and add 4) In a similar way, we can define rules for multiplication 88 by 14, 15,....But, during these multiplications we will have 05324  12 Step III. (double the 3 and add 2) to get four or five times of a digit, which is sometimes 888 not so easy to carry over. We have an easier method of 05324  12 multiplication for those large values. Step IV. (double the 5 and add 3 (=13), 3888 Can you get similar methods for multiplication by 21 put 3 below the line and carry and 31? It is not very tough to define the rules. Try it. over 1) 05324  12 Multiplication by 9 Last Step. (zero doubled is zero, plus 5 63888 Step I: Subtract the right-hand figure of the long number plus carried-over 1) from 10. This gives the right-hand figure of the answer. The answer is 63,888. If you go through it yourself Step II: Taking the next digit from right, subtract it from you will find that the calculation goes very fast and is 9 and add the neighbour on its right. very easy. Step III: At the last step, when you are under the zero in Practice Question: Solve the following: front of the long number, subtract one from the neighbour 1) 35609 × 12 2) 11123009 × 12 and use that as the left-hand figure of the answer. 3) 456789 × 12 4) 22200007 × 12 Ex. 1: 8576 × 9 = ? 5) 444890711 × 12 Ans: 1) 427308 2) 133476108 08576  9 Soln: 3) 5481468 4) 266400084 77184 5) 5338688532 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 Multiplication by 13 2) and add the neighbour 6; the result is 8. To multiply any number by 13, we Step III. (9 – 5) + 7 = 11; put 1 under the line and carry “Treble each digit in turn and add its right over 1. neighbour.” Step IV. (9 – 8) + 5 + l (carried over) = 7, put it down. This is the same as multiplying by 12 except that Step V (Last step). We are under the left-hand zero, so now we “treble” the “number” before we add its we reduce the left-hand figure of 8576 by one, and 7 is “neighbour”. the left-hand figure of the answer. If we want to multiply 9483 by 13, we proceed like Thus answer is 77184. this: Here are a few questions for you. 09483 13 1) 34 × 9 = ? 2) 569 × 9 = ? Step I. (treble the right-hand figure and 9 3) 1328 × 9 = ? 4) 56493 × 9 = ? write it down as there is no 5) 89273258 × 9 = ? neighbour on the right) Answers: 09483 13 1) 306 2) 5121 Step II. 79 (8 × 3 + 3 = 27, write down 7 3)11952 4) 508437 and carry over 2) 5) 803459322 09483 13 We don’t suggest you to give much emphasis on this Step III. (4×3+8+2 = 22, write down 2 rule. Because it is not very much easy to use. Sometimes 279 it proves very lengthy also. and carry over 2) Multiplication 13 Another method: Step III. In the last step we multiply the left-hand Step I: Put a zero at the right end of the number; i.e., figures of both multiplicand and multiplier. write 85760 for 8576. 1 2 Step II: Subtract the original number from that number. Like 85760 – 8576 = 77184 1 3 1 5 6 ( 1 × 1) Multiplication by 25 (2) 17 × 18 = ? Suppose you are given a large number like Step I. 1 7 125690258. And someone asks you to multiply that 1 8 number by 25, What will you do? Probably you will do 6 (7 × 8 = 56, write down 6 carry nothing but go for simple multiplication. Now, we suggest over 5) you to multiply that number by 100 and then divide by 4. Step II. 1 7 To do so remember the two steps: 1 8 Step I: Put two zeroes at the right of the number (as it 0 6 (1 × 8 + 7×1+5 = 20, write has to be multiplied by 100). down 0 and carry over 2) Step II: Divide it by 4. Step III. 1 7 So, your answer is 12569025800 ÷ 4 = 3142256450. Is 1 8 it easier than your method? 3 0 6 (1 × 1 + 2 = 3, write it down) (3) 87 × 92 = ? General Rule for Multiplication Step I. 8 7 Having dealt in fairly sufficient detail with the 9 2 application of special cases of multiplication, we now 4 (7 × 2 = 14, write down 4 and proceed to deal with the “General Formula” applicable to carry over 1) all cases of multiplication. It is sometimes not Step II. 8 7 very convenient to keep all the above cases and their steps 9 2 in mind, so all of us should be very much familiar with 04 (8 × 2 + 9 × 7 + 1 = 80, write “General Formula” of multiplication. down 0 and carry over 8) Step III. 8 7 Multiplication by a two-digit number 9 2 Ex. 1: Solve (1) 12 × 13 = ? 8004 (8 × 9 + 8 = 80) (2) 17 × 18 = ? Practice questions: (3) 87 × 92 = ? 1) 57 × 43 2) 51 × 42 3) 38 × 43 Soln: (1) 12 × 13 = ? 4) 56 × 92 5) 81 × 19 6) 23 × 99 Step I. Multiply the right-hand digits of multiplicand and 7) 29 × 69 8) 62 × 71 9) 17 × 37 multiplier (unit-digit of multiplicand with unit-digit of the 10) 97 × 89 multiplier). Answers: 1 2 l) 2451 2) 2142 3) 1634 4) 5152 5) 1539 6) 2277 1 3 7) 2001 8) 4402 9) 629 6 (2 × 3) 10) 8633 Step II. Now, do cross-multiplication, i.e., multiply 3 Ex 2. Solve (1) 325 × 17 = ? (2) 4359 × 23 = ? by 1 and 1 by 2. Add the two products and write down Soln: (1): to the left of 6. Step I. 3 2 5 1 2 1 7 1 3 5 (5 × 7 = 35, put down 5 and carry 56 (3 × 1 + 2 × 1) over 3) 14 Quicker Maths Step II. 3 2 5 We can write all the steps together: 4 3 5 9 1 7 2 3 2 5 (2×7+5×1+3 = 22, put down 2 and carry over 2) 4 2 / 4 3  2 3/ 3 3  5 2 / 5 3  9 2 / 9 3 = 10 20 22 35 27 Step III. 3 2 5 = 100257 Or, we can write the answer directly without writing 1 7 the intermediate steps. The only thing we should keep in 5 2 5 (3 × 7 + 2 × 1 + 2 = 25, put down 5 mind is the “carrying numbers”. and carry over 2) 4359 4359 Note: Repeat the cross-multiplication until all the 23 or, 23 consecutive pairs of digits exhaust. In step II, we cross-multiplied 25 and 17 and in step III, we cross- 10 2 0 2 2 35 2 7 100257 multiplied 32 and 17. Note: You should try for this direct calculation. It saves a lot of time. It is a very systematic calculation and is Step IV. 3 2 5 very easy to remember. Watch the above steps again and again until you get that systematic pattern of cross- 1 7 multiplication. 5 5 2 5 (3 × 1 + 2 = 5, Put it down) Multiplication by a three-digit number (2) Step I. Ex: 1. Solve (1) 321 × 132 = ? (2) 4562 × 345 = ? 4 3 5 9 (3) 69712 × 641 = ? Soln: 2 3 (1) Step I. 3 2 1 7 (9 × 3 = 27, put down 7, carry over 2) Step II. 4 3 5 9 1 3 2 2 3 2 (1 × 2 = 2) 5 7 (5 × 3 + 9 × 2 + 2 = 35, put down 5 and carry over 3) Step II. 3 2 1 Step III. 4 3 5 9 2 3 1 3 2 2 5 7 (3×3+5×2+3 = 22, put down 7 2 (2 × 2 + 3 × 1= 7) 2, carry over 2) Step IV. 4 3 5 9 Step III. 3 2 1 2 3 1 3 2 0 2 5 7 (4×3+3×2+2 = 20, put down 0, carry over 2) 3 7 2 (2 × 3 + 3 × 2 + 1 × 1 = 13, Step V. 4 3 5 9 write down 3 and carry over 1) 3 2 1 Step IV. 2 3 100257 (4 × 2 + 2 = 10, put it down) 1 3 2 2 3 7 2 Multiplication 15 Step V. 3 2 1 V. 6 9 7 1 2 VI. 6 9 7 1 2 VII. 6 9 7 1 2 641 641 1 3 2 641 4 2 3 7 2 (1 × 3 + 1 = 4) For each of the groups of figures, you have to cross- or, multiply. 3 21 Multiplication by a four-digit number 13 2 Example: Solve 1) 4325 × 3216 2) 646329 × 8124 1 3 / 3  3  1 2 / 2  3  3  2  11/ 2  2  3  1/1 2 Soln: 1) 4325 × 3216 = ? Step I. = 4 12 13 7 2 = 42372 4 3 2 5 Soln: (2): 4562 3 2 1 6 345 0 (5 × 6 = 30, write down 0 and carry over 3) 4  3/ 4  4  3 5 / 5 4  4  5  3 6 / 5 5 Step II.  4 6  3 2 / 5 6  4  2 / 2 5 4 3 2 5 = 15 37 6 3 5 8 3 9 1 0 = 1573890 3 2 1 6 Soln: (3): 0 0 (2×6+1×5+3 = 20, write down 0 and carry over 2) 6 9 712 6 41 Step III. 4 3 2 5 6  6 / 4  6  6  9 / 1 6  4  9  6  7 /1 9  4  7  6 1/ 1 7  4  1  6  2 /11  4  2 /1 2 3 2 1 6 = 44 86 88 45 23 9 2 2 0 0 = 44685392 Note: Did you get the clear concept of cross- Step IV. 4 3 2 5 multiplication and carrying-cross-multiplication? Did you mark how the digits in cross-multiplication increase, remain constant, and then decrease? Take a sharp look at 3 2 1 6 question (3). In the first row of the answer, if you move 9 2 0 0 from right to left, you will see that there is only one multiplication (1×2) in the first part. In the second part Step V. 4 3 2 5 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 3 2 1 6 part two (4 × 6 and 6 × 9) and in the last part only one (6 0 9 2 0 0 × 6) multiplication. The participation of digits in cross- Step VI. multiplication can be shown by the following diagrams. 4 3 2 5 I. 6 9 7 1 2 II. 6 9 7 1 2 III. 6 9 7 1 2 IV. 6 9 7 1 2 3 2 1 6 641 641 641 9 0 9 2 0 0 (4× 2 + 3 × 3 + 2 = 19, write down 9 and carry over 1) 641 16 Quicker Maths Step VII. Practice Problems: Solve the following: 1) 234 × 456 2) 336 × 678 4 3 2 5 3) 872 × 431 4) 2345 × 67 5) 345672 × 456 6) 569 × 952 7) 458 × 908 8) 541 × 342 3 2 1 6 9) 666 × 444 10) 8103 × 450 1390 9 2 0 0 (4 × 3 + 1 = 13, write it down) 11) 56321 × 672 12) 1278 × 569 Ans: 13909200 13) 5745 × 562 14) 4465 × 887 Soln: 1) 646329 × 8124 = ? 15) 8862 × 341 Try this question yourself and match your steps with Answers: the given diagrammatic presentation of participating digits. 1) 106704 2) 227808 3) 375832 4) 157115 Step I. 6 4 6 3 2 9 5) 157626432 6) 541688 7) 415864 8) 185022 8124 9) 295704 10) 3646350 Step II. 6 4 6 3 2 9 11) 37847712 12) 727182 13) 3228690 14) 3960455 8124 15) 3021942 Step III. 6 4 6 3 2 9 Checking of Multiplication Ex 1: 15 × 13 = 195 digit-sum: 6 × 4 = 6 8124 or, 24 = 6 Step IV. 6 4 6 3 2 9 or, 6 = 6. Thus, our calculation is correct. 8124 Ex. 2: 69712 × 641 = 44685392 Step V. 6 4 6 3 2 9 digit-sum: 7 × 2 = 5 or, 14 = 5 or, 5 =5 8124 Therefore, our calculation is correct. Step VI. 6 4 6 3 2 9 Ex. 3: 321 × 132 = 42372 digit-sum: 6 × 6 = 9 8124 or, 36 = 9 Step VII. 6 4 6 3 2 9 or, 9 =9 Thus, our calculation is correct. But if someone gets the answer 43227, and tries 8124 to check his calculation with the help of digit-sum rule, Step VIII. 6 4 6 3 2 9 see what happens: 321 × 132 = 43227 8124 digit sum: 6×6 =9 Step IX. 6 4 6 3 2 9 or, 36 = 9 or, 9 =9 This shows that our answer is correct, but it is 8124 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 3 division which is based on long-established Vedic process 7 38 39 38 2 of mathematical calculations. It is capable of immediate 5 3 application to all cases and it can be described as the Step V. Now our dividend is 38. From this we subtract “crowning gem of all” for the universality of its the product of the index (3) and the 2nd quotient- applications. digit (3), ie, 3 × 3 = 9. The remainder 29 is our next To understand the at-sight mental one-line method of actual dividend and divide that by 7. We get 4 as the division, we should take an example and its explanation. quotient and 1 as the remainder. We put them at their respective places. DIVISION BY A 2-DIGIT NUMBER Ex 1. Divide 38982 by 73. 3 7 38 3 9 38 12 Soln: 5 3 4 Step I. Out of the divisor 73, we put down only the first digit, ie, 7 in the divisor-column and put the other Step VI. Our next dividend is 12 from which, as before, digit, ie, 3 “on top of the flag”, as shown in the we deduct 3 × 4 ie, 12 and obtain 0 an the remainder chart below. 3 7 38 39 38 12 7 3 38 9 8 2 The entire division will be by 7. 5 3 4 0 Step II. As one digit (3) has been put on top, we allot one Thus we say : Quotient (Q) is 534 and Remainder place at the right end of the dividend to the remainder (R) is 0. position of the answer and mark it off from the And thus finishes the whole procedure; and all of it is digits by a vertical line. one-line mental arithmetic in which all the actual division is done by the single-digit divisor 7. 3 7 38 9 8 2 The procedure is very simple and needs no further exposition and explanation. A few more illustrations with Step III. As the first digit from the left of dividend (3) is running comments will be found useful and helpful and less than 7, we take 38 as our first dividend. When are therefore given below : we divide 38 by 7, we get 5 as the quotient and 3 as Ex 2: Divide 16384 by 128 (As 12 is a small number to the remainder. We put 5 down as the first quotient- handle with, we can treat 128 as a two-digit digit and just prefix the remainder 3 before the 9 of number). the dividend. 8 12 16 43 118 64 3 Soln: 7 38 3 9 8 2 1 2 8 0 5 Step I. We divide 16 by 12. Q = 1 & R = 4. Step IV. Now our dividend is 39. From this we, however, Step II. 43 – 8 × 1 = 35 is our next devidend. Dividing it deduct the product of the indexed 3 and the first by 12, quotient-digit (5), ie, 3×5 = 15. The remainder 24 is Q = 2, R = 11. our actual net-dividend. It is then divided by 7 and Step III. 118 – 8 × 2 = 102 is our next dividend. Dividing gives us 3 as the second quotient-digit and 3 as the it by 12, remainder, to be placed in their respective places as Q = 8, R = 6 was done in third step. Step IV. 64 – 8 × 8 = 0 18 Quicker Maths Then our final quotient = 128 & remainder = 0 The vertical line separating the remainder from Ex 3: Divide 601325 by 76. quotient part may be demarcation point for decimal. 6 Therefore, ans = 18.2054 7 60 111 63 22 25 3 Soln : 2 71 1 7 9 1 2 13 (=25 - 6×2) (ii) 08. 15 18 09 Step I. Here, in the first division by 7, if we put 8 down 3 1 2 4 3 0 as the first quotient-digit, the remainder then left Ans = 31.2430 will be too small for the subtraction expected at the next step. We get -ve dividend in next step which is (iii) absurd. So, we take 7 as the quotient-digit and prefix 7 4 8 5 3 2 the remainder 11 to the next dividend-digit. Ans = 7.4853 All the other steps are similar to the previously Ex 7 : Divide 7.3 by 53 mentioned steps in Ex 1 & 2. Our final quotient is 7912 and remainder is 13. 5 3 7. 23 50 6 0 40 40 If we want the values in decimal, we go on dividing Soln. as per rule instead of writing down the remainder. 1 3 7 7 3 5 Such as; Ans = 0.137735 76 60 1 3 2 5 60 50 DIVISION BY A 3-DIGIT NUMBER 7 9 1 2 1 7 1 Ex 8 : Divide 7031985 by 823. Ans = 7912.171 Soln: Note: The vertical line separating the remainder from the Step I. Here the divisor is of 3 digits. All the difference quotient part may be the demarcating point for which we make is to put the last two digits(23) of decimal. divisor on top. As there are two flag-digits (23), we Ex 4: Divide 7777777 by 38 will separate two digits (85) for remainder. Soln: 23 8 70 3 1 9 85 38 7 17 17 57 77 87 77 Step II. We divide 70 by 8 and put down 8 and 6 in their 2 0 4 6 7 8 13 (=77 - 8×8) proper places. Q = 204678, R = 13 23 8 70 63 1 9 85 You must go through all the steps of the above solution. Try to solve it. Did you find some 8 difference? Step III. Now, our gross dividend is 63. From that we Ex 5: Divide 8997654 by 99. Try it step by step. subtract 16, the product of the tens of the flag- Ex 6:(i) Divide 710.014 by 39 (to 4 places of decimals) digits, ie 2, and the first quotient-digit, ie 8, and get (ii) 718.589 ÷ 23 = ? the remainder 63 – 16 = 47 as the actual dividend. (iii) 718.589 ÷ 96 = ? And, dividing it by 8, we have 5 and 7 as Q & R Soln. (i) Since there is one flag-digit the vertical line is respectively and put them at their proper places. drawn such that one digit before the decimal comes under remainder portion. 8 23 70 63 71 9 85 9 8 5 3 7 41 80. 20 21 64 70 Step IV. Now our gross dividend is 71, and we deduct 1 8 2 0 5 4 the cross-products of two flag-digits 23 and the For the last section, we had 64 - 45 = 19 as our two quotient digits (8 & 5) ie 2 × 5 + 3 × 8 = 10 dividend, divided by 3 we choose 4 as our suitable + 24 = 34; and our remainder is 71 – 34 = 37. We quotient. If we take 5 as a quotient it leaves 4 as remainder then continue to divide 37 by 8. We get Q = 4 & R (19 – 15). Now the next dividend will be 40 – 9 × 5 = –5, =5 which is not acceptable. Division 19 Q = 1432, Remainder = 521 – 10 (Cross-multiplication 8 23 70 63 71 59 85 of 43 & 32) – 3×2 = 521 – 10 × 17 – 6 = 345 8 5 4 For decimals : Step V. Now our gross dividend is 59. And actual dividend 7 43 10 3 6 44 43 52 71 70 60 50 is equal to 59 minus cross-product of 23 and 54, ie, 1 4 3 2 4 6 4 3 59 – (2 × 4 + 3 × 5) = 59 – 23 = 36. Dividing 36 by 8, our Q = 4 and R = 4. Ans = 1432.4643 Ex 10: Divide 888 by 672 ( to 4 places of decimals). 8 23 70 63 71 59 485 72 6 8 2 8 38 30 40 5 0 8 5 4 4 1 3 2 1 4 2 Step VI. Actual dividend = 48 – (3 × 4 + 2 × 4) = 48 – 20 = 28. Ans = 1.3214 Dividing it by 8, our Q = 3 & R = 4 Note : Vertical line separating the remainder from