Aptitude

Aptitude
Contents
Chapters Topic Page No.
Chapter-1 Introduction 1
Theory at a glance 1
1. Local Value 1
2. Place Value 1
3. Absolute Value 1
Number Types in Numerical Aptitude Tests 1
1. Even Numbers 1
2. Odd Numbers 1
3. Natural Numbers 1
4. Rational Numbers 2
5. Irrational Numbers 2
6. Real Numbers 2
7. Complex Numbers 2
8. Whole Numbers 2
9. Prime Numbers 2
10. Composite Numbers 2
Prime Numbers 2
Properties of Prime Numbers 2
Process to Check A Number is Prime or not 3
Example 239 is prime or not? 3
Composite Numbers 3
Co-Primes 3
Face value 3
Tests of Divisibility 3
Basic Formulae 4
Division Algorithm 4
Multiplication by Short Cut Methods 4
Progression 5
Problems 5
Chapter-2 Ratio and Proportion 11
Problem 11

India’s No. 1 Aptitude
IES Academy Contents

Chapters Topic Page No.
Chapter-3 Partnership 15
Ratio of Division of Gains: 15
Formulae 15
15
Short cuts
16
Solved Problems
17
Problems on Partnership
Chapter-4 Percentages 23
Profit and Loss 30
Formulae: 30
Solved Problems 30

Chapter-5 Simple Interest 37
Important Facts and Formulae: 37
Compound Interest 42
Formulae: 42
Solved Examples 42
More solved problem 44
Chapter-6 Time & Work 49
1. General Rules 49
Solved Problems 49
More Solved Problems 51
2. Pipes and Cisterns 60
Important Facts 60
Complex Problems 61
Problems on Time and Work 64
Answers Key 67
Exercise on Pipes and Cisterns 68
Answers Key 70
Chapter-7 Time Speed and Distance 71
Solved Problems 71
1. Boats & Streams 74
Important Points 74
Solved problems 74
2. Trains 75
General Concepts 75
Solved Examples 76
More Solved Problems 77
Problems 80
Answers 86
Problems on Boats and Streams 87
Answers 89
Problems on Trains 90
Answers 94

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Chapter-8 Geometry 95
Areas in Geometry 95
Volumes in Geometry 97
Surface area in Geometry 99
Properties of Triangles 99
Solved Examples 100
Problems related to Area 102
Chapter-9 Permutation and Combination 109
Permutation 110
Examples 111
Circular Permutations 112
Restricted – Permutations 113
Restricted – Combinations 114
Restricted Permutations 115
Restricted – Combinations 116
Chapter-10 Probability 119
Basic Concepts 119
Part (1) : Foundation Level 120
Important types of Events: 121
Category - B 126
Category - C 130
Part (2) : (Total Probability) 131
Examples 132
Chapter-11 Reasoning 135
Classification Type 135
Comparison Type Questions 137
Selection Based on Given Conditions 138
140
Family Based Problems
144
Coding Decoding
146
Letter and Numerical Coding
147
Problems
150
Answers
Chapter-12 Calendar 151
Chapter-13 Clocks 159
General Concepts 159
Important points 159
Solved Problems 159
Simple Problems 160
Chapter-14 Puzzles 165

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India’s No. 1 Numbers
IES Academy Chapter 1

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
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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.

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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.

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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

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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.

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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.

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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)

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= 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)
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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)

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Students Notes

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