REASONING

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)

 

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Maha Gupta

Maha Gupta

Founder of www.examscomp.com and guiding aspirants on SSC exam affairs since 2010 when objective pattern of exams was introduced first in SSC. Also the author of the following books:

1. Maha English Grammar (for Competitive Exams)
2. Maha English Practice Sets (for Competitive Exams)