REASONING

# QUESTIONS ON REASONING (PART-5)

#### 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*2 = 784; 784 = 28*28

RHS
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’). क्योंकि 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 of alternative letters 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 after leaving two zeroes on the right 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 is 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 square root of the first number to get the 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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