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# Number Series Questions For IBPS RRB PO Set-2 PDF

Download Top-20 IBPS RRB PO Number Series Questions PDF. Number Series questions based on asked questions in previous year exam papers very important for the IBPS RRB PO (Officer Scale-I, II & III) exam

Download IBPS RRB PO Previous Papers PDF

Instructions

Select the option which replaces (?)

Question 1: 7, 11, 23, 51, 103, ?

a) 187

b) 193

c) 195

d) 185

e) 173

Question 2: 16, 23, 28, 38, ?, 62

a) 51

b) 47

c) 53

d) 45

e) 49

Question 3: 2, 6, 30, 60, 130, ?

a) 220

b) 210

c) 200

d) 190

e) 180

Question 4: 23, 39, 71, 119, ?

a) 192

b) 181

c) 174

d) 183

e) 185

Instructions

Find the next term in the series.

Question 5: 11, 26, 51, 76, 115, 174, ?

a) 245

b) 243

c) 217

d) 219

e) 231

Question 6: 221, 222, 223, 225, 228, 233, 241, 254, ?

a) 263

b) 296

c) 308

d) 275

e) 302

Question 7: 462, 506, 552, 600, 650, ?

a) 806

b) 718

c) 750

d) 812

e) 702

Question 8: 972, 648, 432, 288, ?

a) 186

b) 192

c) 212

d) 188

e) 200

Question 9: 17, 34, 31, 124, 119, 714, ?

a) 4284

b) 743

c) 707

d) 711

e) 728

Instructions

Identify the number which will replace the question mark (?).

Question 10: 11, 13, 16, 21, 28, 39, 52, ?

a) 89

b) 83

c) 69

d) 67

e) 72

Question 11: 23, 31, 29, 35, 35, 39?

a) 39

b) 44

c) 47

d) 41

e) 37

Question 12: 123, 124, 120, 129, 113, 138, ?

a) 102

b) 152

c) 172

d) 113

e) 165

Question 13: 67, 68, 76, ?, 167

a) 112

b) 103

c) 97

d) 107

e) 113

Question 14: 17, 25, 32, 37, ?, 58

a) 51

b) 47

c) 43

d) 49

e) 45

Instructions

Find the missing number in the series.

Question 15: 14, 30, 63, 130, 265, ?

a) 428

b) 447

c) 489

d) 536

e) 545

Question 16: 31, 44, 88, 176, 847, 1595, ?

a) 5586

b) 2534

c) 6537

d) 1892

e) 7546

Question 17: 2, 12, 36, 80, 150, ?

a) 212

b) 272

c) 252

d) 220

e) 290

Question 18: 13, 18, 25, 34, 45, ?

a) 58

b) 54

c) 67

d) 59

e) 65

Question 19: 113, 118, 128, 139, 152, ?

a) 172

b) 159

c) 166

d) 160

e) 162

Question 20: Select the option that logically continues the series:

-2, -3, 8, 95, ?

a) 673

b) 684

c) 691

d) 672

e) 671

Taking the difference of consecutive numbers, we have,
11 – 7 = 4
23 – 11 = 12
51 – 23 = 28
103-51 = 52
Takin difference of the difference of numbers, we have,
12- 4 =8
23-12 = 16
52-28 = 24
Accordingly, 52 + 32 = 84
So, the next number must be
103 + 84 = 187
Hence, option A is the right choice.

16 + 1 +6 = 23
23 + 2 + 3 = 28
28 + 2 + 8 = 38
38 + 3 + 8 = 49
49 + 4+ 9 = 62
Hence, option E is the correct option.

$1^3$ + 1 = 2
$2^3$ – 2 = 6
$3^3$ + 3 = 30
$4^3$ – 4 = 60
$5^3$ + 5 = 130
$6^3$ – 6 = 210
Hence, option B is the correct option.

23+16 = 39
39 + 16*2 = 71
71 + 16*3 = 119
119 + 16*4 = 183
Hence, option D is the correct option.

The continuous prime numbers are multiplied by 1, 2, 3 and so on

11
13 * 2 = 26
17 * 3 = 51
19 * 4 = 76
23 * 5 = 115
29 * 6 = 174
31 * 7 = 217

The difference forms a Fibonacci series
1, 1, 2, 3, 5, 8, 13, 21
The series is:
221
221 + 1 = 222
222 + 1 = 223
223 + 2 = 225
225 + 3 = 228
228 + 5 = 233
233 + 8 = 241
241 + 13 = 254
254 + 21 = 275

Hence, 254 + 21 = 275 is the answer.

The series follows the following pattern:
21 * 22 = 462
22* 23 = 506
23 * 24 = 552
24*25 = 600
25*26 = 650
26*27 = 702

The series follows the following pattern:
972 * (2/3) = 648
648 * (2/3) = 432
432 * (2/3) = 288
288 * (2/3) = 192

The series follows the following pattern:
17 * 2 = 34
34 – 3 = 31
31 * 4 = 124
124 – 5 = 119
119 * 6 = 714
714 – 7 = 707

Consecutive prime numbers are being added in successive terms.
11 + 2 =13
13 + 3 = 16
16 + 5 = 21
21 + 7 = 28
28 + 11 = 39
39 + 13 = 52
Hence, the next term will be
52 + 17 = 69

There are two series in the given set of numbers.
The numbers at odd places form 1 series which is given by
23, 29, 35
The numbers at the even places form the second sequence which is given by
31, 35, 39
Hence the next term would be 35+6 = 41

The pattern being followed is
$123 + 1^{2} = 124$
$124 – 2^{2} = 120$
$120 + 3^{2} = 129$
$129 – 4^{2} = 113$
$113 + 5^{2} = 138$
$138 + 5^{2} = 138$
So the next term would be
$138 – 6^{2} = 102$

The difference between the two consecutive terms is the cubes of natural numbers. For example
67+1 = 68
68+8 = 76
76 + 27 = 103
103 + 64 = 167
Hence, the missing term is 103
Thus, option B is the correct answer.

The sum of the digits of the previous number is being added to the number to get the new number.
For example:
17 + 1 + 7 = 25
25 + 2 + 5 = 32
32 + 3 + 2 = 37
37 + 3 + 7 = 47
47 + 4 + 7 = 58
Hence, the missing number is 47.

The series follows the following pattern:
14
14*2 + 2 = 30
30 * 2 + 3 = 63
63 * 2 + 4 = 130
130*2 + 5 = 265
265*2 + 6 = 536

The series follows the following pattern:
31 + (13) = 44 (reverse of the number is added)
44 + 44 = 88
88 + 88 = 176
176 + 671 = 847
847 + 748 = 1595
1595 + 5951 = 7546

The series follows the following pattern:
$1^2 + 1^3 = 2$
$2^2 + 2^3 = 12$
$3^2 + 3^3 = 36$
$4^2 + 4^3 = 80$
$5^2 + 5^3 = 150$
$6^2 + 6^3 = 252$

The series follows the following pattern:
13 + 5 = 18
18 + 7 = 25
25 + 9 = 34
34 + 11 = 45
45 + 13 = 58

The series follows the following pattern:
113
113 + (1 + 1 + 3) = 118
118 + (1 + 1 + 8) = 128
128 + (1 + 2 + 8) = 139
139 + (1 + 3 + 9) = 152
152 + ( 1 + 5 + 2) = 160

The given numbers are of the form $n! – n^2$.
$2!-2^2$ = $2-4$=$-2$
$3!-3^2$ = $6-9$=$-3$
$4!-4^2$ = $24-16$=$8$
$5!-5^2$ = $120-25$=$95$
$6!-6^2$ = $720-36$=$684$