Most Expected Number Series Questions for IBPS RRB Clerk PDF
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Instructions
Find the wrong number in given series sequence.
Question 1: 1, 4, 15, 64, 325, 1955
a) 15
b) 64
c) 325
d) 1955
e) None of these
Question 2: 6, 12, 21, 33, 49, 66
a) 21
b) 33
c) 49
d) 66
e) None of these
Question 3: 6, 11.5, 19, 28.5, 41
a) 6
b) 11.5
c) 41
d) 28.5
e) None of these
Question 4: 5, 26, 82, 214, 401, 702
a) 26
b) 82
c) 214
d) 401
e) None of these
Question 5: 5,20,73,274,1049
a) 20
b) 73
c) 274
d) 1049
e) None of these
Question 6: Which number comes in the place of the question mark (?)16, 25, 36, 49, 64, ?
a) 121
b) 100
c) 81
d) 144
e) 196
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Question 7: Which number comes in the place of the question mark (?)1, 27, 125, 343, ?
a) 512
b) 1000
c) 729
d) 216
e) 1024
Question 8: Which number comes in the place of the question mark (?)11, 44, 99, 176, ?
a) 274
b) 266
c) 275
d) 265
e) 300
Question 9: Which number comes in the place of the question mark (?)5, 25, 61, 113, ?
a) 144
b) 181
c) 165
d) 200
e) 245
Question 10: Which number comes in the place of the question mark (?)2, 6, 12, 20, 30, ?
a) 50
b) 44
c) 45
d) 42
e) 55
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Instructions
What should come in place of the question mark (?) in the following number series ?
Question 11: 353 354 351 356 349 ?
a) 348
b) 358
c) 338
d) 385
e) 340
Question 12: 1 5 13 29 ? 125 253
a) 83
b) 69
c) 61
d) 65
e) 81
Question 13: 45 57 81 117 165 ?
a) 235
b) 215
c) 205
d) 245
e) 225
Question 14: 17 18 26 53 117 ? 458
a) 342
b) 142
c) 257
d) 262
e) 242
Question 15: $\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 1\frac{1}{4}, 1\frac{1}{2}, 1 \frac{3}{4}, ?$
a) $2$
b) $4$
c) $6$
d) $1\frac{1}{5}$
e) $1\frac{2}{3}$
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Answers & Solutions:
1) Answer (D)
4 = 1*2+2 ; 15 = 4*3+3 ; 64 = 15*4+4 ; 325 = 64*5+5 ; 1956 = 325*6+6
The nth term is of the form,
$T_n=(T_{n-1}\times n)+n$
The last term does not follow the pattern and is thus the wrong number in the sequence.
2) Answer (C)
6+6 = 12
12+9 = 21
21+12 = 33
33+15 = 48
48+18 = 66.
Hence, 49 is wrong number and it should be 48.
3) Answer (C)
1.5*2 + 3 = 6
2.5*3 + 4 = 11.5
3.5*4 + 5 = 19
4.5*5 + 6 = 28.5
5.5*6 + 7 = 40
4) Answer (C)
1*2+3=5 ; 4*5+6 = 26 ; 8*9+10 = 82 ; 13*14+15 = 212 ; 19*20 + 21 =401 ; 26*27 +28 = 702
All the numbers except 214 are following a pattern of $n\times (n+1) + (n+2)$. Hence, 214 is wrong number.
5) Answer (C)
1^2 + 4= 5 ; 2^2 + 16 = 20 ; 3^2 + 64 = 73 ; 4^2 + 256 = 272 ; 5^2 + 1024 = 1049
6) Answer (C)
$4^2 = 16$ ; $5^2=25$ ; $6^2 = 36$ ; $7^2=49$ ; $8^2 = 64$ ; $9^2 = 81$
7) Answer (C)
$1^3 = 1$ ; $3^3 = 27$ ; $5^3 = 125$ ; $7^3=343$ ; $9^3 = 729$
8) Answer (C)
$1 \times11 = 11$ ; $2 \times 22=44$ ; $3 \times 33=99$ ; $4 \times 44=176$ ; $5 \times 55=275$
9) Answer (B)
$1^2 + 2^2 = 5 $; $3^2 + 4^2 = 25$ ; $5^2 +6^2 = 61$ , $7^2 + 8^2 = 113$ ; $9^2 + 10^2 = 181$
10) Answer (D)
2+4=6 ; 6+6 = 12 ; 12+8 = 20 ; 20+10=30 ; 30+12 = 42
11) Answer (B)
There are two arithmetic progression series,
With 353,351,349… having a difference of -2
and 354, 356… having a difference of 2
Hence, the required term is 358
12) Answer (C)
The generalized formula for nth term is, previous term + 2^n
Here the required term is 5th in the series.
Hence, 29 + 2^5 = 61
13) Answer (E)
For the series the nth term is ‘last term + 12*(n-1)’
In the series, we are required to find, 6th term.
Hence, 6th term is 165+12*(5) = 225.
14) Answer (E)
Here, each sucessive nth term is ‘previous term + (n-1)³’
We are supposed to find 6th term.
Hence, 6th term = 117 + (6-1)^3 = 117 + 125 = 242
15) Answer (A)
The given series is Arithmetic Progression with common difference $\frac{1}{4}$.
Hence, the next term will be $1\frac{3}{4}$ + $\frac{1}{4}$ = 2
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