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3 Preface (iii) Quantitative Aptitude forms a very important part of preparation of MBA aspirants. Not just the Quant section but it forms the backbone of the Data Interpretation, Data Sufficiency and Reasoning. Disha s Quantitative Aptitude for CAT/ XAT/ IIFT/ CMAT/ MAT/ Bank PO/ SSC is a book focused on mastering techniques to crack these examinations. The book starts from a foundation level and moves to an expert level. Structure of the book: The book comprises of 5 Units (Numbers, Arithmetic, Algebra, Geometry and Counting Principles) which have been further divided into 22 chapters followed by 3 Mock Tests. Each chapter consists of Theory with Illustrations Foundation Level Exercise Standard Level Exercise Expert Level Exercise Solutions to the 3 levels of exercises Solutions to Salient Features of the Book: Each chapter includes detailed review of all the concepts involved with exhaustive number of well discussed Illustrations. The theory is followed by 3 levels of exercises Foundation Level, Standard Level and Expert Level. The detailed solution to each and every question has been provided immediately at the end of the 3 exercises. Foundation Level : Here the focus is to expose the students to solve problems based on the concepts they have learned in theory part. The student develops a good foundation and is ready for the Standard level. Standard Level : The Standard level is a collection of excellent quality problems which will test a student on the application of the concepts learned in various real-life situations. The problems provide a good platform to develop a very good problem solving aptitude so as to take up the expert level confidently. Expert Level : This is the toughest part of the book and involves the trickiest questions on the concepts involved. Here most of the problems will pose good challenge to the students. The book contains 22 Chapter-wise Tests - on the basis of latest CAT pattern after the exercises in each chapter. The students must attempt these tests in specified time limits and conditions. At the end of the book 3 Mock Tests are provided based on the exact pattern of latest CAT exams. The solutions to the test are provided at the end of the tests. The book contains questions of past exams of CAT/ XAT/ IIFT/ SNAP/ NMAT/ ATMA/ FMS in the various exercises and Illustrations. We would like to thank the DTP team at Disha, especially Mr. Amit Kumar Jha, who have worked really hard to bring the book to the present shape. Although we have taken utmost care while preparing the book but errors might have crept in. We would like to request our readers to highlight these errors. Authors

4 (iv) Unit-I : Numbers 1. Fundamentals 1-28 Shortcuts for Addition and Subtraction Shortcuts for Multiplication Rounding off and Its Uses BODMAS Rule Brackets Factorial Roman Numbers Important Conversion Absolute Value or Modulus of a Number Properties of a Modulus Powers or Exponents Algebraic Identities Squares Properties of Squares Square Roots Cubes Explanation of 2. Number System Tabular Classification of Numbers Concept of Number Line (or Number Line) Conversion of Rational Number of the Form Non-terminating Recurring Decimal into the Rational Number of the form p/q Division Tests of Divisibility Prime Numbers Complex Numbers, Real Numbers and Imaginary Numbers General or Expanded Form of 2 and 3 Digits Numbers Sum of Numbers Formed with given Different Digits Factorisation Number of Factors of a Composite Number Number of ways of Expressing a Composite Number as a Product of Two Factors Sum of Factors (or Divisors) of a composite Number Sum of Unit Digits The Last Digit From Left (i.e., unit digit) of Any Power of a Number Concept of Remainders To Find the Last Digits of the Expression like a 1 a 2 a 3... a n Contents Last Two Digits of a Number with Large Power Number of Zeroes in an Expression like a b c..., where a, b, c,... are Natural Numbers Powers of a Number Contained in a Factorial Base System Successive Division Factors and Multiples Highest Common Factor (HCF) or Greatest Common Divisor (GCD) Least Common Multiple (LCM) Greatest Integral Value Explanation of Unit-II : Arithmetic 3. Averages Average Position of the Average on the Number Line Weighted Average Properties of Average (Arithmetic Mean) Explanation of 4. Alligations Alligation Solving the Problems of Alligations Using Alligation Formula Graphical Representation of Alligation-Cross Method The Straight Line Approach to Solve the Problems Related to Alligations Recognition of Different Situations Where Alligation can be Used A Typical Problem Explanation of 5. Percentages Basic Definition of Percentage Percentage Increase, Percentage Decrease and Percentage Change Percentage Point Change and Percentage Change Calculation of Percentage Value Through Addition

5 (v) Effect of Percentage Change in the Numerator on the Value of a Ratio Percentage Change Graphic Application of Percentage Change Graphic (PCG) Calculation of Multiplication by Numbers Like 2.14, 1.04, 0.35, 0.94 and so on Using Percentage Explanation of 6. Profit, Loss and Discount Total Cost Price (CP) Selling Price (SP) Profit (or Gain) and Loss Use of PCG (Percentage Change Graphic) in Profit and Loss Marked Price, List Price, Discount and Successive Discounts Contribution Margin (CM) Break-Even Point and Break-Even Sales Explanation of 7. Interest Interest Simple Interest (S.I.) Compound Interest (C.I.) Explanation of 8. Ratio, Proportion and Variation Ratio Decimal and Percentage Value of a Ratio Properties of Ratios Uses of Ratios Comparison of Ratios Calculation of Percentage Change in Ratio Using PCG (Percentage Change Graphic) Proportion Properties of Proportion Variations Types of Variations Compound Variations Explanation of 9. Time and Work Concept of Efficiency Concept of Negative Work Concept of Man-days Work Done Work Done Equation Work in Terms of Volume (special case as building a wall) Extension of the Concept of Time and Work Explanation of 10. Time, Speed and Distance Motion or Movement Conversion of kmph (kilometer per hour) to m/s (metre per second) and vice-versa Direct and Inverse Proportionality Between any Two of the Speed (S), Time (T) and Distance (D) When the Third One is Constant Average Speed Relative Speed To and Fro Motion in a Straight Line Between Two Points A and B Uniform Acceleration and Uniform Deceleration Application of Alligation in the Problems Related to Time, Speed and Distance Concept Related to Motion of Trains Boats and Streams Basic Terminology Related to Races Circular Motion Clocks Explanation of Unit-III : Algebra 11. Progressions Arithmetic Progressions (A.P.) n th Term of an A.P. Sum of First n Terms of an A.P. Special Cases of A.P.s in which Sum upto Different Terms are the Same Arithmetic Mean of n Numbers Geometric Progression (G.P.) Considering the Terms in a G.P. Geometric Mean of n Numbers Harmonic Progression (H.P.) Relations between Arithmetic Mean (A.M.), Geometric Mean (G.M.) and Harmonic Mean (H.M.) Useful Results

6 (vi) Explanation of 12. Linear Equations Linear Equations Steps to be Followed to Solve a Word Problem Using Linear Equation(s) Explanation of 13. Functions Function Rules for Finding the Domain of a Function Methods of Representation of Functions Some Special Functions Shifting of Graphs Combination of Shifting of a Graph Explanation of 14. Quadratic & Cubic Equations Quadratic Polynomials Quadratic Equations Graph of a Quadratic Expression Geometrical Meaning of Roots or Solutions of a Quadratic Equation Sign of a Quadratic Expression Sum and Product of Roots Formation of an Equation with Given Roots Greatest and Least Value of a Quadratic Expression Cubic Equations Bi-quadratic equation Explanation of 15. Inequalities Inequality Types of Inequalities Some Properties of Inequality Important Results Solution of an Inequality Equivalent Inequalities Notation and Ranges Solutions of Linear Inequalities in one Unknown Solutions of Quadratic Inequalities Solution of System of Inequalities in one Variable Inequalities Containing a Modulus Explanation of 16. Logarithms Definition Laws of Logarithm Some Important Properties Characteristics and Mantissa Very Useful Results Explanation of 17. Set Theory Sets Representations of Sets Standard Symbols of Some Special Sets Types of Sets Subsets Intervals as Subsets of a Set of Real Numbers (R) Power Set of a Set Universal Set Venn Diagrams Operation on Sets Disjoint Sets Cardinal Number Situation Based Venn Diagrams Explanation of Unit-IV : Geometry 18. Geometry Points, Lines, Line Segment, Ray and Plane Lines and Angles Polygons Triangles Basic Properties and Some Important Theorems of Triangles Important Terms Related to a Triangle Congruency of Two Triangles Similarity of Two Triangles Quadrilaterals Circles Basic Pythagorean Triplets

7 (vii) Determination of Nature of Triangle Important Points Locus Sine and Cosine Rule Explanation of 19. Mensuration Basic Conversion of Units Plane Figures Area of a Triangle Area of a Quadrilateral Area of a Regular Hexagon Area of Irregular Plane Figures Paths Area Related to a Circle Surface area and Volume of Solids Euler s Rule Circle Packing in a Square Circles Packing in a Circle Some Other Important Concepts Explanation of 20. Coordinate Geometry Rectangular Coordinate Axes Sign Conventions in the xy-plane Quadrants of xy-plane and Sign of x and y-coordinate of a Point in Different Quadrants Plotting a Point Whose Coordinates are Known Distance Formula Applications of Distance Formula Section Formula Coordinates of Some Particular Points Area of Triangle and Quadrilateral Transformation of Axes Image of a Point Equation of Straight Line Parallel to An Axis Inclination of a Straight Line Slope (or Gradient) of a Straight Line Equation of Straight Lines Different Forms of the Equation of a Straight Line Point of Intersection of Two Lines Position of a Point Relative to a Line Angle Between Two Straight Lines Equation of Parallel and Perpendicular Lines Distance of a Line from a Point Distance Between Two Parallel Lines Explanation of Unit-V : Counting Principles 21. Permutations and Combinations Fundamental Principle of Counting Factorials Meaning of Permutation and Combination Counting Formula for Linear Permutations Number of Linear Permutations Under Certain Conditions Circular Permutations Counting Formula for Combination Division and Distribution of Objects Dearrangement Important Results about Points Finding the Rank of a Word Explanation of 22. Probability Concept of probability Basic Terms Mathematical Definition of Probability Odds Against and Odds in Favour of an Event Addition Theorem Independent Events Conditional Probability Geometrical Applications Probability Regarding n Letters and Their Envelopes Expectation ` Explanation of Mock Tests Mock Test Mock Test Mock Test (Mock Test - 1 to 3)

8 UNIT-I Numbers Chapter 1 Chapter 2 Fundamentals Number System

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10 1 FUNDAMENTALS l l Shortcuts For Addition and Subtraction l Shortcuts for Multiplication l Rounding off and Its Uses l BODMAS Rule l Brackets l Factorial l Roman Numbers l Important Conversion l Absolute Value or Modulus of a Number l Properties of a Modulus l Powers or Exponents l Algebraic Identities l Squares l Properties of Squares l Square Roots l Cubes INTRODUCTION In the CAT and the likes of competitions, 25% to 35% questions are based on numeracy. So, to save the time for other questions in competitions, it is essential to command over shortcuts of addition, subtraction and multiplications given in this chapter. SHORTCUTS FOR ADDITION AND SUBTRACTION Addition is the mother of all calculations, which gives you an extra edge that makes your calculations faster. Subtraction is the extension of addition. I. Addition of smaller number to larger number is easier than addition of larger number to smaller number. For example addition in the order is easier than the addition in the order Hence to add the numbers, it is better to first arrange them in decreasing order and then add them. II. To find the sum like , first add the thousands and then hundreds, tens and once in order. Thus = ( = 8000) (= 8300) (= 8700) (= 9500) + 40 (= 9540) + 70 (= 9610) + 5 (= 9615) + 6 (9621) + 2 = 9623 III. To find the sum of large numbers like First add the thousands like (= 107) + 94 (= 201) + 8 (= 209) + 7 = 216 At this stage you know the answer would be (a maximum of 5000), as there are five numbers whose last 3 digits numbers are not added. If the range from to is sufficient to choose the correct option, then no need to add further otherwise add the hundredth digits of given numbers = 13. At this stage you know the answer would be (a maximum of 500) If the range from to is sufficient to choose the correct option, then no need to add further otherwise add the last two digits of numbers = 162. Hence the correct sum will be = There are two advantages of process of addition (i) No need to get final sum as in this process of addition, you could choose the correct option at earlier stage also. (ii) In the entire calculation, you have not gone above two digits additions. IV. Sometimes you have to add so many large numbers. In that case you can find the required sum using the following methods. (A) Column Form Write the given numbers one below the other with right align if the given numbers are whole numbers and with decimal point align if the given numbers are decimal numbers as we write in conventional method of addition.

11 2 l Quantitative Aptitude Right align Decimal point align (i) Addition of Whole Numbers To add the whole numbers with right align, we start adding the digits in the right most column by going down but when the running total becomes 10 or higher than 10, then we reduce it by 10 and go ahead with reduced number. As we do so, we make a small dash at the right top corner of the digit that makes our total 10 or higher than 10 as given below for right most column = 13, which is more than 10, so we subtract 10 from 13 and mark a dash at the right top corner of the digit 8 and start adding again = 10, so we subtract 10 from it and mark a dash at the right top corner of the digit 7 and start adding again = = = = 15, which is more than 10, so we subtract 10 to from 15 and mark a dash at the right top corner of the digit 6 and start adding again = = 14, which is greater than 10, so we subtract from it and mark a dash at the right top corner of the digit 5 and start adding again = 13, which is greater than 10, so we subtract from it and mark a dash at the right top corner of the digit 9. The dashes and the final figure 3 will be written under the first column from right as Now we count the dashes marked in the first column from right. Number of dashes in this column is 5. Now add the number of dashes 5 in the top digits 2 of the second column from right, then start adding this column as we add the first column from right. In the same way, we add the other columns one by one from right. After adding the left most column, write the number of dashes in this column in the left of the total of this column as given below Number of dashes in the just right column The advantage of this process is that the entire calculation is done only by adding one digit numbers. (ii) Addition of Decimal Numbers Addition of decimal numbers with decimal point align is the same as addition of whole numbers with right align. In addition of decimal numbers, we put a decimal point in the sum total align with decimal in the given numbers as given below. Illustration 1: Find the sum of the following numbers using column form , , 4.213, 56.8, and Solution:

12 Fundamentals l 3 (B) Row Form To find the sum of numbers, it is not necessary to write them one below the other with align i.e., column form. You can find the sum of numbers written in a row form using the same method discussed above for column form but there is a problem of alignment. To overcome this problem of alignment, we use the method of column form in slightly different way as discussed below. This method of addition is very important. If you get command over it, you can stop wasting time in writting the numbers in column form. (i) Addition of Decimal Numbers Suppose you have to find the sum (a) Put zeros to the right of the last digit after decimal to make the number of digits after decimal equal in each number. For example, the above addition may be written as (b) Start adding the last digit from right of all the numbers. During running total, don t exceed 10. That is, when you exceed 10, mark a tick with pencil anywhere near about your calculation and go ahead with the number exceeding = 7 5 plus 0 is 5; 5 plus 9 is 14, mark a tick in rough area and carry over 4; 4 plus 3 is 7; 7 plus 0 is 7, so write down 7. During addition we strike off all the digits which are added. It saves us from confusion and duplication. (c) Add the number of ticks (marked near by calculation in rough) with the digits at 2nd places from right and erase that tick from rough = 97 1 (number of tick) plus 2 is 3; 3 plus 2 is 5; 5 plus 0 is 5; 5 plus 4 is 9 and 9 plus 0 is 9; so write down 9 at the second place from right in the sum. (d) = plus 8 is 11; mark a tick in rough and carry over 1; 1 plus 0 is 1; 1 plus 9 is 10, mark another tick in rough and carry over zero; 0 plus 6 is 6, so put down 6 at the third place from right in the sum. (e) Following the same way get the result: = Illustration 2: Find the sum of the following numbers using row form , , 4.213, 56.8, and Solution: = (ii) Addition of Whole Numbers Suppose you have to find the sum Follow the steps mentioned in steps (b), (c), (d) and (e) of section (B) (i) (addition of decimal numbers in row form) above (without considering the decimal). Thus, = = = = = = Illustration 3: Find the sum of the following numbers using row form. 5834, 96182, 459, 2128, and 735 Solution: = V. Single Step Solution for Addition and Subtration in a Single Row: Digit-Sum Method To understand this method, let us find the value of First of all, we check that the required value or number will +ve or ve by just looking at the given numbers with signs. In the case of +ve required number, the digits of the required number will be zero or +ve integer and in the case of ve required number, the digits of the required number will be zero or ve integer. Clearly the required number will be +ve. Hence, digits of required number will be zero or +ve.

13 4 l Quantitative Aptitude Now to find the unit digit of the required number, add and subtract the digits at units places of these given numbers according to the sign attached with these numbers as = 6 Since required number will be +ve, therefore its unit digit can not be ve. To make ( 6) positive, we borrow from tens of largest given positive number. You should remember that we can t borrow from negative given number if required number is +ve. So, we borrow 1 from tens digit 0 of Now, we add 10 to ( 6), this can be shown as ( 1) ( 1) = 4 Now, we add and subtract the digits at tens places of given numbers according to the sign attached with these numbers = 5 Since 5 is positive, hence 5 is the tens digit of the required number. This can be shown as ( 1) ( 1) = 54 Now add the digit at hundredth places as (9 1) 2 = 7, which is positive. Hence hundredth digit of required number will be 7. Now add the digit at thousand places as = 11 Thus the last two digits of the required number are 11. Hence ( 1) ( 1) = The same above method is used for decimal numbers also after making the equal number of digits after decimal in all the given numbers by putting zero(s) at the end of the number after decimals. This method requires some practice. But after some practice, you will find it is faster method. Illustration 4: =? Solution: ( 1) (+1) = 5962 After adding and subtracting the digits at tens places according to the sign attached with the respective numbers, we get 16, which has two digits. So, 6 is written at tenth place in the required number and 1 is added to 5 (hundredth digit) of Illustration 5: =? Solution: ( 2) = 2060 After adding and subtracting the digits at tens places according to the sign attached with the respective numbers, we get 14. Since 14 is more than 10 but not more than 20. Therefore to make 14 as a single positive digit we have to borrow 2 from hundredth digit i.e., 6 of Now = 6, therefore tens digit of the required number is 6. Illustration 6: =? Solution: (+1) = 4060 After adding and subtracting the digits at tens places according to the sign attached with the respective numbers, we get 16. So we take 6 as tenth digit of the required number and add 1 to the hundreth digit i.e., 5 of Illustration 7: =? Solution: ( 1) = 330 Illustration 8: =? Solution: ( 1) = 6388 By observing the given numbers with signs, it is clear that the required number or value will be ve. Hence digits of the required number will be zero or negative integer. Now = 8, so unit digit of required number is 8 (without the sign) = 2, which is +ve. To make 2 negative, borrow 1 from hundredth digit of the largest given ve number i.e. borrow 1 from 7 of Now subtract 10 from 2, which gives 8. So 8 is the tens digit of required number. Similarly, we find the hundredth and thousand digit of the required number as 3 and 6 respectively. Since required number will be ve, therefore we put a ve sign before 6388 which gives 6388 as required number. Illustration 9: =? Solution: First equate the number of digits after decimals by putting zero(s) at the end. So, ( 1) ( 1) ( 1) ( 1) (+1) = = SHORTCUTS FOR MULTIPLICATION 1. Line Segment Method of Multiplications of Two Whole Numbers of any Number of Digits To clearly understand this method, we will discuss some examples. (i) Consider the multiplication of two digit numbers, The digit of the different places of the required product will be found out as follows. (a) Finding the Units Place Digit To Find the unit s digit of the product of any two numbers, we always find the product their unit s digits. Here product of unit digits = 6 9 = 54 Unit s digit 4 of 54 is the unit s digit of the required product. Tenth digit 5 of 54 will be carry over to the tens place. Thus carry over to the tens place

14 Fundamentals l 5 (b) Finding the Tens Place Digit = = (from carry over) = 92 Here unit s digit 2 of 92 is the tens place digit of the required product. Tens digit 9 of 92 will be carry over to the hundred s place digit. Thus carry over to the hundred s place. (c) Finding the Hundred s Place Digit = (from carry over) = 37 Since 7 and 4 are the last digits on the left in both the given numbers, so this is the last calculation in this multiplication and hence we can write 37 for the remaining 2 digits in the required product. Thus (ii) Consider the multiplication of more than 2 digits numbers, Study the following table which explains the process of finding the digit of different places of the required product. Finding the digit Diagram showing the calculation process Calculation Required digit(s) Carry on to the next place digit Explanation of the diagram showing the calculation process Unit digit Tens digit Hundred digit Thousand digit Ten thousand digit = Multiplication between unit s digit of both the number shows by line segment between 2 and = (carry over) = Multiplication of tens digit 0 of 5402 by unit s digit 5 of 315 shows by line segment between 0 and 5, then rotate this line segment in clockwise direction about their midpoint to find the next pair of digits to be multiplied = Multiplication of hundred s digit 4 of 5402 by unit s digit 5 of 315 shows by line segment between 4 and 5, then rotate this line segment in clockwise direction about their mid-point to find the next pair of digits to be multiplied = (carry over) = = (carry out) = Similar explanation as for given above for hundred digit but there is no digit in the left of 3 in 315, so the unit digit 2 of 5402 will not be multiplied by any digit. 0 2 Since unit digit 5 of 315 is multiplied by left most digit 5 of 5402 in finding the thousand digit. Hence tens digit 1 of 315 multiplies the left most digit 5 of 5402 and rotate the line segment in clockwise direction between 1 and 5 about their mid-point to find the next pair of digits to be multiplied but there is no digit in the left of 3 in 315, so further rotation of line segment between 3 and 4 in clockwise direction will not find any two digits to be multiplied and hence the ten s and unit s digit of 5402 will not be multiplied by any digit.

15 6 l Quantitative Aptitude Finding the digit Diagram showing the calculation process Calculation Required digit(s) Carry on to the next place digit Explanation of the diagram showing the calculation process Last digit(s) 3 5 = (carry over) = 17 Hence required product = Since ten s digit 1 of 315 is multiplied by left most digit 5 of 5402 in finding the ten thousand s digit, so hundred digit 3 of 315 multiplies the left most digit 5 of Since there is no digit in the left of 3 in 315, so rotation of line segment between 3 and 5 about their mid-point in anticlockwise direction will not find any two digits to be multiplied further and hence hundred, tens and unit digits of 5402 will not be multiplied by any digit. In CAT and CAT like competitions large multiplications might not be required but it might be required to find any specific digit of the product of large multiplication, then the above method of multiplication is quite useful. 2. Multiplication of Two Numbers Using Formulae (a b) (a + b) = a 2 b 2 If the difference between two numbers x and y is a small even number, then the smaller is express as (a b) whereas larger is expressed as (a + b), then the product of x and y is found out by the formulae x. y i.e., (a b) (a + b) = a 2 b 2 Here a should be such that a 2 is very easily calculated. For example: (i) = (40 2) (40 + 2) = (40) 2 (2) 2 = = 1596 (ii) = (70 4) (70 + 4) = (70) 2 (4) 2 = = 4884 (iii) = (2100 6) ( ) = (2100) 2 (6) 2 = = If the difference between the two numbers is not even, still this method is used by modify as = = (50 3) (50 + 3) + 47 = (50) 2 (3) = = Multiplying Two Numbers Close to 100, 1000, 10000, , etc To multiply two numbers close to 100, 1000, or ; we can use a specific method which is discussed in the following illustrations. (i) Let us multiply 92 and 97. Step (a): Calculate the difference from 100 of both the numbers and write them as follows: Step (b): Initial digits of the required product is found out by cross addition as 92 +( 3) or 97 +( 8) =89 24 Last two digits of the required product ( 8) ( 3) =24 Thus, =8924 (ii) Let us multiply 1008 and 994. Difference from Initial digits Last three digits Here we first find the initial digits by cross addition as ( 6) or = 1002 Now write 1002 as initial digits and write last three digits as 000, (i.e., last three zeroes of 1000) which means numbers value is Now in add the product 8 ( 6) = 48, which gives the required product i.e., Illustration 10: Find the product Solution: Difference from Last digits First two digits Hence = 11232

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