Aptitude

Aptitude Contents

Chapters Topic Page No.

Chapter-1 Introduction Theory at a glance 1. Local Value 2. Place Value 3. Absolute Value Number Types in Numerical Aptitude Tests 1. Even Numbers 2. Odd Numbers 3. Natural Numbers 4. Rational Numbers 5. Irrational Numbers 6. Real Numbers 7. Complex Numbers 8. Whole Numbers 9. Prime Numbers 10. Composite Numbers Prime Numbers Properties of Prime Numbers Process to Check A Number is Prime or not Example 239 is prime or not? Composite Numbers Co-Primes Face value Tests of Divisibility Basic Formulae Division Algorithm Multiplication by Short Cut Methods Progression Problems

1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 5 5

Chapter-2 Ratio and Proportion Theory at a glance Problem

11 11 11

India’s No. 1 Aptitude IES Academy Contents

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Chapters Topic Page No.

Chapter-3 Partnership Theory at a glance Ratio of Division of Gains: Formulae Short cuts Solved Problems Problems on Partnership

15 15 15 15 15 16 17

Chapter-4 Percentages Theory at a glance Profit and Loss Formulae: Solved Problems

23 23 30 30 30

Chapter-5 Simple Interest Theory at a glance Important Facts and Formulae: Compound Interest Formulae: Solved Examples More solved problem

37 37 37 42 42 42 44

Chapter-6 Time & Work Theory at a glance 1. General Rules Solved Problems More Solved Problems 2. Pipes and Cisterns Important Facts Complex Problems Problems on Time and Work Answers Key Exercise on Pipes and Cisterns Answers Key

49 49 49 49 51 60 60 61 64 67 68 70

Chapter-7 Time Speed and Distance Theory at a glance Solved Problems 1. Boats & Streams Important Points Solved problems 2. Trains General Concepts Solved Examples More Solved Problems Problems Answers Problems on Boats and Streams Answers Problems on Trains Answers

71 71 71 74 74 74 75 75 76 77 80 86 87 89 90 94


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Chapter-8 Geometry Theory at a glance Areas in Geometry Volumes in Geometry Surface area in Geometry Properties of Triangles Solved Examples Problems related to Area

95 95 95 97 99 99

100 102

Chapter-9 Permutation and Combination Theory at a glance Permutation Examples Circular Permutations Restricted – Permutations Restricted – Combinations Restricted Permutations Restricted – Combinations

109 109 110 111 112 113 114 115 116

Chapter-10 Probability Theory at a glance Basic Concepts Part (1) : Foundation Level Important types of Events: Category - B Category - C Part (2) : (Total Probability) Examples

119 119 119 120 121 126 130 131 132

Chapter-11 Reasoning Theory at a glance Classification Type Comparison Type Questions Selection Based on Given Conditions Family Based Problems Coding Decoding Letter and Numerical Coding Problems Answers

135 135 135 137 138 140 144 146 147 150

Chapter-12 Calendar Theory at a glance

151 151

Chapter-13 Clocks Theory at a glance General Concepts Important points Solved Problems Simple Problems

159 159 159 159 159 160

Chapter-14 Puzzles Theory at a glance

165 165


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India’s No. 1 Numbers IES Academy Chapter 1

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1. Numbers

Theory at a Glance (For GATE & PSU)

Introduction

1. Local Value A local value of a number is the face value of that number. For example the local value of 4 in 3248 is 4.

2. Place Value A place value of a number is the value of the place it occupies times its local value. For example the place of 4 in 3248 is forty.

3. Absolute Value The absolute value of a number is the number that remains after neglecting the sign of the given number. Absolute value of a number x is denoted by |x|. So, |x|= x if x is positive |x|= - x if x is negative For example |35| = 35 and |- 35| = + 35

Number Types in Numerical Aptitude Tests There are ten types of number generally used in numerical aptitude tests. They are:

1. Even Numbers All numbers which can be divided by 2 are called even numbers. For example 2, 4, 6, 8… are even numbers.

2. Odd Numbers All numbers which can’t be divided by 2 are called odd numbers, e.g. 1, 3, 5, 7, 9, … are called odd numbers.

3. Natural Numbers The natural flow of numbers starting from 1 are called natural numbers e.g. 1, 2, 3, 4 and so on.

4. Rational Numbers Any number in the form of p/q, where p and q are integers and q ≠ 0 and p & q are in lowest terms, is called a rational number. The set of all rational numbers is denoted by Q. Q = {x:x = p/q; p, q € 1, q ≠ 0} Every integer becomes a rational when we write it in the form of p/q. For example 3 = 3/1, - 8 = - 8/1



5. Irrational Numbers A number which can not be expressed in the form of p/q is called an irrational number. Numbers √2, √3, √5, √7, √11 etc are examples of the irrational numbers.

6. Real Numbers When rational and irrational numbers are combined together, the numbers in the series are called real numbers.

7. Complex Numbers When you do square root of negative numbers, the resulting numbers are not real numbers. Such numbers are called imaginary numbers. You can devote √-1 by ‘I’ then: √-5 = √-1 √5 = i√5 √-64 = √-1 √64 = 8i A number of the form a + ib where i = √-1 and a and b are real numbers, is called a complex number. The ‘a’ is known as its real part and ‘ib’ as its imaginary part. Thus, 2-3i is a complex number where real part is 2 and imaginary part is -3i.

8. Whole Numbers The whole numbers are all numbers in series starting from 0 as 0, 1, 2, 3, 4, 5,… and so on.

9. Prime Numbers The natural numbers which can be divided by either 1 or themselves only, are called prime number. For example the numbers 2, 3, 5, 7, 11, 12, 17, 19, 23, … and so on are called prime numbers.

10. Composite Numbers The numbers which are not prime and are divisible by at least one smaller natural number other than 1 are called composite numbers. For examples the composite numbers include 4, 6, 8, 9, 10, 12… and so on.

Prime Numbers A natural number larger than unity is a prime number if it does not have other divisors except for itself and unity. Note:- Unity i.e. 1 is not a prime number.

Properties of Prime Numbers • The lowest prime number is 2. • 2 is also the only even prime number. • The lowest odd prime number is 3. • The remainder when a prime number p ≥ 5 s divided by 6 is 1 or 5. However, if a

number on being divided by 6 gives a remainder 1 or 5 need not be prime. • The remainder of division of the square of a prime number p ≥ 5 divide by 24 is 1. • For prime numbers p>3, p²-1 is divided by 24. • If a and b are any 2 odd primes then a²-b² is composite. Also a²+b²is composite. • The remainder of the division of the square of a prime number p ≥ 5 divided by 12 is 1.



Process to Check A Number is Prime or not Take the square root of the number. Round of the square root to the next highest integer call this number as Z. Check for divisibility of the number N by all prime numbers below Z. If there is no numbers below the value of Z which divides N then the number will be prime. Example 239 is prime or not? √239 lies between 15 or 16.Hence take the value of Z = 16. Prime numbers less than 16 are 2, 3, 5, 7, 11 and 13. 239 is not divisible by any of these. Hence we can conclude that 239 is a prime number.

Composite Numbers The numbers which are not prime are known as composite numbers. Co-Primes Two numbers a an b are said to be Co-primes, if their H.C.F is 1. Example: (2,3), (4,5), (7,9), (8,11)..... Place value or Local value of a digit in a Number:

Place value Example : 689745132 Place value of 2 is (2 × 1) = 2 Place value of 3 is (3 × 10) = 30 and so on.

Face value It is the value of the digit itself at whatever place it may be. Example : 689745132 Face value of 2 is 2. Face value of 3 is 3 and so on.

Tests of Divisibility Divisibility by 2:- A number is divisible by 2, if its unit’s digit is any of 0, 2, 4, 6, 8. Example: 84932 is divisible by 2, while 65935 is not. Divisibility by 3:- A number is divisible by 3, if the sum of its digits is divisible by 3. Example1. 592482 is divisible by 3, since sum of its digit 5 + 9 + 2 + 4 + 8 + 2 = 30 which is divisible by 3. Example 2. 864329 is not divisible by 3, since sum of its digits 8 + 6 + 4 + 3 + 2 + 9 = 32 which is not divisible by 3. Divisibility by 4:- A number is divisible by 4, if the number formed by last two digits is divisible by 4.



Example1. 892648 is divisible by 4, since the number formed by the last two digits is 48 divisible by 4. Example 2. But 749282 is not divisible by 4, since the number formed by the last two digits is 82 is not divisible by 4. Divisibility by 5:- A number divisible by 5,if its unit’s digit is either 0 or 5. Example : 20820, 50345 Divisibility by 6:- If the number is divisible by both 2 and 3. Example: 35256 is clearly divisible by 2 sum of digits by 6. =3 + 5 + 2 + 5 + 21, which is divisible by 3 Thus the given number is divisible. Divisibility by 8:- A number is divisible by 8 if the last 3 digits of the number are divisible by 8. Divisibility by 11:- If the difference of the sum of the digits in the odd places and the sum of the digits in the even places is zero or divisible by 11. Example : 4832718 (8 + 7 + 3 + 4) - (1 + 2 + 8) =11 which is divisible by 11. Divisibility by 12:- All numbers divisible by 3 and 4 are divisible by 12. Divisibility by 7, 11, and 13:- The difference of the number of its thousand and the remainder of its division by 1000 is divisible by 7, 11, and 13.

Basic Formulae • (a + b)² = a² + b² + 2ab • (a - b)² = a² + b² - 2ab • (a + b)² - (a - b)² = 4ab • (a + b)² + (a - b)² = 2(a² + b²) • a² - b² = (a + b)(a - b) • (a - + b + c)² = a² + b² + c² + 2(ab + bc + ca) • a³ + b³ = (a + b)(a²+ b² - ab) • a³ - b³ = (a - b)(a² + b² + ab) • a³ + b³ + c³ - 3a bc = (a + b + c)(a² + b² + c² - ab - bc - ca) • If a + b + c = 0 then a³ + b³ + c³ = 3abc

Division Algorithm If we divide a number by another number, then Dividend = (Divisor × quotient) + Remainder

Multiplication by Short Cut Methods 1. Multiplication by distributive law: a) a (b + c) = ab + ac b) a (b - c) = ab - ac



2. Multiplication of a number by 5n:- Put n zeros to the right of the multiplicand and divide the number so formed by 2n.

Progression A succession of numbers formed and arranged in a definite order according to certain definite rule is called a progression. 1. Arithmetic Progression

If each term of a progression differs from its preceding term by a constant. This constant difference is called the common difference of the A.P. The n th term of this A.P is Tn = a(n - 1) + d. The sum of n terms of A.P is Sn = n/2[2a + (n - 1)d]. Important Results: a. 1 + 2 + 3 + 4 + 5...................... = n (n + 1)/2. b. 12 + 22 + 32 + 42 + 52......................=n(n + 1)(2n + 1)/6. c. 13 + 23 + 33 + 43 + 53......................=n2(n + 1)2/4

2. Geometric Progression A progression of numbers in which every term bears a constant ratio with its preceding

term. i.e. a, a r,ar2,ar3............... In G.P Tn = arn-1 Sum of n terms Sn=a(1-rn)/1-r

Problems 1. What could be the maximum value of Q in the following equation? 5PQ + 3R7 + 2Q8 = 1114

Solution: 5 P Q 3 R 7 2 Q 8 11 1 4 2 + P + Q + R = 11 Maximum value of Q = 11 – 2 = 9 (P = 0, R = 0)

2. Which of the following is a prime number? A. 241 B. 337 C. 391 Solution: A. 241 16 >√241. Hence take the value of Z = 16. Prime numbers less than 16 are 2, 3, 5, 7, 11 and 13. 241 is not divisible by any of these. Hence we can conclude that 241 is a prime number.

B. 337 19 > √337. Hence take the value of Z = 19. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13 and 17. 337 is not divisible by any of these. Hence we can conclude that 337 is a prime number.



C. 391 20 > √391. Hence take the value of Z = 20. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13, 17 and 19. 391 is divisible by 17. Hence we can conclude that 391 is not a prime number.

3. Find the unit’s digit n the product 2467 153 * 34172 ? Solution: Unit’s digit in the given product = Unit’s digit in 7 153 * 172 Now 7 4 gives unit digit 1 7 152 gives unit digit 1 7 153 gives 1*7 = 7. Also 172 gives 1 Hence unit’s digit in the product = 7*1 = 7.

4. Find the total number of prime factors in 411 *7 5 *112? Solution: 411 7 5 112 = (2*2) 11 *7 5 *112 = 222 *7 5 *112 Total number of prime factors = 22 + 5 + 2 = 29

5. What least value must be assigned to * so that that number 197*5462 is divisible by 9? Solution: Let the missing digit be x Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 + 2) = 34 + x For 34 + x to be divisible by 9, x must be replaced by 2 The digit in place of x must be 2.

6. What least number must be added to 3000 to obtain a number exactly divisible by 19? Solution: On dividing 3000 by 19 we get 17 as remainder Therefore number to be added = 19 – 17 = 2.

7. Find the smallest number of 6 digits which is exactly divisible by 111? Solution: Smallest number of 6 digits is 100000 On dividing 10000 by 111 we get 100 as remainder Number to be added = 111-100 =11. Hence, required number = 10011.

8. On dividing 15968 by a certain number the quotient is 89 and the remainder is 37. Find the divisor? Solution: Divisor = (Dividend-Remainder)/Quotient = (15968 - 37)/89 = 179.

9. A number when divided by 342 gives a remainder 47. When the same number is divided by 19 what would be the remainder? Solution: Number = 342K + 47 = 19*18K + 19*2 + 9 = 19(18K + 2) + 9. The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder.



10. Find the remainder when 231 is divided by 5? Solution: 210 = 1024.unit digit of 210 * 210 * 210 is 4 as 4*4*4 gives unit digit 4 unit digit of 231 is 8. Now 8 when divided by 5 gives 3 as remainder. 231 when divided by 5 gives 3 as remainder.

11. How many numbers between 11 and 90 are divisible by 7? Solution: The required numbers are 14, 21, 28,..........., 84 This is an A.P with a = 14, d = 7. Let it contain n terms then T = 84 = a + (n - 1)d = 14 + (n - 1)7 = 7 + 7n 7n = 77 = > n =11.

12. Find the sum of all odd numbers up to 100? Solution: The given numbers are 1, 3, 5.........99. This is an A.P with a = 1, d = 2. Let it contain n terms 1+ (n - 1)2 = 99 = > n = 50 Then required sum = n/2(first term +last term) = 50/2(1 + 99) = 2500

13. How many terms are there in 2, 4, 6, 8.........., 1024? Solution: Clearly 2, 4, 6........1024 form a G.P with a = 2, r = 2 Let the number of terms be n then 2 × 2n - 1 = 1024 2n - 1 = 512 = 29 n - 1 = 9 n = 10. 14. 2 + 22 + 23 + 24 + 25.......... + 28 = ?

Solution: Given series is a G.P with a = 2,r = 2 and n = 8. Sum Sn = a(1 - rn)/1 – r = Sn = 2(1 - 28)/1 - 2. = 2 × 255 = 510.

15. A positive number which when added to 1000 gives a sum, which is greater than when it is multiplied by 1000. The positive integer is? (a) 1 (b) 3 (c) 5 (d) 7 Solution: 1000 + N > 1000N Clearly N = 1.

16. The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by the sum of the original three numbers is equal to? (a) 18 (b) 22 (c) 36 (d) none Solution: Let the one digit numbers x,y,z Sum of all possible two digit numbers = (10x + y) + (10x + z) + (10y + x) + (10y + z) + (10z + x) + (10z + y)



= 22(x + y + z) Therefore sum of all possible two digit numbers when divide by sum of one digit numbers gives 22.

17. The sum of three prime numbers is 100. If one of them exceeds another by 36 then one of the numbers is? (a) 7 (b) 29 (c) 41 (d) 67 Solution: x + (x + 36) + y = 100 2x + y = 64 Therefore y must be even prime which is 2 2x + 2 = 64 = > x = 31. Third prime number = x + 36 = 31 + 36 = 67.

18. A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder. The number is? (a) 1220 (b) 1250 (c) 22030 (d) 220030 Solution: Number = (555 + 445) × (555 - 445) × 2 + 30 = (555 + 445) × 2 × 110 + 30 = 220000 + 30 = 220030.

19. The difference between two numbers s 1365. When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15. The smaller number is? (a) 240 (b) 270 (c) 295 (d) 360 Solution: Let the smaller number be x, then larger number = 1365 + x Therefore 1365 + x = 6x + 15 5x =1350 = > x = 270 Required number is 270.

20. In doing a division of a question with zero remainder, a candidate took 12 as divisor instead of 21. The quotient obtained by him was 35. The correct quotient is? (a) 0 (b) 12 (c) 13 (d) 20 Solution: Dividend = 12× 35 = 420. Now dividend = 420 and divisor = 21. Therefore correct quotient = 420/21 = 20.

21. In a garden, there are some trees arranged in certain number of rows. ist row has i trees and each tree of that row has i2 flowers. The total number of flowers in the garden is 55 times the total number of trees in it. Find the number of rows in the garden. (a) 9 (b) 10 (c) 11 (d) 12

Ans. (b)

22. A total of 1540 steel balls are stacked in a pile. The top layer has 1 ball. The layer below it has 3 balls. The layer below it has 6 balls. The layer below it has 10 balls and so on. How many horizontal layers are there in the pile? (a) 18 (b) 20 (c) 22 (d) 24

Ans. (b)



23. A number between 100 and 5000 has the sum of its distinct prime factors equal to 10. It has an odd number of factors. If it can be expressed as a product of two numbers co-prime to each other in the minimum number of ways. How many values can it assume? (a) 1 (b) 2 (c) 3 (d) 4

Ans. (b)

24. A group of 105 children’s are standing in N rows for a group photograph. Each row had three children less than the row in front of it. Which of the following cannot be a possible value of N? (a) 3 (b) 4 (c) 5 (d) 6

Ans. (b)

25. The cost of 10 biscuits, 12 chocolates and 15 ice creams is Rs. 315. The cost of 12 biscuits, 15 chocolates, 19 ice creams is Rs. 389. The cost of 56 biscuits and x ice creams is Rs. 1797. Find the value of x for which a unique value for the cost of each item cannot be determined. (a) 68 (b) 77 (c) 81 (d) 87

Ans. (d)



Students Notes

Aptitude

Documents

numbers theory

irrational numbers

real numbers

number x

natural flow of numbers

number types

types of number

absolute value