Number Series Questions for IBPS PO and RRB Prelims

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Number Series Questions for IBPS PO and RRB Prelims

Here you can download a free Number Series questions PDF with answers for IBPS PO and IBPS RRB PO 2022 by Cracku. These are some tricky questions in the IBPS PO and IBPS RRB PO 2022 exam that you need to find the Number Series for the given questions. These questions will help you to do practice and solve the Number Series questions in the IBPS PO and IBPS RRB PO exams. Utilize this best PDF practice set which is included answers in detail. Click on the below link to download the Number Series MCQ PDF for IBPS PO and IBPS RRB PO 2022 for free.

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Instructions

What will come in place of question mark (?) in the following series.

Question 1: 62, ?, 514, 1032, 4130, 8264, 33058

a) 132

b) 136

c) 140

d) 128

e) None of the above

1) Answer (D)

Solution:

$62\times2+4 = 124+4 = 128$
$128\times4+2 = 512+2 = 514$
$514\times2+4 = 1028+4 = 1032$
$1032\times4+2 = 4128+2 = 4130$
$4130\times2+4 = 8260+4 = 8264$
$8264\times4+2 = 33056+2 = 33058$

Here 128 will come in place of question mark.
Hence, option d is the correct answer.

Question 2: 180, 90, 135, ?, 1181.25, 5315.625

a) 332.5

b) 348.5

c) 359.5

d) 351.5

e) 337.5

2) Answer (E)

Solution:

$180\times0.5 = 90$
$90\times1.5 = 135$
$135\times2.5 = 337.5$
$337.5\times3.5 = 1181.25$
$1181.25\times4.5 = 5315.625$

Here 337.5 will come in place of question mark.
Hence, option e is the correct answer.

Question 3: 326, ?, 295, 124, 268, 149, 245

a) 120

b) 135

c) 95

d) 80

e) None of the above

3) Answer (C)

Solution:

$326-21\times11 = 326-231 = 95$
$95+20\times10 = 95+200 = 295$
$295-19\times9 = 295-171 = 124$
$124+18\times8 = 124+144 = 268$
$268-17\times7 = 268-119 = 149$
$149+16\times6 = 149+96 = 245$

Here 95 will come in place of question mark.
Hence, option c is the correct answer.

Question 4: 6.2, 8.2, 12.7, 20.2, 30.7, ?

a) 47.2

b) 45.8

c) 44.2

d) 48.8

e) None of the above

4) Answer (A)

Solution:

$6.2+1.5\times2 = 6.2+3 = 8.2$
$8.2+1.5\times3 = 8.2+4.5 = 12.7$
$12.7+1.5\times5 = 12.7+7.5 = 20.2$
$20.2+1.5\times7 = 20.2+10.5 = 30.7$
$30.7+1.5\times11 = 30.7+16.5 = 47.2$

Here 47.2 will come in place of question mark.
Hence, option a is the correct answer.

Question 5: 183, 1911, 580, 1580, 851, ?, 1020

a) 339

b) 351

c) 1347

d) 1363

e) None of the above

5) Answer (D)

Solution:

$183+12^{3} = 183+1728 = 1911$
$1911-11^{3} = 1911-1331 = 580$
$580+10^{3} = 580+1000 = 1580$
$1580-9^{3} = 1580-729 = 851$
$851+8^{3} = 851+512 = 1363$
$1363-7^{3} = 1363-343 = 1020$

Here 1363 will come in place of question mark.
Hence, option d is the correct answer.

Instructions

Identify the wrong numbers in series I and series II and answer the question accordingly.
series I : 149, 66.5, 130.5, 45, 694.5, 4536, 74304
series II : 78, 183, 276, 364, 436, 493

Question 6: Find out the difference between the wrong numbers of series I and II.

a) 496.5

b) 475

c) 418.5

d) 362

e) 351

6) Answer (C)

Solution:

series I :
$149\times0.5-2^{3} = 74.5-8 = 66.5$
$66.5\times1+4^{3} = 66.5+64 = 130.5$
$130.5\times2-6^{3} = 261-216 = 45$
$45\times4+8^{3} = 180+512 = 692$
$692\times8-10^{3} = 5536-1000 = 4536$
$4536\times16+12^{3} = 72576+1728 = 74304$
Here 694.5 is the wrong number in the series.

series II :
$78+15\times7 = 78+105 = 183$
$183+16\times6 = 183+96 = 279$
$279+17\times5 = 279+85 = 364$
$364+18\times4 = 364+72 = 436$
$436+19\times3 = 436+57 = 493$
Here 276 is the wrong number in the series.

Difference between the wrong numbers of series I and II = 694.5-276
= 418.5
Hence, option c is the correct answer.

Instructions

Find out the missing number in each series and answer the questions as per the options.

Question 7: Series I : 3000, 1200, (A), 1152, 1843.2, 3686.4
Series II : (B), 960, 3765, 11385, 22665, 22785
Series III : 1220, 309, 150.5, 154.5, (C), 1224, 9788
What is the value of (A+B+C)?

a) 1375

b) 1445

c) 1480

d) 1465

e) 1360

7) Answer (B)

Solution:

Series I :
$3000\times0.4 = 1200$
$1200\times0.8 = 960$
$960\times1.2 = 1152$
$1152\times1.6 = 1843.2$
$1843.2\times2 = 3686.4$

Series II :
$180\times5+60 = 900+60 = 960$
$960\times4-75 = 3840-75 = 3765$
$3765\times3+90 = 11295+90 = 11385$
$11385\times2-105 = 22770-105 = 22665$
$22665\times1+120 = 22665+120 = 22785$

Series III :
$1220\times0.25+4 = 305+4 = 309$
$309\times0.5-4 = 154.5-4 = 150.5$
$150.5\times1+4 = 150.5+4 = 154.5$
$154.5\times2-4 = 309-4 = 305$
$305\times4+4 = 1220+4 = 1224$
$1224\times8-4 = 9792-4 = 9788$

value of (A+B+C) = 960+180+305
= 1445
Hence, option b is the correct answer.

Question 8: Series I : (P), 9484, 16324, 21220, 24580, 26764
Series II : 24, (Q), 960, 3000, 7560, 16464
Series III : 3265920, 322560, 35280, 4320, (R), 96

a) R>Q>P

b) Q>R>P

c) R>P>Q

d) P>Q>R

e) P>Q=R

8) Answer (C)

Solution:

Series I :
$244 + 21^{3} – 21 = 244 + 9261 – 21 = 9484$
$9484 + 19^{3} – 19 = 9484 + 6859 – 19 = 16324$
$16324 + 17^{3} – 17 = 16324 + 4913 – 17 = 21220$
$21220 + 15^{3} – 15 = 21220 + 3375 – 15 = 24580$
$24580 + 13^{3} – 13 = 24580 + 2197 – 13 = 26764$

Series II :
$2^{5}-2^{3} = 32-8 = 24$
$3^{5}-3^{3} = 243-27 = 216$
$4^{5}-4^{3} = 1024-64 = 960$
$5^{5}-5^{3} = 3125-125 = 3000$
$6^{5}-6^{3} = 7776-216 = 7560$
$7^{5}-7^{3} = 16807-343 = 16464$

Series III :
10!-9! = 3628800-362880 = 3265920
9!-8! = 362880-40320 = 322560
8!-7! = 40320-5040 = 35280
7!-6! = 5040-720 = 4320
6!-5! = 720-120 = 600
5!-4! = 120-24 = 96

R>P>Q
600>244>216
Hence, option c is the correct answer.

Instructions

Study the following information carefully.
A series is given and certain conditions are given below for it. Observe the conditions carefully and answer the questions.

J, K, L, M, 4125, 4414, __, Y

$J = P^{2}+Q$
$K = (T-1)^{2}+J$
$L = S^{2}+K$

Note ::
(i) The value of Q is equal to the largest root of the equation $y^{2}-6y-3712 = 0$.
(ii) The difference between S and Q is 35.
(iii) LCM of P and T is 1184 where P is a prime number.
(iv) The value of S is not the multiple of 3.

Question 9: Which of the following statements is/are true?
(i) The value of M is the multiple of 7.
(ii) The difference between the values of P and T is five.
(iii) The immediate next term after Y in the given series has 91 as a factor.

a) Only (i)

b) Only (ii)

c) Only (iii)

d) Only (i) and (ii)

e) Only (ii) and (iii)

9) Answer (B)

Solution:

(i) The value of Q is equal to the largest root of the equation $y^{2}-6y-3712 = 0$.
$y^{2}-(64-58)y-3712 = 0$
$y^{2}-64y+58y-3712 = 0$
y(y-64)+58(y-64) = 0
(y-64) (y+58) = 0
y = -58, 64
largest root of the equation = Q = 64
(ii) The difference between S and Q is 35.
S-64 = 35 ⇒ S = 35+64 = 99
64-S = 35 ⇒ S = 64-35 = 29
(iv) The value of S is not the multiple of 3.
S = 29 (99 is the multiple of 3. So it will not be possible.)
(iii) LCM of P and T is 1184 where P is a prime number.
PT = $37\times32$
So P = 37 and T = 32
$J = P^{2}+Q = 37^{2}+64 = 1369+64 = 1433$
$K = (T-1)^{2}+J = (32-1)^{2}+1433 = 31^{2}+1433 = 961+1433 = 2394$
$L = S^{2}+K = 29^{2}+2394 = 841+2394 = 3235$
From these three numbers, now we can get the pattern from which we can obtain the value of Y.
M = $23^{2}+3235 = 529+3235 = 3764$
$19^{2}+3764 = 361+3764 = 4125$
$17^{2}+4125 = 289+4125 = 4414$
$13^{2}+4414 = 169+4414 = 4583$
Y = $11^{2}+4583 = 121+4583 = 4704$
$7^{2}+4704 = 49+4704 = 4753$
Hence, option b is the correct answer.

Question 10: What is the value of ‘Y’?

a) 4754

b) 4736

c) 4762

d) 4708

e) None of the above

10) Answer (E)

Solution:

(i) The value of Q is equal to the largest root of the equation $y^{2}-6y-3712 = 0$.
$y^{2}-(64-58)y-3712 = 0$
$y^{2}-64y+58y-3712 = 0$
y(y-64)+58(y-64) = 0
(y-64) (y+58) = 0
y = -58, 64
largest root of the equation = Q = 64
(ii) The difference between S and Q is 35.
S-64 = 35 ⇒ S = 35+64 = 99
64-S = 35 ⇒ S = 64-35 = 29
(iv) The value of S is not the multiple of 3.
S = 29 (99 is the multiple of 3. So it will not be possible.)
(iii) LCM of P and T is 1184 where P is a prime number.
PT = $37\times32$
So P = 37 and T = 32
$J = P^{2}+Q = 37^{2}+64 = 1369+64 = 1433$
$K = (T-1)^{2}+J = (32-1)^{2}+1433 = 31^{2}+1433 = 961+1433 = 2394$
$L = S^{2}+K = 29^{2}+2394 = 841+2394 = 3235$
From these three numbers, now we can get the pattern from which we can obtain the value of Y.
M = $23^{2}+3235 = 529+3235 = 3764$
$19^{2}+3764 = 361+3764 = 4125$
$17^{2}+4125 = 289+4125 = 4414$
$13^{2}+4414 = 169+4414 = 4583$
Y = $11^{2}+4583 = 121+4583 = 4704$
Hence, option e is the correct answer.

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Instructions

Identify the missing numbers in Series I and Series II and answer question.
Series I :: 469, 3629269, 3266389, 3306709, 3301669, ?, 3302269
Series II :: 109, 125, 206, 831, 3232, 17873, ?

Question 11: Which of the following statements are true about the missing numbers in the given series.
(i) In Series II, the digits on the unit and tens place of the missing number are the same.
(ii) The sum of the digits of the missing number in Series I is 26.
(iii) The missing number in Series II is the multiple of three.

a) Only (i)

b) Only (ii) and (iii)

c) Only (i) and (ii)

d) Only (ii)

e) Only (iii)

11) Answer (E)

Solution:

Series I
469 + 10! = 469 + 3628800 = 3629269
3629269 – 9! = 3629269 – 362880 = 3266389
3266389 + 8! = 3266389 + 40320 = 3306709
3306709 – 7! = 3306709 – 5040 = 3301669
3301669 + 6! = 3301669 + 720 = 3302389
3302389 – 5! = 3302389 – 120 = 3302269
Here 3302389 is the missing number in the given series.

Series II
$109+2^{4} = 109+16 = 125$
$125+3^{4} = 125+81 = 206$
$206+5^{4} = 206+625 = 831$
$831+7^{4} = 831+2401 = 3232$
$3232+11^{4} = 3232+14641 = 17873$
$17873+13^{4} = 17873+28561 = 46434$
Here 46434 is the missing number in the given series.

Hence, option e is the correct answer.

Instructions

Identify the wrong number in Series I and correct that number which will be the value of ‘A’. In Series II, find out the missing value in place question mark which will be the value of ‘B’.
Series I :: 526, 275.5, 39.75, 101.75, 167.5, 686, 5484
Series II :: 1248, 1591, 1898, 2171, ?, 2623
Note :: both the series are following a different pattern.

Question 12: Find out the difference between the values of ‘A’ and ‘B’.

a) 1758.25

b) 2712.25

c) 1936.25

d) 2374.25

e) None of the above

12) Answer (D)

Solution:

Series I
$526\times0.25 + 144 = 131.5+144 = 275.5$
$275.5\times0.5 – 100 = 137.75-100 = 37.75$
$37.75\times1 + 64 = 37.75+64 = 101.75$
$101.75\times2 – 36 = 203.5-36 = 167.5$
$167.5\times4 + 16 = 670+16 = 686$
$686\times8 – 4 = 5488-4 = 5484$
Here 37.75 should come instead of 39.75 to make the series correct. So A = 37.75

Series II
$1248+19^{2}-18 = 1248+361-18 = 1591$
$1591+18^{2}-17 = 1591+324-17 = 1898$
$1898+17^{2}-16 = 1898+289-16 = 2171$
$2171+16^{2}-15 = 2171+256-15 = 2412$
$2412+15^{2}-14 = 2412+225-14 = 2623$
The missing number is 2412.

difference between the values of ‘A’ and ‘B’ = 2412-37.75
= 2374.25
Hence, option d is the correct answer.

Instructions

Find out the wrong number in the following series.

Question 13: 3150, 2028, 1210, 648, 272, 100

a) 648

b) 1210

c) 272

d) 3150

e) 100

13) Answer (C)

Solution:

$15^{3}-15^{2} = 3375-225 = 3150$
$13^{3}-13^{2} = 2197-169 = 2028$
$11^{3}-11^{2} = 1331-121 = 1210$
$9^{3}-9^{2} = 729-81 = 648$
$7^{3}-7^{2} = 343-49 = 294$
$5^{3}-5^{2} = 125-25 = 100$

Here 272 is the wrong number.
Hence, option c is the correct answer.

Question 14: 464, 687, 1374, 3435, 10305, 36067.5

a) 687

b) 10305

c) 3435

d) 464

e) 1374

14) Answer (D)

Solution:

$458\times1.5 = 687$
$687\times2 = 1374$
$1374\times2.5 = 3435$
$3435\times3 = 10305$
$10305\times3.5 = 36067.5$

Here 464 is the wrong number.
Hence, option d is the correct answer.

Question 15: 1186, 1378, 1578, 1818, 2090, 2396

a) 1818

b) 2396

c) 1378

d) 2090

e) 1578

15) Answer (C)

Solution:

$1186+13\times14 = 1186+182 = 1368$
$1368+14\times15 = 1368+210 = 1578$
$1578+15\times16 = 1578+240 = 1818$
$1818+16\times17 = 1818+272 = 2090$
$2090+17\times18 = 2090+306 = 2396$

Here 1378 is the wrong number.
Hence, option c is the correct answer.

Question 16: 164, 180, -76, 1124, -2876, 7124, -13612

a) -76

b) 180

c) -2876

d) 164

e) 1124

16) Answer (E)

Solution:

$164+2^{4} = 164+16 = 180$
$180-4^{4} = 180-256 = -76$
$-76+6^{4} = -76+1296 = 1220$
$1220-8^{4} = 1220-4096 = -2876$
$-2876+10^{4} = -2876+10000 = 7124$
$7124-12^{4} = 7124-20736 = -13612$

Here 1124 is the wrong number.
Hence, option e is the correct answer.

Instructions

Find out the wrong number in the following series.

Question 17: 312, 368, 259, 480, 32, 928

a) 928

b) 480

c) 368

d) 32

e) 259

17) Answer (E)

Solution:

$312+56=368$
$368-112=256$
$256+224=480$
$480-448=32$
$32+896=928$
$\therefore$ The wrong number in the series is 259
Hence, the correct answer is Option E

Question 18: 286, 288, 290, 296, 320, 440

a) 286

b) 320

c) 288

d) 296

e) 290

18) Answer (A)

Solution:

The pattern here is
$287+1!=287+1=288$
$288+2!=288+2=290$
$290+3!=290+6=296$
$296+4!=296+24=320$
$320+5!=320+120=440$
$\therefore$ The wrong number in the series is 286
Hence, the correct answer is Option A

Question 19: 928, 332, 285, 431, 1031, 4320

a) 431

b) 332

c) 1031

d) 285

e) 4320

19) Answer (D)

Solution:

The pattern here is
$928\times0.25+100=232+100=332$
$332\times0.5+121=166+121=287$
$287\times1+144=287+144=431$
$431\times2+169=862+169=1031$
$1031\times4+196=4124+196=4320$
$\therefore$ The wrong number in the series is 285
Hence, the correct answer is Option D

Instructions

Find out the missing number in place of a question mark.

Question 20: 180, 126, ?, 124.74, 162.162, 243.243

a) 115.4

b) 114.3

c) 114.5

d) 113.4

e) 112.3

20) Answer (D)

Solution:

$180\times0.7 = 126$
$126\times0.9 = 113.4$
$113.4\times1.1 = 124.74$
$124.74\times1.3 = 162.162$
$162.162\times1.5 = 243.243$

Here 113.4 is the missing number.
Hence, option d is the correct answer.

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