REASONING

# QUESTIONS ON REASONING (PART-5)

#### QUESTIONS ON REASONING (PART-5)

Most of these questions are taken from the previous examinations conducted by the Staff Selection Commission (SSC) of the General Intelligence and Reasoning section of the following exams as well as other exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Tier-I

4. Stenographers Exam

5. Grade-II DASS Exam conducted by Delhi Staff Subordinate Services (DSSSB)

#### QUERY 81

121     156     105
145     187     126
115     190       ?

255
305
231
225

MAHA GUPTA
156 – 121 = 35; 35*3 = 105
187 – 145 = 42; 42*3 = 126

Therefore the required number is 190 – 115 = 75; 75*3 = 225 (option ‘4’)

#### QUERY 82

3 : 7 :: 15 : ?

45
49
30
35

MAHA GUPTA
LHS:

Second term = First term*2 + first term/3 = 3*2 + 3/3 = 7
RHS:
Second term = First term*2 + first term/3 = 15*2 + 15/3 = 35 (option ‘4’)

#### QUERY 83

583 : 488 :: 293 : ?

378
487
581
291

MAHA GUPTA
LHS:
(4+8+8) – (5+8+3) = 4
So we have to see among the answer options which is that number whose digit sum is more by 4 that of the first number.

Obviously it’s 378 (option ‘1’)
(3+7+8) – (2+9+3) = 4

#### QUERY 84

1, 2, 2, 4, 8, ?

8
9
16
32

MAHA GUPTA
Take the first two numbers i. e. 1 and 2; then multiply them to get the third number, and so on. See how

1*2 = 2
2*2 = 4
2*4 = 8

Therefore the required number = 4*8 = 32 (option’4′)

Also see this
1 x 2 = 2 (12)
2 x 2 = 4 (1224)
2 x 4 = 8 (12248)
4 x 8 = 32 (1224832)

#### QUERY 85

24 : 60 :: 120 : ?

160
220
300
108

MAHA GUPTA
LHS
24*2 +24/2 = 60

Therefore RHS
120*2 + 120/2 = 300 (option ‘3’)

6 : 222 :: 7 : ?

Md Imran
LHS
6^3 + 6 = 222

Therefore RHS
7^3 + 7 = 350

#### QUERY 87

If 12*7 = 408
9*8 = 207
then 13*7 = ?

MAHA GUPTA
Insert ‘0’ between the digits in reverse order of the product
Therefore 13*7 = 109

#### QUERY 88

122 : 170 :: 290 : ?

MAHA GUPTA
LHS is addition of 1 to squares of consecutive prime numbers 11 and 13; see how
11^2 + 1 = 122
13^2 + 1 = 170

LHS
The first number is addition of 1 to the square of prime number 17, i.e. 17^2 + 1 = 290

Likewise the required number will be addition of 1 to the square of the next prime number 19, i.e. 19^2 + 1 = 362

#### QUERY 89

If 2+3+5=30, 3+4+6=72, 5+6+2=60; then 5+4+0=?

MAHA GUPTA
In every set the RHS is equal to the product of all the numbers on LHS

Therefore the required number is 5*4*0 = 0

#### QUERY 90

392 : 28 :: 722 : ?

18
28
38
48

MAHA GUPTA
LHS
392; 392*2 = 784; 784 = 28*28; 28

RHS
722; 722*2 = 1444; 1444 = 38*38
Therefore the desired number is 38 (option ‘3’)

#### QUERY 91

Choose odd number
(A) 240     (B) 304     (C) 272     (D) 210

SHIV KISHOR
304 (option ‘B’). क्यूंकि 15*15 -15 = 210; 16*16 -16 = 240; 17*17 – 17 = 272 लेकिन 304 का ऐसा कोई सम्बन्ध नही है।

#### QUERY 92

a k, e o, i s, …….. , q a, u e

MAHA GUPTA
Each letter in every set is increasing by 4; so m w is the right answer.

#### QUERY 93

11 × 12 × 13 = 234, 24 × 23 × 35 = 658; then 31 × 43 × 54 = ?

497
974
749
479

MAHA GUPTA
Every digit of RHS is equal to the sum of every number of LHS in order
So 31 × 43 × 54 = ? ==> 3+1, 4+3; 5+4 = 479 (option ‘4’)

#### QUERY 94

8 : 18 : : 24 : ?

38
32
44
43

Krishan Sharma
38 (option ‘1’)
3 ka squre-1 : 4 ka squre+2 :: 5 ka squre-1 : 6 ka squre+2

#### QUERY 95

3, 5, 8, 13, 20, 31, ?

MAHA GUPTA
The difference in numbers is increasing in consecutive prime numbers; so it will be 31+13 = 44 (answer)

See the difference: 2, 3, 5, 7, 11. So it will be 13 next.

#### QUERY 96

5       1       25
6       2       18
10     4      25
3       3       ?

10
9
4
3

MAHA GUPTA
The last number in each row is equal to the product of the first number by the quotient of first number divided by the second; see how

5*(5/1) = 25
6*(6/2) = 18
10*(10/4) = 25

Hence the required number = 3*(3/3) = 3 (option ‘4’)

#### QUERY 97

600 : 2000 : : ? : 9000

5400
5600
3000
3600

MAHA GUPTA
If you see every number comprising leaving two zeroes on the right here is sum of squares of a number with addition to the same number; and the difference in those numbers is increasing by 2 in the second number. See how
(2^2 + 2)*100 = 600
(4^2 + 4)*100 = 2000

(9^2 + 9)*100 = 9000
Therefore the required number is (7^2 + 7)*100 = 5600 (option ‘2’)

#### QUERY 98

2, 7, 9, 3, 8, 11, 4, 9, 13, ?

MAHA GUPTA
The series actually a series of the set of first three numbers i.e. 2, 7, 9. The first number is increasing by 1 in each set, the second number is again increasing by 1 and the third number by two. But here we need the first number of the set of three, so it must be more than 4 by 1.

Hence it’s 4+1 = 5

#### QUERY 99

If 54 + 43 = 2, 60 + 51 = 10; then
62 + 72 = ?

9
18
20
30

MAHA GUPTA
Add the subtraction of the digits of each number to get the RHS; see how
5 – 4 = 1; 4 – 3 = 1; RHS = 1 + 1 = 2
6 – 0 = 6; 5 – 1 = 4; RHS = 6 + 4 = 10

Therefore the required number:
6 – 2 = 4; 7 – 2 = 5; RHS = 4 + 5 = 9 (option ‘1’)

#### QUERY 100

If 16 – 2 = 2, 9 – 3 = 0, 81 – 1 = 8; then 64 – 4 = ?

2
4
6
8

MAHA GUPTA
Subtract the second number from the root of the first number to get RHS; see how
root16 = 4; RHS = 4 – 2 = 2
root9 = 3; RHS = 3 – 3 = 0
root81 = 9; RHS = 9 – 1 = 8

Therefore the desired number
root 64 = 8; RHS = 8 – 4 = 4 (option ‘2’)

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