# 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’)

QUERY 86

**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’)