ICAI Foundation Paper 3 Quantitative Aptitude PDF

SARANSH Last Mile Referencer for The Institute of Chartered Accountants of India Board of Studies (Set up by an Act of Parliament) FOUNDATION PAPER 3 QUANTITATIVE...

SARANSH Last Mile Referencer for The Institute of Chartered Accountants of India Board of Studies (Set up by an Act of Parliament) FOUNDATION PAPER 3 QUANTITATIVE APTITUDE www.icai.org I https://boslive.icai.org  Saransh - QUANTITATIVE APTITUDE While due care has been taken in preparing this booklet, if any errors or omissions are noticed, the same may be brought to the notice of the HoD, BoS. The Council of the Institute is not responsible in any way for the correctness or otherwise of the matter published herein. © 2024. The Institute of Chartered Accountants of India (Also referred to as ICAI) All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form, or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior permission, in writing, from the publisher. Published in May 2024 by: Board of Studies The Institute of Chartered Accountants of India ‘‘ICAI Bhawan” A-29, Sector-62, Noida 201 309 ©ICAI B O S  The Board of Studies (BoS) is dedicated to delivering outstanding services to its PREFACE students, tirelessly striving for the highest standards of education and support. It imparts quality academic education through its value added study materials, that explain concepts clearly and in simple language. Illustrations and Test Your Knowledge Questions contained therein facilitate enhanced understanding and application of concepts learnt. Revision Test Papers provides Questions & Answers to help students to update themselves to revise the concepts by solving questions contained therein. Suggested Answers containing the ideal manner of answering questions set at examination which also helps students to revise for the forthcoming examination. Mock Test Papers empower students to gauge their readiness ahead of each examination, ensuring confidence and clarity. Additionally, BoS offers engaging live virtual classes led by distinguished faculty, reaching students far and wide across the nation. To effectively engage with its students, the Board of Studies (BoS) has been publishing subject specific capsules in its monthly Students’ Journal, "The Chartered Accountant Student," since 2017. These capsules are aimed at facilitating efficient revision of concepts covered across various topics at the Foundation, Intermediate, and Final levels of the Chartered Accountancy Course. This initiative underscores the BoS's commitment to enhancing learning and comprehension among its students through accessible and attractive educational materials. Each issue of the journal includes a capsule relating to specific topic(s) in one subject at each of the three levels. In these capsules, the concepts and provisions are presented in attractive colours in the form of tables, diagrams and flow charts for facilitating easy retention and quick revision of topics. The Board of Studies (BoS) is now introducing a comprehensive booklet titled 'Saransh - Last Mile Referencer for Foundation Paper 3–Quantitative Aptitude'. This booklet aims to consolidate significant concepts across various topics in Business mathematics, Logical Reasoning and Statistics, particularly those included in the syllabi at the Foundation level of CA Course. It features diagrams, flow charts, tables, and illustrated journal entries, providing a one-stop repository for key Quantitative Aptitude topics. However, the readers are advised to refer the study material for comprehensive study and revision. Under no circumstances, this booklet substitutes the detailed study of the material provided by the Board of Studies. Further, the readers are advised to enhance their ability to address the issues and solve the problems based on fundamentals of Accounting, illustrations and questions given in the study material, revision test papers and mock test papers. By capturing essential points in a concise format, 'Saransh' will facilitate better understanding and retention of critical accounting principles, enhancing the learning experience for students preparing for their examinations and beyond. It will indeed serve as a valuable ready reckoner for readers, enabling them to grasp the essence of the subject comprehensively. Happy Reading! ©I CAI BOS S ARANSH President Message It is with immense pride that I introduce the Saransh booklets, a meticulously curated resource available across the Foundation, Intermediate, and Final levels of the Chartered Accountancy course. ICAI has always been dedicated to providing our students with the best possible resources to succeed in their studies and careers, and Saransh is a demonstration of this commitment. The Saransh — Last Mile Referencers have been thoughtfully designed by the Board of Studies (BoS) to serve as an invaluable companion for your studies and exam preparation. Our aim is to simplify complex concepts and provisions, making them easier to understand, memorize, and revise. However, Saransh is not a substitute for the detailed BoS study material but a supplementary tool to complement your in-depth study. The newly revamped Saransh booklets have been updated not only in content but also in their presentation. With a more logical and organized structure, enhanced visual appeal, and a userfriendly layout, these booklets are now more effective in aiding your studies. We have extended the Saransh series to cover all core areas of the Chartered Accountancy course. Whether you are studying Direct Tax Laws and International Taxation, Indirect Tax Laws, Accounting Standards, Indian Accounting Standards, Auditing, Cost and Management Accounting, Strategic Cost Management and Performance Evaluation, Company Law, or Financial Management and Strategic Management, you will find a Saransh booklet for each subject. Saransh is designed not only to help you grasp and recall essential concepts but also to guide you in approaching each subject strategically. The insights provided in these booklets will help you develop a structured approach to your studies, ensuring that you are well-prepared for your examinations. I urge you to make the most of the Saransh booklets. While these booklets will support you, it is your dedication, perseverance, and hard work that will ultimately determine your success. I wish each of you the very best in your studies and future careers. Warm regards, CA. Ranjeet Kumar Agarwal President, ICAI ©ICAI B O S  The Institute of Chartered Accountants of India (Set up by an Act of Parliament) Board of Studies Saransh Foundation Paper 3: Quantitative Aptitude INDEX Chapters Pg No. Part A: Business Mathematics Chapter 1 1 Unit 1: Ratio 1 Unit 2: Proportions 2 Unit 3: Indices 12 Unit 4: Logarithms 16 Chapter 2 - Equations 30 Chapter 3 - Linear Inequalities 43 Chapter 4 - Mathematics of Finance 54 Chapter 5 - Permutations and Combinations 63 Chapter 6 - Sequence and Series, Arithmetic and Geometric Progressions 78 Chapter 7 - Sets, Relations and Functions 80 Limits and Continuity functions 91 Chapter 8 - Application of Differential and Integral Calculus 95 Part B: Logical Reasoning Chapter 9 - Number Series, Coding, Decoding and odd man out 102 Chapter 10 - Direction Tests 106 ©I CAI BOS S ARANSH The Institute of Chartered Accountants of India (Set up by an Act of Parliament) Board of Studies Saransh Foundation Paper 3: Quantitative Aptitude INDEX Chapters Pg No. Chapter 11 - Seating arrangements 109 Chapter 12 - Blood Relations 117 Part C: Statistics Chapter 13 122 Unit 1 Statistical Description of Data 122 Unit 2 Sampling 142 Chapter 14 151 Unit 1 Measures of Central Tendency 151 Unit 2 Measures of Dispersion 166 Chapter 15 - Probability 184 Chapter 16 - Theoretical Distributions 192 Chapter 17 - Correlation and Regression 203 Chapter 18 - Index Numbers 220 ©ICAI B O S  SA R AN S H The subject “Quantitative Aptitude” has been designed specifically for the Before We Begin students who are aiming to pursue CA course, keeping in view the relevance of subjects after they become full-fledged professional. Mathematics and Statistics applications are very important for the students of Chartered Accountancy Course. As professional work in future will demand quantitative and analytical skills. Logical Reasoning has been included in the syllabus to test analytical and mental ability skills which will help them in honing their interpretative skills while pursuing and thereafter CA course. Students will be equipped with the knowledge to absorb various concepts of other subjects of the chartered accountancy course like accounting, auditing and assurance, financial management, cost and management accounting, strategic cost management, etc. Understand the Syllabus of Quantitative Aptitude Quantitative Aptitude (100 Marks) Business Mathematics Logical Reasoning Statistics Business Mathematics (40 Marks): Mathematics of Finance Ratio, Propotions, Idices and Logarithms Equations Airthmetic and Geometric Progressions Sets, Relations, Funtions, Basics of Limts and Continuity Linear Inequalities Permutations and Combinations Applications of Differential and Integral Calculus ©I CAI BOS S ARANSH Logical Reasoning (20 Marks): Before We Begin Number Series, Coding, Decoding and Odd man out Direction Tests Seating Arrangements Blood Relations Statistics (40 Marks): Statistical Description of Data and Sampling Measures of Central Tendency and Dispersion Correlation and Regression Index Numbers Probability Theoretical Distributions The Saransh Quantitative Aptitude has been developed in a simple and easy to understand language. Every concept has been explained with the help of solved examples. A number of illustrations have been incorporated in each chapter to explain various concepts and related computational techniques. The diagrams have been drawn neatly in a such way that the students have the complete understanding of the problem by perusing them. The Quantitative Aptitude paper is tested on multiple choice questions or objective type of questions pattern only. Quantitative Aptitude is a crucial and scoring paper in the CA Foundation course. To excel in this paper, it's essential to have a clear understanding of the concepts, regular practice, and a disciplined study routine. By following tips such as focusing on conceptual clarity, practicing a variety of problems, managing time effectively during the exam, and staying consistent with preparation by focusing on these strategies student will crack the CA Foundation Paper 3 Quantitative Aptitude. Happy Reading and Best Wishes! ©ICAI B O S  SA R AN S H Chapter -1: Ratio and Portion, Indices and Logarithms Unit 1: Ratio a If a and b are two non-zero number of same kind then fraction is called RATIO of a to b. It is denoted by a : b. b Here a is called ANTECEDENT and b is called CONSEQUENT. For example, Ratio of two numbers 4 and 6 is 2 : 3. (i) the ratio between 5 ,and 6 is 5 : 6 (ii) the ratio between 150gm and 2 kg. is 3 : 40 (iii) the ratio between 25 minutes and 45 Seconds is 100 : 3 Inverse Ratio If a:b then b:a is called inverse ratio of a:b, Example : The inverse ratio of 11: 15 is 15: 11 Duplicate Ratio: If a : b is a given ratio then Duplicate Ratio is a2:b2 , Example : The duplicate ratio of 2: 3 4: 9 Sub Duplicate Ratio If a : b is given ratio then a : b is called Sub Duplicate Ratio, Example : The Sub Duplicate Ratio of 25 : 16 is 5 : 4 Triplicate Ratio If a : b is given ratio then a3 :b3 is called Triplicate Ratio, Example : Triplicate Ratio of 3 : 2 is 27 : 8 Sub-Triplicate Ratio 3 3 If a : b is a given ratio then a : b is called Sub Triplicate Ratio of a : b Example: Triplicate Ratio of 125 : 27 is 5 : 3 Compounded Ratio The Ratio Compound of a : b and c : d is ac : bd. Compounded Ratio of (a) 5 : 6 and 2 : 3 is 5 : 9 (b) 2 : 5, 5 : 7, and 6 : is 3 : 14 Question 1: Ratio of 5 kg and 15 kg is 5kg 1 Solution Ratio = = = 1:3 15kg 3 They have same units, so it is a ratio. ©I CAI BOS 1 S ARANSH Ratio and Portion, Indices and Logarithms Question 2 Ratio of 5m and 200 cm; is 5×100cm 5 Solution Since 1 metre = 100 cm, Ratio of 5 m & 200 cm = = = 5:2 200cm 2 Question 3 Ratio of the height 170 cm and weight 60 kg of a person is Cm is unit of length and kg is unit of mass. So, they cannot be converted in same units. Solution 170 cm : 60 kg is not a ratio. 1 1 Question 4 Which ratio out of the following ratios is greater. 2 :3 and 3.6:4.8 3 3 3.6 Solution 7:10 = 0.70 = 0.75 Clearly 3.6 : 4.8 is greater ratio. 4.8 Unit II Proportion AN EQUALITY OF TWO RATIO IS CALLED A PROPORTION If a, b, c, d are four numbers then a : b = c : d [Also written as a : b : : c : d] is called Proportion of four numbers. a:b=c:d a c ∴ = b d ∴ ad = bc [Product of 1st and 4th] = [Product of 2nd and 3rd] Note : If a : b = c : d then ‘d’ is called ‘Fourth Proportion’ 2 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Mean Proportion : If a, b, c are in Continuous Proportion Then Middle Term b is called The Mean Proportion. b2 = ac then b = ac Mean Proportion Note : Mean Proportion of two number x and y is xy Types Of Proportion 1. INVERTENDO PROPORTION If a : b=c : d then b : a=d : c is called INVERTENDO PROPORTION 2. ALTERNENDO PROPORTION If a : b=c : d then a : c=b : d is called ALTERNENDO PROPORTION 3. COMPONENDO PROPORTION If a : b=c : d then (a+b): b=(c+d):d is called COMPOUND PROPORTION 4. DIVIDENDO PROPORTION If a : b=c : d then (a-b): b=(c-d): d is called DIVIDENDO PROPORTION 5. COMPOUNDED & DIVIDENDO PROPORTION If a : b=c : d then (a+b):(a-b)=(c+d):(c-d) is CALLED COMPOUNDED and DIVIDENDO PROPORTION 6. ADDENDO PROPORTION If a : b = c : d=e : f ………. then (a+c+e ……..) : (b+d+f+ ……..) is called ADDENDO PROPORTION Continuous Proportion a c If a, b, c are Continuous Proportion then = or b2 = ac b d i.e. Ratio of 1st of 2nd = Ratio of 2nd to 3rd i.e. Square of middle term = Product of 1st and last Note : If x, y, z a, b, c............. are Continuous Proportion then x y z a b = = = = = ……………………………. y z a b c Question 5 If A: B= 2 : 3 ; B : C = 4 : 5 and C : D = 6 : 7 Then A : D is equal to Solution A B C 2 4 6 x x = x x A : D = 16 : 35 B C D 3 5 7 Two vessels contain equal quantity of mixtures of milk and water in the ratio 9 : 5 and 4 : 3 Question 6 respectively. Both the mixtures are mixed find the ratio of milk to water in the new mixture. Tricks Milk : Water sum Vessel I 9 : 5 ] 14 × 1 = 14 Vessel II 4 : 3 ] 7 ×2 = 14 Vessel III (9 + 4×2) : 5 + 3x2] (Adding corresponding terms of the ratio) Take LCM of 14 & 7 = 14. Then make it of equal quantity (i.e. = 14) ∴ Ratio of Milk to Water in the new vessel = 17 : 11. ©I CAI BOS 3 S ARANSH Ratio and Portion, Indices and Logarithms Two vessels containing water and milk in the ratio 2 : 3 and 4 : 5 are mixed in the ratio 1 :2. Question 7 The ratio of milk and water in the resulting mixture. Tricks Step 1. Milk : Water sum Step 2. Vessel I 3:2]5×9=45 ∴Vessel I 27: 18] 45x1 Vessel II 5 : 4 ] 9 x 5 = 45 Vessel II 25 : 20 ] 45 x 2 Step 3. Mixing Vessel I & Vessel II mixtures, we get Vessel III (27 + 50) : (18+40) = 58 : 77 Take LCM of 5 & 9 = 45. Making same quantities of Vessel I & II i.e. 45. We have to mix the mixtures of vessel I & II in the ratio 1 : 2 ∴ Multiply I by 1 and II by 2 ,Then we add. We get the answer. Rules of Alligation Sugar at rate Rs. 15 per kg is mixed with sugar at rate Rs. 20 per kg in the ratio 2:3. Find the Question 8 price per kg of the mixture. Solution : Tricks 20 - x 2 = x = Rs. 18. x - 15 3 Varun mixes milk costing Rs.6.92 per litre with milk costing Rs. 7.77 per litre and sells the Question 9 mixture at Rs. 8.80 per litre and earns a profit of 17.5% on his sales price. In what proportion does he mix them? Solution Selling price per litre = Rs. 8.80 Less profit 17.5% of 8.80 = Rs. 1.54 Cost per litre Rs. 7.26 = 3:2 Two numbers are in the ratio 2 : 3 and the difference of their squares is 320. The number Question 10 are : (a) 12,18 (b) 16,24 (c) 14,21 (d) None Solution : Tricks Go by choices (a); (b) & (c) all are in ratio 2:3 But For option (a) 182 -122 ≠ 320 (b) 242 -162 =( 24 + 16) (24 - 16) = 40 x 8 = 320 4 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 11 If p : q is the sub-duplicate ratio of p - x2: q - x2, then x2 is : p-x2 p Solution Detail Method = - = q-x2 q p-x2 p2 Squaring on both side; we get : = q-x2 q2 or pq2-q2 x2 = p2 q-p2 x2 or p2 x2-q2 x2 = p2 q-pq2 or x2 (p2-q2) = Pq(P-q) or x2 (p+q)(p-q〉=pq(p - q) pq or x2 = p+q Question 12 An alloy is to contain copper and zinc in the ratio 9:4. The zinc required to melt with 24 kg of copper is: Solution Let Zinc = x kg ∴ 9 = 24 ∴ x= 4×24 = 32 = 10 2 kg 4 x 9 3 3 Question 13 Two numbers are in the ratio 7: 8. If 3 is added to each of them, their ratio becomes 8 : 9. The numbers are: Solution Let x is common in the ratio ∴ Numbers are 7x & 8x Now 7x+3 = 8 8x+3 9 or 64x + 24 = 63x + 27 or 64x - 63x = 27-24 or x = 3 1st number = 7x = 7×3 = 21 2nd number = 8x = 8x3 = 24 Question 14 A box contains Rs. 56 in the form of coins of one rupee, 50 paise and 25 paise. The number of 50 paise coin is double the number of 25 paise coins and four times the numbers of one rupee coins. The numbers of 50 paise coins in the box is : Solution Let No. of 50 Paise coins = x ∴ No. of Re. 1 coins = x and No. of 25 Paise coins = x 4 2 x x ∴ Total Value = x 1 + x 0.50 + x 0.25 = 56 4 2 or 0.25x + 0.50x + 0.125x or 0.875x = 56 56 or x = 0.875 = 64 ©I CAI BOS 5 S ARANSH Ratio and Portion, Indices and Logarithms Question 15 A bag contains Rs. 187 in the form ofl rupee, 50 paise and 10 paise coins in the ratio 3:4:5. Find the number of each type of coins: Solution Let x is common in the ratio No. of 1 Rupee ; 50 Paise and 10 Paise Coins are 3x ; 4x and 5x 3x×1+4x×0.50+5x×0.10=187 187 x= = 34 5.50 ∴ No. of 1 Rupee coins = 3x = 3 × 34 = 102 No. of 50 Paise coins = 4x= 4x34 = 136 No. of 10 Paise coins = 5x = 5x34 = 170 Question 16 Ratio of earnings of A and B is 4 : 7. If the earnings of A increase by 50% and those of B decrease by 25%, the new ratio of their earning becomes 8 : 7. What is A’s earning ? Solution Let x is common in the ratio ∴ A’s and B’s present earnings are 4x and 7x respectively From question 4x + 4x x 0.50 8 x= = 7x - 7x x 0.25 7 6 8 OR; = 5.25 7 x cannot be found. ∴ Data is inadequate Question 17 P, Q and R are three cities. The ratio of average temperature between P and Q is 11 : 12 and that between P and R is 9: 8. The ratio between the average temperature of Q and R is : Solution P : Q = 11 : 12 Q : P = 12 : 11 Q P 12 9 x = x P R 11 8 Q 27 = 1 Q : R = 27.22 R 221 Question 20 Rs. 407 are to be divided among A, B and C so that their shares are in the ratio 1/4:1/5:1/6 The respective shares of A, B, C are: Solution 1 1 1 A:B:C= ; ; ] x LCM of denominator = 60 = 15 : 12 :10 4 5 6 407 A’s share = 15+12+10 = Rs. 165 407 B’s share = x 12 = Rs. 132 37 407 C’s share = x 10 = Rs,110 37 6 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 21 The incomes of A and B are in the ratio 3 :2 and their expenditures in the ratio 5 : 3. If each saves Rs. 1,500, then B’s income is: Solution Detail Method Let x is common in the ratio. ∴ A’s income = 3x B’s income = 2x 3x - 1500 5 2x - 1500 = 3 or 10x - 7500 = 9x - 4500 or 10x - 9x = 7500 - 4500 or x = 3000 :- B’s income = 2x = 2x3000 = Rs 6000. Question 22 In 40 litres mixture of glycerine and water, the ratio of glycerine and water is 3:1. The quantity of water added in the mixture in order to make this ratio 2:1 is: 40 Solution Glycerine = x 3 = 30 litres. 3+1 40 Water = x 1 = 10 litres. 4 Let x liters of water is added to the mixture 30 2 Then = or 2x + 20 = 30 or x = 5 10 + x 1 Question 23 The third proportional between (a2-b2) and (a+b)2 is : (Meanprop.)2 Solution 3rd Proportion = IstProp {(a+b)2}2 (a+b)4 (a+b)3 = = = a -b 2 2 3(a+b)(a+b) a-b Question 24 If A, B and C started a business by investing Rs. 1,26,000, Rs. 84,000 and Rs. 2,10,000. If at the end of the year profit is Rs. 2,42,000 then the share of each is : Solution Investment ratio is A : B : C = 126,000 : 84,000 : 2,10,000 ] ÷14,000 = 9:6:15] +3 = 3: 2 : 5 Rs242,000 A’s share = x 3=Rs72,600 3+2+5 242,000 B’s share = ×2 = RS. 48,400 20 242,000 C’s share = x 5 = Rs. 1,21,000 10 ©I CAI BOS 7 S ARANSH Ratio and Portion, Indices and Logarithms p 2 2p+q Question 24 If = - then the value of is: q 3 2p-q Solution p -2 ∵ = 2p+q 2(-2)+3 -4+3 -1 1 q 3 Tricks = = = = 20-q 2(-2)-3 -4-3 -7 7 Question 26 Fourth proportional to x, 2x, (x +1) is: x+1 Solution Let Fourth Proportional is K. ∴ x = 2x K or k.x = 2x (x + 1) or k = 2 (x + 1) = 2x +2 Question 27 What must be added to each term of the ratio 49 : 68 so that it becomes 3.: 4 ? 49+x 3 Solution Let Fourth Proportional is K. Then = = 68+8 4 or 196 +4x = 204 + 3x or 4x - 3x = 204 - 196 or x = 8 Question 28 If A : B= 2:5, then (10A + 3B): (5A+2B) is equal to Solution It A : B = 2 : 5 Then 10A+3B 10×2+3×5 35 7 = = = = 7:4 5A+2B 5×2+2×5 20 4 Question 29 In a film shooting, A and B received money in a certain ratio and B and C also received the money in the same ratio. If A gets Rs. 1,60,000 and C gets Rs.2,50,000. Find the amount received by B ? Solution A:B=B:C So, B2 = AC ; So, B = AC = 1,60,000×2,50,000 = 400 × 500 = 2,00,000 Question 30 The ratio compounded of 4:5 and sub-duplicate of “a” : 9 is 8:15. Then value of “a” is Solution 4 x a = 8 4 a = 8 a = 2 ⇒ a=4 5 9 15 or 5 x 9 15 8 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 31 The ratio compounded of 4:5 and sub-duplicate of “a” : 9 is 8:15. Then value of “a” is 4 x a= 8 Solution 5 9 15 4 a 8 or 5 x 3 = 15 ∴ a = 2 ⇒ a=4 Question 32 If X varies inversely as square of Y and given that Y =2 for X = 1, then the value of X for Y =6 will be 1 1 k Solution x= =X=K 2 ⇒X= 2 y2 y y When x = 1 Then y = 2 4 x= y2 4 1 When y = 6 Then x = = 62 9 1 ∴x= 9 Question 33 Triplicate ratio of 4 : 5 is (a) 125 : 64 (b) 16 : 25 (c) 64 : 125 (d) 120 : 46 Solution (c) Triplicate ratio of 4:5 = 43 :53 = 54:125 Question 34 The mean proportion between 24 and 54 is _______ (a) 33 (b) 34 (c) 35 (d) 36 Solution (d) Mean - Proportion = 24 X 54 = 36 Question 35 If P is 25% less than Q and R is 20% higher than Q the Ratio of R and P (a) 5:8 (b) 8:5 (c) 5:3 (d) 3:5 Solution (b) is correct Let Q = 100, So, P = 100-025 = 75 &R= 100 + 20= 120 R 120 8 = = P 75 5 ©I CAI BOS 9 S ARANSH Ratio and Portion, Indices and Logarithms Question 36 A person has assets worth Rs. 1,48,200. He wish to divide it amongst his wife, son and daughter in the ratio 3 : 2 : 1 respectively. From this assets the share of his son will be 2 Solution Share of son = X 1,48,200 3+2+1 = Rs.49,400 Question 37 If x : y = 2 : 3 then (5x+2y): (3x -y) = Solution 5x + 2y 5 x 2 + 2 x 3 16 = = 3y - y 3x2-3 3 Question 38 The first, second and third month salaries of a person are in the ratio 2:4:5. The difference between the product of the salaries of first 2 months & last 2 months is Rs. 4,80,00,000. Find the salary of the second month Solution Let x be common in the ratio. 1st, 2nd, and 3rd month salaries of a person = 2x ; 4x ; 5x :- From this question 4x×5x-2x×^x = 4,80,00,000. or,12x2 -4,80,00,000. or,x2 = 4000000 x = 2000. 2nd month salary = 4x = 4x 2000 = Rs. 8000 Question 39 (2p2 - q2) = 7pq, where p, q are positive then p : q (a) 5:6 (b) 5:7 (c) 3:5 (d) 3:7 Solution (a) is correct 15(2p2 - q2) = 7pq Tricks: Go by choices For (a) put p = 5; q = 6 we get 15|2×52 -62 |= 3×5×6 or 15x14 = 210 or 210 = 210 10 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 40 Find the ratio of third proportional of 12,30 and mean proportional of 9, 25: (a) 7:2 (b) 5 : 1 (c) 9:4 (d) None of these 302 Solution 3rd proportional = = 75 12 Mean Proportional = 9 X 25 = 15 75 Ratio = = 5:1 15 ∴(b) is correct Question 41 What must be added to each of the numbers 10, 18,22, 38 to make them proportional: (a) 5 (b ) 2 (c) 3 (d) 9 Solution (b) is correct let x be added. 10 + x 22 + x = 18 + x 38 + x Tricks: Go by choices. ∴ x = 2 satisfies it. Question 42 x, y, z together starts a business. If x invests 3 times as much as y invests and y invests two third of what z invests, then the ratio of capitals of x, y, z is (a) 3:9:2 (b) 6:3:2 (c) 3:6:2 (d) 6 : 2 : 3 Solution (d) Tricks: Go by choices 2 6 = 3×2 and 2 = 3 x 3 Question 43 If a:b = 2:3,b:c = 4:5,c:d=6:7 then a: d is _____ Solution 2 4 6 16 Multiply all ratios. = x x = 3 5 7 35 Question 43 1 1 1 1 If , , , are in proportion then x = 2 3 5 x Solution Product of middle two terms = Product of extremes 1 1 = ; x = 15/2 2x 15 ©I CAI BOS 11 S ARANSH Ratio and Portion, Indices and Logarithms Unit III Indices If n is a positive integer, and ‘a’ is a real number, i.e. n ∈ N and a ∈ R (where N is the set of positive integers and R is the set of real numbers), ‘a’ is used to denote the continued product of n factors each equal to ‘a’ as shown below: an = a x a x a …………………………………..to n factors. Here an is a power of “a” whose base is “a” and the index or power is “n”. For example, in 3 x 3 x 3 x 3 =34, 3 is base and 4 is index or power. Laws of Indices 1. aman = (a)m+n 2. a n = am-n m a 3. a-n = (a)m+n m a 4. a0 = 1 5. [am]n = amn 1 m 6. [am]n = (a)n m n 7. am = (a)n 8. [ab]m = (a)m (b)m [ ab ] = bam m 9. 1 10. a-m = am 11. If ax = ay ∴x = y 12. If(a)x = (b)x ∴x = y 13. If ax = 1 ∴x = 0 Solved Questions Question 1: Value of (a1/8 + a-1/8)(a1/8 - a-1/8) Solution [a1/8 + a-1/8][a1/8 - a-1/8][a1/4 + a-1/4][a1/2 + a-1/2] [Use Formula (a+b)(a-b) = a2 - b2] = [(a1/8)2-〈a1/8 )2][a1/4+a-1/4][a↓/z+a-1/2] = (a1/4-a-1/4)(a1/4+a-1/4)(a1/2 +a-1/2) = [(a1/4)2-(a-1/4)2][a1/2+a-1/2] = (a1/2-a-1/2)(aι/2+a-1/2) = (a1/2)2-(a-1/2)Z=a-a-1=a- a1 12 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 2 If4X=5y=20z then z is equal to: Solution Let 4x=5y =20z =k or 4=kI/X;5=k1/y; 20=k1/8 :. 20=4×5 1 y or k1/Z = k1/xk 1 1 + k1/Z = k k y 1 1 1 1 y+x y = x + y = z = xy xy z = x+y 5/2 7/2 Question 3 3 9 x 9 is equal to….. 9 3 3 Solution 3 5/2 9 7/2 x9 9 3 3 31/2 5 32 7 1/2 =[( 1 ( 1/2 ) ] x 32 32 33 1 1 =[3( -2) 5 x 3(2 - 1 - ) 7]1/2.32 2 2 -15 7/2 1/2 2 -15 7 1/2 2 =[3 2. 3 ] x 3 = (3 2 + 2 ).3 =(3-4)112.32=3-2.32=3-2+2=30=i Question 4 If 2x-2x-1 = 4, then the value of xx is: Solution 2x-2x-1 = 4 or 2x-1 〈2-1)=4 or 2x-1 x 1=22 :x-1=2 :x=3 ∴xX=33=27 Question 5 If x = 31/3+3-1/3 then fmd value of 3x3-9x Solution If x=31/3 +3-1/3 By Cubing on both sides; we get x3=(31/3 〉3+(3-1/3〉3+3.31/3.3-1/3〈31/3 +3-1/3) = 3 + 3-1 + 3 x 1/xx (From (1)) 1 or x3 = 3 + + 3x 3 or x3 - 3x = 9 + 1 3 or 3x3-9x=10 ©I CAI BOS 13 S ARANSH Ratio and Portion, Indices and Logarithms 1 Question 6 Find the value of. [1-{1-(1-x2 )-1}1 ] 3 is Solution [1-{1-(1-x2 〉-1}-1]1/2 1 [1-{1- }-1]-1/2 1-x2 1 - x2 - 1 -1 -1/2 = [1-{ } ] 1-x2 -x2 -1 -1/2 = [1-{ } ] 1-x2 1 - x2 -1/2 1 - x2 1/2 = [1- ] = [1+ ] -x 2 x2 x2 + 1 - x2 -1/2 =[ ] x2 1 1 -1/2 2 =( ) = (x-2) = x x2 2n+2n-1 Question 7 2n+1-2n 1 2n+2n-1 2n(20+2-1) 1+ 2 3 Solution = = = 2n+1-2n 2n(21-20) 2- 1 2 Question 8 Simplify 2x 1/2 3x-1 if x = 4 Solution We have 2x 1/2 3x -1 = 6x 1/2 x-1 = 6x1/2-1= 6x 1/2 6 6 6 6 = = 1/2 = 2 1/2 = =3 x1/2 4 [2 ] 2 Question 9 Simplify 6ab2c3 × 4b–2c–3d. Solution 6ab2c3 × 4b–2c–3d = 24 × a × b2 × b–2 × c3 × c–3 × d = 24 × a × b2+(–2) × c3+(–3) × d = 24 × a × b2 – 2 × c3 – 3 × d = 24a b0 × c0 × d = 24ad Question 10 Simplify (xa.y–b)3. (x3 y2)–a Solution (xa.y–b)3. (x3 y2)–a = (xa)3. (y–b)3. (x3)–a. (y2)–a = x3a–3a. y–3b–2a. = x0. y–3b–2a. = 1/y 3b+2a 14 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 11 Find x, if x = (x x )x Solution x (x) 1/2 = xx.xx/2 or, x1+ 1/2 = xx + x/2 or, x3/2 = x3x/2 [If base is equal, then power is also equal] 3 3x 3 2 i.e = or x =1 2 2 2 2 ∴x = 1 Question 12 Find the value of k from ( 9 )-7 x ( 3 )-5 = 3k Solution ( 9 )-7 x ( 3 )-5 = 3k or, (32 × 1/2) –7 × (3½) –5 = 3k 3-7 x 3-5/2 = 3k or 3-7.-5/2 = 3k or, 3 –19/2 = 3k or, k = –19/2 ©I CAI BOS 15 S ARANSH Ratio and Portion, Indices and Logarithms Unit IV Logarithms At the foundation level the concepts Logarithms is used in accounting and finance. Here in this Unit Logarithms an attempt is made for solving and understanding the concepts of with the help of following questions with solutions. if ab = c; Where a ≠ 1 and a; c > 0 (positive) Then b is said to be the logarithm of the number c to the base “a” and expressed as Logac = b; Where a ≠ 1. Types of Logarithms (I) Natural Logarithm: The Logarithm of a number to base “e” is called Natural Logarithm. i.e. Logex where x = a number e = 2.7183 (ii) Common Logarithm: Logarithm of a number to the base 10 is called common Logarithm. i.e. Log10x where x = A number Note; If base is not given then in arithmetical or commercial work; base is always taken as 10. Remember Some Formulae 1. If ab = c Logac = b; Where a ≠ 1. 2. axlog ab=bx 3. logaα = l 4. loga 1=0 1 5. logba= ⇒ logba⋅logab=1 (loga b) 6. (i)logb a=logb xlogX a=logx a logb x (ii) logb a=logx a.logy x.logz y….iogb k logb a=logb x.logX y.logyz loglς a logX a 7. (i) logb a = logx b logb a (i) logb a = logI b 8. if logb a =x Then (i) log 1 a = _X b 1 (ii) logb a = -x (iii) 3log1 1 + a 16 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H 9. (i) loga (mn) = log am+ log an (ii) log a(mnr...) = log am+ loga n+loga r + …….. 10. loga( m n )=loga m-logan n 11. (i) log(ab)nn = b logam (ii) loga (mn)=nloga m 1 log m (iii) logabm= b a 12. (i) If logam = logbm ⇒ a=b (ii) If logam = logan ⇒ m = n Question 1 log216=? (a) 3 (b) 4 (c) 5 (d) None Solution (b) is correct Let log216 = x ∴2X=16=* 2X=24 ∴ x = 4. Question 2 If loga 16 = 4; then a equals. (a) 4 (b) 16 (c) 2 (d) None Solution (c) is correct Go by Choices a4= 16 Question 3 16log4 5 = ? (a) 5 (b) 16 (c) 25 (d) None Solution (c) is correct 16log4 5= (42)log45 = 42log45 = 52= 25. Question 4 If log gx + log4x + log2x =11, The value of x is Solution log 23X+log22X+log2x=11 ⇒ 1 log X + 1 log X + log X = 11 3 2 2 2 2 ⇒ log X [ 1 + 1 + 1] = 11 2 3 2 ⇒ log X [ 2 + 3 + 6 ] = 11 2 6 ⇒ log X. 11 = 11 2 6 ©I CAI BOS 17 S ARANSH Ratio and Portion, Indices and Logarithms orlog2 x = 11 x 6 = 6 11 or 26=x = x =64 Question 5 The value of log5 5 5 5..... 10 to ∞ is (a) 0 (b) 1 (c) 2 (d) None Solution (b) is correct ∴ 5 5 5......+ upto ∞ = 5 is ∴ log55 = 1 Question 6 The value of log8 25 given log 2 = 0.3010 is (a) 1 (b) 2 (c) 1.5482 (d) None Solution (c) is correct log8 25=log2a52= 3 2 log 5 2 2 10 2 = 3 log2 ( 2 )= 3 [log210-log22] =32 [ 1 log102 - 1] = 2 [ 1 - 1] = 1.5482 3 0.3010 ∴ (c) is correct. 1 1 1 Question 7 + + is equal to 1 + loga(bc) 1 + logb(ca) 1 + logc(ab) (a) 0 (b) 1 (c) 3 (d) -1 Solution (b) is correct 1 1 1 + + logaa+loga(bc) logbb+logb(ca) logcc+logc(ab) 1 1 1 = + + loga (abc) logb (abc) logc (abc) =logabca + logabcb + logabcc =logabcabc = 1 18 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 8 loga logb logc If = = the value of ay+z. bz+x. cx+y is given by y-z z-x x-y (a) 0 (b) 1 (c) -1 (d) None Solution (b) is correct loga logb logc Let = = =k y-z z-x x-y loga=k(y-z)⇒(y+z)loga=k(y-z)(y+z) ⇒logay+z =k(y2-z2) logb =k(z-x)⇒(z+x)logb=k(z-x)(z+x) ⇒logaz+x =k(z2-x2) Similarly log c = k (x - y) => log cx+y = k(x2 - y2) Adding Then log ay+z + log bz+x + log cx+y = k (y2- z2 + z2- x2 +z2- x2 +x2-y2) or log (aγ+z. bz+x. cx+y) = kx 0 = 0 = log 1, ∴ay+z. bz+x ⋅ cx+y = 1 ©I CAI BOS 19 S ARANSH Ratio and Portion, Indices and Logarithms Question 9 The value of the expression aloga b⋅logbc⋅logcd⋅logdt) (a) t (b) abcdt c) (a + b + c + d +1) (d) None. Solution loga b⋅logbc⋅logcd⋅logdt a =a1logat)=t1=t ∴ (a) is correct Question 10 If log10000x= -1 then x is given by : 4 (a) 1/100 (b) 1/10 (c) 1/20 (d) None of these. Solution log 10000x= 41 1 ∴(10000) 4 = x 1 or (104) 4 = x 1 or x=10-1 = 10 ∴(b) is correct 20 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 11 If log 〈 2a- 3b) = log a- log b, then a=? 3b2 3b b2 3b2 (a) (b) (c) (d) 2b-1 2b-1 2b+1 2b+1 a Solution log (2a-3b) = log b a or 2a-3b = b or 2ab-3b2=a or 2ab-a=3b2 or a(2b-1)=3b2 3b2 or a = 2b-1 ∴(a) is correct Question 12 1 1 1 + + is equal to : logϕ 〈abc〉 logbc (abc) logco (abc) (a) 0 (b) 1 (c) 2 (d) -1 Solution 1 1 1 + + logab 〈abc〉 logbc (abc) logca (abc) = logabc(abbcca) =logabc(abc)2 =2logabc(abc) =2×1=2 ∴ (c) is correct Question 13 Number of digits in the numeral for 264 [Given log 2 = 0.30103] : (a) 18 digits (b) 19 digits (c) 20 digits (d) 21 digits Solution (c) Let x = 264 log x = 64 log 2 = 64x0.30103 = 19.26592 x = Anti Log (19.26592) characteristics =19 No. of digits in the number =19 + 1 ∴ (c) is correct = 20 log3 8 Question 14 The value log9 16⋅log4 10 (a) 3 log|0 2 (b) 7 log|03 (c) 3 loge z (d) None Solution log3 8 log323 = log9 16⋅log4 10 log3224×log2210 3log3 2 3 = = 3log102 ∴ (a) is correct 4 4 log2 10 log32 x log2 10 2 2 ©I CAI BOS 21 S ARANSH Ratio and Portion, Indices and Logarithms If x = en -e -n then the value of n is : n -n Question 15 e +e Solution x = en-e-n 1 en+e-n By Componendo and dividendo; we get 1+x en+e-n +en-e-n 2en 1-x = en+e-n -en-e-n = 2e-n 1+x or =en+n 1-x 1+x or e2n = 1-x Taking log on both sides; we ge 1+x loge e2n=loge ( ) 1-x 1+x or 2nlogee=logε ( ) 1-x 1+x or 2n×1= log ( ) 1-x 1+x n = 1 log ( ) 2 1-x Question 16 log 144 is equal to: Solution log I44= log (16×9) = log 16+ log 9 = log 24+ log 32 =4 log 2+2 log 3 Question 17 If log2 [log3 (Iog2 x)]=1, then x equals; Solution log2[log3 (log2 x)] =1 Tricks : 2321 =x or x=232=23×3 =29=512 Detail Method: log2[log3 (log2 x)] =1 or log3 (log2 x)=21=2 or log2 x=32 or log2 x=9 or x=29=512 22 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 18 If log ( a + b )= 1 ( log a+ log b) then: a + b 4 2 b a Solution log ( a + b ) = 1 ( log a+ log b) 4 2 or log ( a + b ) = log (ab)1/2 4 or a + b = ab 4 or a+b=4 ab Squaring on both sides; we get (a+b)2=16ab or a2+b2+2ab=16ab. or a2+b2=14ab or a + b = 14ab [Dividing by ab on both sides] 2 2 ab ab ab or a + b = 14 b a Question 19 Iog (m+n)= log m+ log n, m can be expressed as: Solution Iog(m+n) = log m+ Iog n or log (m+n)= log (mn) or m+n=mn or m-mn =-n or m(1-n) =-n or m = -n = n 1-n n-1 Question 20 log4 (x2+x)-log4 (x+1)=2. find x Solution log4 〈x2+x) -Iog4 (x+1)=2 or log4 x +x = 2 2 x+1 x(x+1) or =42 x+1 or x=16 ©I CAI BOS 23 S ARANSH Ratio and Portion, Indices and Logarithms Question 21 2 log x+2 log x2+2 log x3+...…..+2 log xn will be : - n(n+1) log x (a) (b) n〈n+1) log x 2 (c) n2 log x (d) None of these Solution Detail Method 2 log x+2 log x2+2 log x3+----+2 log xn =2logx +22logx +2.3logx +-+2n.logx =2 logx. [1+2+3+_--+n] n(n+1) =2 logx. =n(n+1) log x 2 =(b) is correct Tricks :-Put n=2 in options directly This should be equal to sum of 1st 2 terms =2 log +2.2logx =6 logx Which gives option (b) (logX 10-3) 11-logX 10 Question 22 Solve: + = 20 2 3 (a) 10-1 (b) 10^2 (c) 10 (d) 103 logX 10-3 11-logx 10 Solution + =2 2 3 Tricks: - Go by choices [Do Mentally] For (a) x=10-1 log(101)10-3 11-log10 -1 10 L.H. S + 2 3 -1-3 11-(-1) = + 2 3 =-2+4=2 (R.H.S) ∴ (a) is correct Question 23 If n =m! where (‘ m’ is a positive integer >2) then the value of: 1 1 1 1 + + +........+ log2n logn3 log4n logmn Solution Given n=m ! 1 1 1 1 = + + +...... + log2n log3n log4n logmn =logn2+logn3+logn4+...……….+lognm =Iogn (2.3.4………..m) =logn (1.2.3.4……..m) log(m1)(m!)=1 24 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 24 If log2 x+log4x=6, then the value ofx is (a) 16 (b) 32 (c) 64 (d) 128 Solution (a) is correct Tricks: Go by choices for (a) if x=16 L.H. S=log2 16+log4 16=4+2=6(RHS) ∴ (a) is correct Detail Method : log2 x+log4x=6 or log2x+log22 ) x=ó or log2 x + 1 log22x=6 2 1 or (1 + 2 ) log2 x=6 or log2 x 6×2 = 4 3 x=24=16 Question 25 If log KY=100 and log2 x=10, then the value of ‘Y’ is: (a) 210 (b) 2100 (c) 21,000 (d) 210000 Solution (c) log2 x=10 x=210 Now logx y=100.=y=x100.: y=(210 )100=21000 ∴ (c) is correct Question 26 Which of the following is true. If 1 + 1 + 1 = 1 ab bc ca abc (a) log (ab+bc+ca)=abc (b) log ( 1 + 1 + 1 )=abc a b c (c) log (abc) =0 (d) log (a+b+c)=0 Solution (d) is correct = 1 + 1 + 1 = 1 ab bc ca abc Multiplying both sides by abc; abc + abc + abc + abc ab bc ca abc or c+a+b=1 or a+b+c=1 Taking Iog on both sides; we get log 〈a+b+c)= logl =0 ©I CAI BOS 25 S ARANSH Ratio and Portion, Indices and Logarithms Question 27 If (log√x 2)2 =logx2 then x= (a) 16 (b〉 32nn (c) 8 (d) 4 Solution (a) is correct (logfK2)2=logK2 or (logx1'22)Z=logx2 1 or [ logK2)2 =logK2 1 2 or 4(logK2)2-logK2=0 or logx 2[4logλ2-1]=0 If logK2=0 (Invalid) 4logK2-1=0 or 4logK2=1 1 or logλ2 = 4 or xI/4=2=x=24 =16 Tricks :- Go by choices For (a) LHS 1 (log√16 2)2=(log4 2)2 1 1 =( log22)2 = 2 4 1 1 RHS =log162=log242 = log22 = 4 4 Question 28 Find Value of [logy x.logzy.logxz]3= (a) 0 (b) -1 (c) (d) 3 Solution (c) is correct [logyx.logz y.logXz]3 = [logX x]3=3=1 Question 29 Find the value of Log4 9Log3 2= (a) 3 (b) 9 (c) 2 (d) 1 Solution (d) is correct Log4 91xg32= log(2l) (32).log32 2 = log23.log32 2 =1×1=1 26 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 30 If X=log24 12:y=log36 24;z=log48 36 then xyz +1=? (a) 2xy (b) 2zx (c) 2yz (d) 2 Solution (c) is correct xyz+1=log24 12.log36 24.log48 36+1 = log48 12+log_48 48 = Iog48 (12×48)=log48 (12×2)2 = 2log48 24=2log36 24.log4836 = 2yz Question 31 If x2+y2=7Xy ιheIl log 1 (x+y)= 3 (a) logx + logy (b) 1 (log x+ log y) 2 1 1 (c) (log x+ log y) (d) (Iogx.log y) 3 3 1 Solution log 1 (x+y) = 2 log { 1 (x + y)} 3 2 3 x2+y2+2xy = 1 log { 1 (x + y)}2 = 1 log ( ) 2 3 2 9 7xy+2xy = 1 10 log ( ) 1 log (xy) 2 9 2 (b) is correct Question 32 If log x=a-b;logy =a+b then log ( ) (a) 1-a-3b (b) a-1+3b (c) a+3b-1 (d) 1-b+3a Solution (a〉 is correct : - log x =a+b; logy =a-b. log ( 10x ) = logI010+ log x-1og y2 y2 =1+a+b-2 log y=1+a+b-2 (a-b) =1+a+b-2a+2b=1-a+3b ©I CAI BOS 27 S ARANSH Ratio and Portion, Indices and Logarithms Question 33 If x=1+logp qr, y=1+logq rp and z=] +logr pq men me value or 1 + 1 + 1 = x y z (a) 0 (b) 1 (c) -1 (d) 3 Solution 1 + 1 + 1 x y z 1 1 1 = + + 1+logpqr 1+logq rp 1+logr pq 1 1 = + logp p+logρqr logqq+logqrp 1 + logrr+logrpq 1 1 1 = + + logp pqr logq pqr logr pqr =logpqr p+logpq,q+logp,,rr = logpqrPqr=1 (b) is correct Question 34 If log x=m+n; logy =m-n theti log ( 10x )= y2 (a) 0 (b) 1 (c) -1 (d) 3 Solution If log x=m+n; logy =m-n Then log ( 10x ) y2 = log 10+ log x-log y2 = 1+ logx -2 log y = 1+(m+n)-2(m-n) = 1+m+n-2m+2n = 1-m+3n Question 35 log35. xlog5 4×log2 3: (a) 2 (b) 5 (c) -2 (d) None ofthese Solution log3 5.log54.log23 =log3 4.log2 3=log2 4=2 (a) is correct Question 36 The integral part of a logarithm is called , and the decimal part of a logarithm is called (a) Mantissa, Characteristic (b) Characteristic, Mantissa (c) Whole, Decimal (d) None of these Solution (b) is correct. 28 ©ICAI B O S Ratio and Portion, Indices and Logarithms SA R AN S H Question 37 The value of 1 + 1 + 1 =- log3 60 log4 60 log5 60 (a) 0 (b) 1 (c) 5 (d) 60 Solution (b) is correct logω 3+log604+log605 =log60(3×4×5)+logó060=1 Question 38 If log 4k2+x+log4(x+1)=2 then the value of x is (a) 2 (b) 3 (c) 16 (d) 8 Solution (c) is correct (x2+x) log4 =2 (x+1) x(x+1) or log4{ }=2 (x+1) or log4x=2⇒x=42=16 ©I CAI BOS 29 S ARANSH Chapter - 2: Equations In this chapter, students are able to understand the concepts of equations such as linear, simultaneous, quadratic, and cubic. They will also know how to solve the different equations using different solution methods. Definition of Equations Applications Equations Simple Equation Simulatneous Equations in two unknowns Methods of Solution Methods of solving three Cross Multiplication Method linear equations with Elimination Method three variables Quadratic Equation Nature of the roots Constrution of Quadratic Equation Roots of the Quadratic Equation Introduction Equation is defined to be a mathematical statement of equality. If the equality is true for certain value of the variable involved, the equation is often called a conditional equation and equality sign ‘=’ is used; while if the equality is true for all values of the variable involved, the equation is called an identity. Determination of value of the variable which satisfies an equation is called solution of the equation or root of the equation. For example: x+2 x+3 + = 3 holds true only for x = 1. 3 2 So it is a conditional. On the other hand, x+2 x+3 5x + 13 + = 3 2 6 is an identity since it holds for all values of the variable x. 30 ©ICAI B O S Equations SA R AN S H An equation in which highest power of the variable is 1 is called a Linear (or a simple) equation. This is also called the equation of degree 1. Two or more linear equations involving two or more variables are called Simultaneous Linear Equations. An equation of degree 2 (Highest Power of the variable is 2) is called Quadratic equation and the equation of degree 3 is called Cubic Equation For Example: 8x+17(x–3) = 4 (4x–9) + 12 is a Linear equation. 3x2 + 5x + 6 = 0 is a Quadratic equation. 4x3 + 3x2 + x–7 = 1 is a Cubic equation. x + 2y = 1, 2x + 3y = 2 are jointly called Simultaneous equations. Simple equation in one unknown x is in the form ax + b = 0. Where a, b are known constants and a ≠ 0 Note: A simple equation has only one root. 4x 14 19 Example 1: - 1= x + 3 15 5 Solution: By transposing the variables in one side and the constants in other side we have 4x 14x 19 (20-14)x 19+5 6x 24 - = + 1 or = or = 3 15 5 15 5 15 5 24 x 15 x= = 12 5x6 Example 2: The denominator of a fraction exceeds the numerator by 5 and if 3 be added to both the fraction 3 becomes. Find the fraction. 4 Solution: x Let x be the numerator and the fraction be x+5 x+3 3 By the question = or 4x + 12 = 3x + 24 or x = 12 x+5+3 4 12 The required fraction is 17 Example 3: If thrice of A’s age 6 years ago be subtracted from twice his present age, the result would be equal to his present age. Find A’s present age. Solution: Let x years be A’s present age. By the question 2x–3(x–6) = x or 2x–3x + 18 = x or –x + 18 = x or 2x = 18 or x=9 A’s present age is 9 years ©I CAI BOS 31 S ARANSH Equations Example 4: A number consists of two digits the digit in the ten’s place is twice the digit in the unit’s place. If 18 be subtracted from the number the digits are reversed. Find the number. Solution: Let x be the digit in the unit’s place. So the digit in the ten’s place is 2x. Thus the number becomes 10(2x) + x. By the question 20x + x– 18 = 10x + 2x or 21x – 18 = 12x or 9x = 18 or x = 2 So the required number is 10 (2 × 2) + 2 = 42. Example 5: For a certain commodity the demand equation giving demand ‘d’ in kg, for a price ‘p’ in rupees per kg. is d = 100 (10 – p). The supply equation giving the supply s in kg. for a price p in rupees per kg. is s = 75(p – 3). The market price is such at which demand equals supply. Find the market price and quantity that will be bought and sold. Solution: Given d = 100(10 – p) and s = 75(p – 3). Since the market price is such that demand (d) = supply(s) we have 100 (10 – p) = 75 (p – 3) or 1000 – 100p = 75p – 225 or – 175p = – 1225. -1225 p= =7 -175 So market price of the commodity is ` 7 per kg. the required quantity bought = 100 (10 – 7) = 300 kg. and the quantity sold = 75 (7 – 3) = 300 kg. Simultaneous linear equation in two unknowns The general form of a linear equations in two unknowns x and y is ax + by + c = 0 where a, b are non-zero coefficients and c is a constant. Two such equations a1x + b1y + c1 = 0 and a2 x + b2 y + c2 = 0 form a pair of simultaneous equations in x and y. A value for each unknown which satisfies simultaneously both the equations will give the roots of the equations. Method of Solution 1. Elimination Method: In this method two given linear equations are reduced to a linear equation in one unknown by eliminating one of the unknowns and then solving for the other unknown. Example 1: Solve: 2x + 5y = 9 and 3x – y = 5. Solution: 2x + 5y = 9 (i) 3x – y = 5 (ii) By making (i) x 1, 2x + 5y = 9 and by making (ii) x 5, 15x – 5y = 25 Adding 17x = 34 or x = 2. Substituting this values of x in (i) i.e. 5y = 9 – 2x we find; 5y = 9 – 4 = 5 y = 1 x = 2, y = 1. 2. Cross Multiplication Method: Let two equations be: a1x + b1y + c1 = 0, a2x + b2y + c2 = 0 We write the coefficients of x, y and constant terms and two more columns by repeating the coefficients of x and y as follows: 1 x 2 y 3 1 4 b1 c1 a1 b1 b2 c2 a2 b2 32 ©ICAI B O S Equations SA R AN S H and the result is given by: x y 1 = = (b1 c2 - b2 c1) (c1 a2 - c2 a1) (a1 b2 - a2 b1) b1 c2 - b2 c1 c1 a2 - c2 a1 so the solution is : x= y= a1 b2 - a2 b1 a1 b2 - a2 b1 Example 2: Solve 3x + 2y + 17 = 0, 5x – 6y – 9 = 0 Solution: 3x + 2y + 17 = 0 (i) 5x – 6y – 9 = 0 (ii) Method of elimination: By (i) x 3 we get 9x + 6y + 51 = 0 (iii) Adding (ii) & (iii) we get 14x + 42 = 0 42 or x =- =–3 14 Putting x = –3 in (i) we get 3(–3) + 2y + 17= 0 8 or, 2y + 8= 0 or, y=– =–4 2 So x = –3 and y = –4 Method of cross-multiplication: 3x + 2y + 17 = 0 5x – 6y – 9 = 0 x y 1 = = 2(-9)-17(-6) 17x(5)-3(-9) 3(-6)-5x(2) x y 1 or, = = 84 112 -28 x y 1 or, = = 3 4 -1 or x = –3, y = –4 ©I CAI BOS 33 S ARANSH Equations Methos of Solving Simultaneous Linear Equation with three variables Example 1: Solve for x, y and z: 2x – y + z = 3, x + 3y – 2z = 11, 3x – 2y + 4z = 1 Solution: (a) Method of elimination 2x – y + z = 3................................................................................................(i) x + 3y – 2z = 11............................................................................................(ii) 3x – 2y + 4z = 1..........................................................................................(iii) By (i) × 2 we get 4x – 2y + 2z = 6........................................................................................(iv) By (ii) + (iv), 5x + y = 17........................................................................(v) [the variable z is thus eliminated] By (ii) × 2, 2x + 6y – 4z = 22.............................................................(vi) By (iii) + (vi), 5x + 4y = 23................................................................(vii) By (v) – (vii), –3y = – 6 or y = 2 Putting y = 2 in (v) 5x + 2 = 17, or 5x = 15 or, x = 3 Putting x = 3 and y = 2 in (i) 2×3–2+z=3 or 6 – 2 + z = 3 or 4 + z = 3 or z = –1 So x = 3, y = 2, z = –1 is the required solution. (Any two of 3 equations can be chosen for elimination of one of the variables) (b) Method of cross multiplication We write the equations as follows: 2x – y + (z – 3) = 0 x + 3y + (–2z –11) = 0 By cross multiplication x y 1 = = -1(-2z-11)-3(z-3) (z-3)-2(-2z-11) 2'3-1(-1) x y 1 = = 20-z 5z + 19 7 20-z 5z + 19 x= , y= 7 7 Substituting above values for x and y in equation (iii) i.e. 3x - 2y + yz = 1, we have 20-z 5z + 19 3 -2 + 4z=1 7 7 or 60 – 3z – 10z – 38 + 28z = 7 or 15z = 7 – 22 or 15z = –15 or z = –1 20-(-1) 21 5-(-1)+19 14 x= = = 3, y= = =2 7 7 7 7 Thus x = 3, y = 2, z = –1 34 ©ICAI B O S Equations SA R AN S H Example 2: Solve for x, y and z : 1 1 1 2 3 4 3 2 1 + + = 5, - - = 11, + - = -6 x y z x y z x y z 1 1 1 Solution: We put u = ; v= ; w= and get x y z u + v + w = 5..........................................................................................................................(i) 2u – 3v – 4w = –11..............................................................................................................(ii) 3u + 2v – w = –6.................................................................................................................(iii) By (i) + (iii) 4u + 3v = –1...............................................................(iv) By (iii) x 4 12u + 8v – 4w = –24............................................(v) By (ii) – (v) –10u – 11v = 13 or 10u + 11v = –13..................(vi) By (iv) × 11 44x + 33v = –11........................................................(vii) By (vi) × 3 30u + 33v = –39....................................................(viii) By (vii) – (viii) 14u = 28 or u = 2 Putting u = 2 in (iv) 4 x 2 + 3v = –1 or 8 + 3v = –1 or 3v = –9 or v = –3 Putting u = 2, v = –3 in (i) or 2–3 + w = 5 or –1 + w = 5 or w = 5 + 1 or w = 6 1 1 1 1 1 1 Thus x = = , y= = , z= = is the solution. u 2 y 3 w 6 Example 3: Solve for x, y and z: xy xy yz =70, =84, =140 x+y x+z y+z Solution: We can write as x+y 1 1 1 1...................................................................................(i) = or + = xy 70 x y 70 x+z 1 1 1 1...................................................................................(ii) = or + = xz 84 z x 84 y+z 1 1 1 1...................................................................................(iii) = or + = yz 140 y z 140 By (i) + (ii) + (iii), we get 1 1 1 1 1 1 14 2 + + = + + = x y z 70 84 140 420 1 1 1 7 1 or + + = =...................................................................................(iv) x y z 420 60 ©I CAI BOS 35 S ARANSH Equations 1 1 1 4 By (iv)–(iii) = - = or x = 105 x 60 140 420 1 1 1 2 By (iv)–(ii) = - = or y = 210 y 60 84 420 1 1 1 By (iv)–(i) = - or z = 420 z 60 70 Required solution is x = 105, y = 210, z = 420 1. If the numerator of a fraction is increased by 2 and the denominator by 1 it becomes 1. Again if the numerator is decreased by 4 and the denominator by 2 it becomes 1/2. Find the fraction. Solution: Let x/y be the required fraction. x+2y x-4 1 By the question = 1, = y+1 y-2 2 Thus x + 2 = y + 1 or x – y = –1.............. (i) and 2x – 8 = y – 2 or 2x – y = 6............. (ii) By (i) – (ii) – x = –7 or x = 7 from (i) 7 – y = –1 or y = 8 So the required fraction is 7/8 2. The age of a man is three times the sum of the ages of his two sons and 5 years hence his age will be double the sum of their ages. Find the present age of the man? Solution: Let x years be the present age of the man and sum of the present ages of the two sons be y years. By the condition x = 3y ………………………….…(i) and x + 5 = 2 (y + 5 + 5)……………………..……(ii) From (i) & (ii) 3y + 5 = 2 (y + 10) or 3y + 5 = 2y + 20 or 3y – 2y = 20 – 5 or y = 15 x = 3 × y = 3 × 15 = 45 Hence the present age of the man is 45 years 3. A number consist of three digit of which the middle one is zero and the sum of the other digits is 9. The number formed by interchanging the first and third digits is more than the original number by 297 find the number. Solution: Let the number be 100x + y. we have x + y = 9.........(i) Also 100y + x = 100x + y + 297......................................................................(ii) From (ii) 99(x – y) = –297 or x – y = –3...........................................................................................................(iii) Adding (i) and (iii) 2x = 6 or x = 3 ∴ from (i) y = 6 Hence the number is 306. 36 ©ICAI B O S Equations SA R AN S H Quadratic Equation: An equation of the form ax2 + bx + c = 0 where x is a variable and a, b, c are constants with a≠ 0 is called a quadratic equation or equation of the second degree. When b=0 the equation is called a pure quadratic equation, when b ≠ 0 the equation is called an affected quadratic. Examples: i) 2x2 + 3x + 5 = 0 ii) x2 – x = 0 iii) 5x2 – 6x –3 = 0 The value of the variable say x is called the root of the equation. A quadratic equation has got two roots. Roots of a quadratic equation: ax2 + bx + c = 0 (a ≠0) b b2 4ac or x = 2a Sum and Product of the Roots: Let one root be α and the other root be β -b -coefficient of x α+β= = a coeffient of x2 c constant term product of the roots = = a coefficient of x2 How to construct a Quadratic Equation For the equation ax2 + bx + c = 0 we have -b c Or x2 + = x+ 0 a a -b c Or x2 - x+ =0 a a Or x2 – (Sum of the roots) x + Product of the roots = 0 b b2 4ac Nature of the Roots x = 2a i) If b –4ac = 0 the roots are real and equal; 2 ii) If b2–4ac >0 then the roots are real and unequal (or distinct); iii) If b2–4ac 0 but not a perfect square the roots are real, irrational and unequal. Since b2–4ac discriminates the roots b2 – 4ac is called the discriminant in the equation ax2 + bx + c = 0 as it actually discriminates between the roots. Note: (a) Irrational roots occur in conjugate pairs that is if (m+√n) is a root then (m-√n) is the other root of the same equation. If one root is reciprocal to the other root then their product is 1 and so c = i.e. c = a a If one root is equal to other root but opposite in sign then. b their sum = 0 and so = 0. i.e. b = 0 a ©I CAI BOS 37 S ARANSH Equations 1. Solve x2 – 5x + 6 = 0 Solution: 1st method : x2 – 5x + 6 = 0 or x2 –2x –3x +6 = 0 or x(x–2) – 3(x–2) = 0 or (x–2) (x–3) = 0 or x = 2 or 3 2nd method (By formula) x2 – 5x + 6 = 0 Here a = 1, b = –5 , c = 6 (comparing the equation with ax2 + bx + c = 0) b b2 4ac (5) 25-24 x= = 2a 2 5±1 6 4 = and ; x = 3 and 2 2 2 2 2. Examine the nature of the roots of the following equations. i) x2– 8x + 16 = 0 ii) 3x2 – 8x + 4 = 0 iii) 5x2 – 4x + 2 = 0 iv) 2x2 – 6x – 3 = 0 Solution: (i) a = 1, b = –8, c = 16 b2 – 4ac = (–8)2 – 4 × 1 × 16 = 64 – 64 = 0 The roots are real and equal. (ii) 3x2 – 8x + 4 = 0 a = 3, b = –8, c = 4 b2 – 4ac = (–8)2 – 4 × 3 × 4 = 64 – 48 = 16 > 0 and a perfect square The roots are real, rational and unequal (iii) 5x2 – 4x + 2 = 0 b2 – 4ac = (–4)2 – 4 × 5 × 2 = 16 – 40 = –24 < 0 The roots are imaginary and unequal (iv) 2x2 – 6x – 3 = 0 b2 – 4ac = (–6)2 – 4 × 2 (–3) = 36 + 24 = 60 > 0 The roots are real and unequal. Since b2 – 4ac is not a perfect square the roots are real irrational and unequal. 3. If α and β be the roots of x2 + 7x + 12 = 0 find the equation whose roots are (α-β)2. Solution Now sum of the roots of the required equation = (α + β)2 + (α - β)2 = (-7)2 + (α + β)2 - 4αβ = 49 + (–7)2 – 4x12 = 49 + 49 – 48 = 50 Product of the roots of the required equation = (α+ß)2 ((α - β)2= 49 (49–48) = 49 Hence the required equation is x2 – (sum of the roots) x + product of the roots = 0 or x2 – 50x + 49 = 0 38 ©ICAI B O S Equations SA R AN S H α2 β2 4. If α, β be the roots of 2x2 – 4x – 1 = 0 find the + value of β α -(-4) -1 Solution: α+β = = 2, αβ = 2 2 α2 β2 α3+ β3 (α+β)3 - 3 αβ (α+β) + = = β α αβ αβ 1 23-3 -.2 1 1 - 1 5. Solve x : 4x – 3.2x+2 + 25 = 0 Solution: 4x – 3.2x+2 + 25 = 0 or (2x)2 – 3.2x. 22 + 32 = 0 or (2x)2 – 12. 2x + 32 = 0 or y2 – 12y + 32 = 0 (taking y = 2x) or y2 – 8y – 4y + 32 = 0 or y(y – 8) – 4(y – 8) = 0 (y – 8) (y – 4) = 0 either y – 8 = 0 or y – 4 = 0 y = 8 or y = 4. ⇒2x = 8 = 23 or 2x = 4 = 22 Therefore x = 3 or x = 2. 2 1 1 1 6. Solve x- +2 x+ =7 x x 4 2 1 1 29 x- +2 x+ = x x 4 2 1