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# Number System Questions For SBI PO

Instructions: In the following number series only one number is wrong. Find out the wrong number.

Question 1: 7, 12, 40, 222, 1742, 17390, 208608

a) 7
b) 12
c) 40
d) 1742
e)208608

Question 2:6, 91, 584, 2935, 11756, 35277, 70558

a) 91
b) 70558
c) 584
d) 2935
e)35277

Find the wrong number in given series sequence.

Question 3: 1, 4, 15, 64 325, 1955

a) 15
b) 64
c) 325
d) 1955
e)None of these

Instructions: In the following number series only one number is wrong. Find out the wrong number.

Question 4: 1, 4, 25, 256, 3125, 46656, 823543

a) 3125
b) 823543
c) 46656
d) 25
e)256

Question 5: 8424, 4212, 2106, 1051, 526.5, 263.25, 131.625

a) 131.625
b) 1051
c) 4212
d) 8424
e)263.25

Question 6: 5531 5506 5425 5304 5135 4910 4621

a) 5531
b) 5425
c) 4621
d) 5135
e)5506

Question 7: 6 7 9 13 26 37 68

a) 7
b) 26
c) 69
d) 37
e)9

Question 8: 1 3 10 36 152 760 4632

a) 3
b) 36
c) 4632
d) 760
e)152

Question 9: 4 3 9 34 96 219 435

a) 4
b) 9
c) 34
d) 435
e)219

Question 10: 157.5 45 15 6 3 2 1
a) 1
b) 2
c) 6
d) 157.5
e)45

(7-1) x 2 = 12
(12 – 2) x 4 = 40
(40 – 3) x 6 = 222
(222 – 4) x 8 = 1744 â‰  1742
(1744 – 5) x 10 = 17390
(17390 – 6) x 12 = 208608

6 x 7 + 72 = 91 , 91 x 6 + 62 = 582 â‰  584 , 582 x 5 + 52 = 2935,…

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,
Tn=(Tnâˆ’1Ã—n)+nT_n=(T_{n-1}\times n)+nTnâ€‹=(Tnâˆ’1â€‹Ã—n)+n
The last term does not follow the pattern and is thus the wrong number in the sequence.

11 = 1 , 22 = 4 , 33 = 27 â‰  25 , 44 = 256 , 55 = 3125 , 66 = 46656 , 77 = 823543

$\frac{8424}{2}$ = 4212

$\frac{4212}{2}$ = 2106

$\frac{2106}{2}$ = 1053 â‰  1051

$\frac{1051}{2}$ = 526.5

$\frac{525.5}{2}$ = 263.25

$\frac{263.25}{2}$ = 131.625

5531 – 52 = 5506
5506 – 92 = 5425
5425 – 112 = 5304
5134 – 132 = 5135
5135 – 152 = 4910
4910 – 172 = 4621

Ever number after 5531 has a difference of a square.
Hence 5531 is the wrong number.

6 + 1 = 7
7 + 2 = 9
9 + 4 = 13
13 + 13 = 26
26 + 11 = 37
37 + 22 = 69
The number to be added is increasing everytime except for 37 where 11 is added after 13.

(1 x 1) + 2 = 3
(3 x 2) + 4 = 10
(10 x 3) + 6 = 36
(36 x 4) + 8 = 152
(152 x 5) + 10 = 770 â‰ 760
(770 x 6) + 12 = 4632

02 + 4 = 4
12 + 2 = 3
32 + 0 = 9
62 – 2 = 34
102 – 4 = 96
152 – 6 = 219
212 – 8 = 433 â‰  435