# QUESTIONS ON REASONING (PART-11)

#### QUESTIONS ON REASONING (PART-11)

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 similar 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 201

**If 3454 = 5, 2332 = 5, then 5245 = ?**

5

10

7

8

**MAHA GUPTA**

RHS = Addition of 5 to the subtraction of the sum of second and fourth digits from the sum of first and third digits from the left; see how

(3+5) – (4+4) + 5 = 5

(2+3) – (3+2) + 5 = 5

Therefore the required number = (5+4) – (2+5) + 5 = 7 (option ‘3’)

QUERY 202

**23, 51, 108, 225, ?, 936**

327

628

459

531

**MAHA GUPTA**

Add sum of the digits to the twice of every number to get the next one, see how

23*2 + (2+3) = 51

51*2 + (5+1) = 108

108*2 + (1+0+8) = 225

225*2 + (2+2+5) = 459 (option ‘3’)

459*2 + (4+5+9) = 936

QUERY 203

**2, 5, 8, 17, 26, 53, ?**

**MAHA GUPTA**

To get the numbers at even places of the sequence add 1 to twice of their previous number, and to get the numbers at odd places of the sequence add difference of its two preceding numbers to it’s previous; see how

Numbers at even places

2*2 + 1 = 5

8*2 + 1 = 17

26*2 +1 = 53

Numbers at odd places

5 + (5 – 2) = 8

17 + (17 – 8) = 26

Therefore the required number

53 + (53 – 26) = 80 (answer)

QUERY 204

**5 : 135::7 : ?**

353

245

273

293

** SHIV KISHOR
**5^3 + 10 = 135

Similarly 7^3 + 10 = 353 (option ‘1’)

QUERY 205

**3, 14, 5, 30, 7, ?**

** KUMAR SAURABH**

See the pattern in every set of two number starting from the beginning

3, 3^2 + 5

5, 5^2 + 5

Therefore the required number

7, 7^2 + 5 = 54 (answer)

QUERY 206

**7, 25, 61, 121, ?**

210

211

212

209

**MAHA GUPTA**

2^3 – 1 = 7

3^3 – 2 = 27

4^4 – 3 = 61

5^3 – 4 = 121

Therefore the required number

6^3 – 5 = 211 (option ‘2’)

QUERY 207

**16 : 4 :: 9 : ?**

4

25

1

2

** SAURABH KUMAR**

4^2 = 16

2^2 = 4

Therefore, the pair on RHS will also be squares of alternate numbers, see how

3^2 = 9

1^2 = 1 (option ‘3’)

QUERY 208

**678 = 366**

** 567 = 225**

** 946 = ?**

** KAPIL KATEWA**

Subtract 20 from the product of the last two digits of LHS and write the left digit to the right of that product to get the number on RHS, see how

8*7 – 20 = 36 => 366

7*6 – 20 = 22 => 225

Therefore the required number

6*4 – 20 4 = 4 => 49

QUERY 209

**5*6 = 35**

** 8*4 = 28**

** 6*8 = ?**

46

34

23

38

** KARTIK RUPANI**

Shift left digit of LHS to the right and write half of the right digit on the left to get the number on RHS; see how

6/2 = 3 => 35

4/2 = 2 => 28

Therefore, the required number

8/2 = 4 => 46 (option ‘A’)

QUERY 210

**81 : 8 :: ? : ?**

128 : 13

136 : 9

124 : 5

144 : 11

** RAWT SAURABH**

(8+1)^2 : 8

Therefore the needed pair

(11+1)^2 : 11

= 144 : 11 (option ‘4’)

QUERY 211

**13 ÷ 14 = 12**

** 15 ÷ 16 = 30**

** 21 ÷ 27 = 28**

** then**

** 31 ÷ 29 = ?**

**LUV KUMAR**

Find the product of the digits on the LHS to get RHS, see how

1*3*1*4 = 12

1*5*1*6 = 30

2*1*2*7 = 28

Therefore the required number

3*1*2*9 = 54 (answer)

QUERY 212

**9/8, 9/4, 27/8, 9/2, ? , ?**

** SHIV KISHOR**

(9/8)*(2/1) = 9/4

(9/4)*(3/2) = 27/8

(27/8)*(4/3) = 9/2

Therefore the next numbers

(9/2)*(5/4) = 45/8

(45/8)*(6/5) = 27/4

QUERY 213

**2, 9, 30, ?, 436, 2195, 13182**

216

105

324

178

**MAHA GUPTA**

9 = (2 + 7)*1

30 = (9 + 6)*2

105 = (30 + 5)*3 (option ‘2’)

436 = (105 + 4)*4

2195 = (436 + 3)*5

13182 = (2195 + 2)*6

QUERY 214

**Find odd one out in following sequence**

** 10, 25, 45, 54, 60, 75, 80**

54

75

10

45

**MAHA GUPTA**

All numbers except 54 are multiples of 5; so 54 (option ‘1’) is correct.

QUERY 215

**95, 115 , 145, 155, ?**

215

175

185

165

**MAHA GUPTA**

Every number in the sequence is product of 5 and a consecutive prime number beginning with 19 see how

19*5 = 95

23*5 = 115

29*5 = 145

31*5 = 155

Therefore the required number

37*5 = 185 (option ‘3’)

QUERY 216

**9 14 22**

** 3 2 11**

** 11 13 44**

** 33 91 x**

**Aarav SINGH**

The first and the last number in each column are 3, 7, and two times of the second and third respectively, see how

9 = 3*3

33 = 11*3

14 = 2*7

91 = 13*7

22 = 11*2

x = 44*2; means x = 88 (answer)

QUERY 217

**200 —–(11)—–324**

** 329 ——(X)—–137**

**MAHA GUPTA**

Some of the digits of the end numbers is equal to the middle number, see how

2 + 0 + 0 + 3 + 2 + 4 = 11

Therefore the required number

3 + 2 + 9 + 1 + 3 + 7 = 25 (answer)

QUERY 218

**Student : books :: postman : ?**

delivery

mail

uniform

bicycle

**MAHA GUPTA**

MAIL (option ‘2’). The primary work of a student is dealing with books, the primary work of a postman is delivering of mail. Si MAIL is correct.

QUERY 219

**If**

** 2 – 3 = 61**

** 4 – 7 = 283**

** 5 – 9 = 454**

** Then**

** 1 – 8 = ?**

**MAHA GUPTA**

To find the number on the RHS, write the difference of the number at the unit digit and put the product of both the numbers on the left of it, see how

3 – 2 = 1; 3*2 = 6 => 61

7 – 4 = 3; 7*4 = 28 => 283

9 – 5 = 4; 9*5 = 45 => 454

Therefore the required number

8 – 1 = 7; 8*1 = 8 => 87 (answer)