Number Series Questions for TISSNET 2022 – Download PDF

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Number Series Questions for TISSNET
Number Series Questions for TISSNET

Number Series Questions for TISSNET 2022 – Download PDF

Download TISSNET 2022 Number Series Questions pdf by Cracku. Very Important Number Series Questions for TISSNET 2022 based on asked questions in previous exam papers. These questions will help your TISSNET exam preparation. So kindly download the PDF for reference and do more practice.

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Question 1: What will come in the place of question mark (?) in the following series ?
2  9  28  65 ?

a) 96

b) 106

c) 126

d) 130

e) None of these

Question 2: What should come in place of the question mark (?) in the following number series ?
1, 5, 17, 53, 161, 485, ?

a) 1168

b) 1254

c) 1457

d) 1372

e) None of these

Question 3: What approximate value should come in place of the question mark (?) in the following question?
$54.786 \div 10.121 \times 4.454 = ?$

a) 84

b) 48

c) 118

d) 58

e) 24

Question 4: What should come in place of the question mark (?) in the following number series?
2 5 11 23 47 95 ?

a) 168

b) 154

c) 191

d) 172

e) None of these

Question 5: What should come in place of the question mark (?) in the following number series?
1 4 14 45 139 422 ?

a) 1268

b) 1234

c) 1272

d) 1216

e) None of these

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Question 6: What would be the compound interest accrued on an amount of Rs. 9,000 at the rate of 11 p.c.p.a. in two years ?

a) Rs. 2089.90

b) Rs. 2140.90

c) Rs. 2068.50

d) Rs. 2085.50

e) None of these

Question 7: 16 8 12 30 ? 472.5

a) 104

b) 103

c) 106

d) 105

e) None of these

Question 8: 2, 5, 12, 27, 58, ?

a) 122

b) 121

c) 123

d) 120

e) None of these

Question 9: 18 19.7 16.3 23.1 9.5 ?

a) 36.5

b) 36.8

c) 36.7

d) 36.9

e) None of these

Question 10: 68, ?, 77, 104, 168, 293

a) 69

b) 70

c) 68

d) 74

e) None of these

Question 11: In how many different ways can the numbers ‘256974’ be arranged, using each digit only once in each arrangement, such that the digits 6 and 5 are at the extreme ends in each arrangement ?

a) 48

b) 720

c) 36

d) 360

e) None of these

Question 12: What will come in place of both the question marks (?) in the following question ?$\frac{(?)^{0.6}}{104}=\frac{26}{(?)^{1.4}}$

a) 58

b) -48

c) -56

d) 42

e) -52

Question 13: Out of the fractions $\frac{1}{2}, \frac{7}{8}, \frac{3}{4}, \frac{5}{6}$, and $\frac{6}{7}$ what is the difference between the largest and smallest fractions ?

a) $\frac{7}{13}$

b) $\frac{3}{8}$

c) $\frac{4}{7}$

d) $\frac{1}{6}$

e) None of these

Question 14: If $(11)^{3}$ is subtracted from $(46)^{2}$ . what will be the remainder ?

a) 787

b) 785

c) 781

d) 783

e) None of these

Question 15: 9, 11, 16, 33, 98, ?

a) 350

b) 355

c) 360

d) 365

e) 370

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Question 16: 65, 70, 63, 74, 61, ?

a) 78

b) 58

c) 72

d) 46

e) 68

Question 17: 13, 14, 30, 93, ?, 1885

a) 358

b) 336

c) 364

d) 376

e) 380

Question 18: 8, 14, 25, 46, 82, ?

a) 132

b) 130

c) 138

d) 128

e) 142

Question 19: 14, 8, 7, 11.5, 22, ?

a) 54

b) 64

c) 62

d) 58

e) 56

Question 20: $\frac{1}{8}\times121+\frac{1}{5}\times76-?=25$

a) 5

b) 45

c) 15

d) 35

e) 65

Question 21: $\sqrt{628}\times17.996\div15.04=?$

a) 30

b) 10

c) 5

d) 20

e) 25

Question 22: 80.04% of 150.16 + 60.02% of 50.07 = ?

a) 150

b) 125

c) 210

d) 175

e) 213

Question 23: $17.98^{2} + 4.05 \times 90.11 \div 4.98 = ?$

a) 396

b) 336

c) 242

d) 423

e) 816

Question 24: 90.05 + 281 ÷ 4 -151.06 = $\sqrt[3]{?}$

a) 27

b) 343

c) 216

d) 729

e) 176

Question 25: What should come next in the number series ?
1 8 3 6 5 4 7 2 9 1 8 3 6 5 4 7 2 1 8 3 6 5 4 7 1 8 3 6 5 4

a) 1

b) 2

c) 4

d) 8

e) None of these

Question 26: 26, 12, 11, 15.5, 30, ?

a) 72

b) 68

c) 74

d) 82

e) 78

Question 27: 8, 9.4, 12.2, 17.8, 29, ?

a) 53.6

b) 51.4

c) 52.1

d) 48.6

e) 49.8

Question 28: 17, 16, 30, 87, 344, ?

a) 1735

b) 1760

c) 1660

d) 1685

e) 1715

Question 29: 13, 13, 20, 37.5, 83, ?

a) 233

b) 216

c) 234

d) 235

e) 239

Question 30: 29, 31, 37, 49, 69, ?

a) 108

b) 99

c) 94

d) 103

e) 88

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Answers & Solutions:

1) Answer (C)

Each number is of the form $(n^3+1)$ where $n$ is a natural number

$1^3+1$ = 2

$2^3+1$ = 9

$3^3+1$ = 28

$4^3+1$ = 65

$5^3+1$ = 126

=> Ans – (C)

2) Answer (C)

The pattern here followed is :

1 * 3 + 2 = 5

5 * 3 + 2 = 17

17 * 3 + 2 = 53

53 * 3 + 2 = 161

161 * 3 + 2 = 485

485 * 3 + 2 = 1457

3) Answer (E)

Expression : $54.786 \div 10.121 \times 4.454 = ?$

= $\frac{55}{10} \times 4.5$

= $24.75 \approx 24$

4) Answer (C)

The pattern here followed is :

2 * 2 + 1 = 5

5 * 2 + 1 =  11

11 * 2 + 1 = 23

23 * 2 + 1 = 47

47 * 2 + 1 = 95

95 * 2 + 1 = 191

5) Answer (C)

The pattern here followed is :

1 * 3 + 1 = 4

4 * 3 + 2 = 14

14 * 3 + 3 = 45

45 * 3 + 4 = 139

139 * 3 + 5 = 422

422 * 3 + 6 = 1272

6) Answer (E)

$C.I. = P [(1 + \frac{R}{100})^T – 1]$

= $9000 [(1 + \frac{11}{100})^2 – 1]$

= $9000 [(1.11)^2 – 1]$

= $9000 \times (1.2321 – 1)$

= $9000 \times 0.2321$ = Rs. $2,088.90$

7) Answer (D)

Odd multiples of $\frac{1}{2}$ are multiplied

16 $\times \frac{1}{2}$ = 8

8 $\times \frac{3}{2}$ = 12

12 $\times \frac{5}{2}$ = 30

30 $\times \frac{7}{2}$ = 105

105 $\times \frac{9}{2}$ = 472.5

8) Answer (B)

Each number is multiplied by 2 and then consecutive natural numbers are added

2 $\times 2 + 1$ = 5

5 $\times 2 + 2$ = 12

12 $\times 2 + 3$ = 27

27 $\times 2 + 4$ = 58

58 $\times 2 + 5$ = 121

9) Answer (C)

The pattern is :

18 $+ 1.7 \times 2^0$ = 19.7

19.7 $- 1.7 \times 2^1$ = 16.3

16.3 $+ 1.7 \times 2^2$ = 23.1

23.1 $- 1.7 \times 2^3$ = 9.5

9.5 $+ 1.7 \times 2^4$ = 36.7

10) Answer (A)

Cubes of consecutive natural numbers are added

68 $+ 1^3$ = 69

69 $+ 2^3$ = 77

77 $+ 3^3$ = 104

104 $+ 4^3$ = 168

168 $+ 5^3$ = 293

11) Answer (A)

Case 1 : 6 at left end and 5 is at right end : 6 _ _ _ _ 5

Now, four empty places can be filled by 2,9,7 and 4 in = $4!$ ways

= $4 \times 3 \times 2 \times 1 = 24$

Case 2 : 6 at right end and 5 at left end : 5 _ _ _ _ 6

Similarly, no. of ways = $4!$

= $4 \times 3 \times 2 \times 1 = 24$

$\therefore$ Total no. of ways = $24 + 24 = 48$

12) Answer (E)

$\frac{(x)^{0.6}}{104}=\frac{26}{(x)^{1.4}}$

${(x)^{0.6}} * {(x)^{01.4}}$ = 104*26

${(x)^{2}}$ = 104*26

x = ±52

13) Answer (B)

Given values are ,
$\frac{1}{2}$ = 0.5

$\frac{7}{8}$ = 0.87

$\frac{3}{4}$ = 0.75

$\frac{5}{6}$ = 0.83

$\frac{6}{7}$ = 0.86

∴ Required difference = $\frac{7}{8}$ – $\frac{1}{2}$ = (7-4)/8 = 3/8

14) Answer (B)

Here

$(46)^2$ = 2116

$(11)^3$ = 1331

So, 2116 – 1331 = 785

15) Answer (B)

Number of the form $(2^n + 1)$ are added where n is even number

9 $+ (2^0 + 1)$ = 11

11 $+ (2^2 + 1)$ = 16

16 $+ (2^4 + 1)$ = 33

33 $+ (2^6 + 1)$ = 98

98 $+ (2^8 + 1)$ = 355

16) Answer (A)

We can note that consecutive prime numbers are either added or subtracted starting from 5.

65+5 = 70

70-7 = 63

63+11 = 74

74-13 = 61

61+17 = 78

Option A is the right answer.

17) Answer (D)

13*1 + 1 = 14

14*2 +2 = 28+2 = 30

30*3 + 3 = 90+3 =93

93*4 + 4 = 372+4 = 376

Option D is the right answer.

18) Answer (C)

14 – 8 = 6

25 – 14 =11

46 – 25 = 21

82 – 46 = 36

Let us take a look at the second order difference.

11 – 6 = 5

21 – 11 = 10

36 – 21 = 15

The next term will be 20. The difference will be 36 + 20 = 56.

The required term is 82 + 56 = 138.

Option C is the right answer.

19) Answer (E)

14*0.5 + 1 = 7+1= 8

8*1 – 1 = 8-1 = 7

7*1.5 + 1 = 10.5 + 1 = 11.5

11.5*2 – 1 = 23 – 1 = 22

22*2.5 +1 = 55+1 =56

Option E is the right answer.

20) Answer (A)

The given equation can be written as $\frac{120}{8}+\frac{75}{5} – 25 =x$

$15+15-25=5$

Option A is the right answer.

21) Answer (A)

The given equation can be written as$ \frac{\sqrt{625}*18}{15}$

=$\frac{25*6}{5}$

= 30.

Option A is the right answer.

22) Answer (A)

80.04 ~ 80

150.16 ~ 150

60.02 ~ 60

50.07 ~ 50

Now, (80.04% of 150.16 + 60.02% of 50.07) is equivalent to (80% of 150 + 60% of 50) = 120 +30 = 150

23) Answer (A)

The given equation can be written as $18^{2} + \frac{4*90}{5}$

= $324 + 4*18$

= $396$

Option A is the right answer.

24) Answer (D)

The given statement can be written as $90 + 60 – 151 = \sqrt[3]{?}$

$9 =\sqrt[3]{?}$

=> ? = 729

Option D is the answer.

25) Answer (A)

The series can be shown as below

1 8 3 6 5 4 7 2 9
1 8 3 6 5 4 7 2
1 8 3 6 5 4 7
1 8 3 6 5 4

After this, the series has to repeat itself and hence, the next term will be 1

26) Answer (C)

$\frac{26}{2/1}-1$ = 12

$\frac{12}{2/2}-1$ = 11

$\frac{11}{2/3}-1$ = 15.5

$\frac{15.5}{2/4}-1$ = 30

$\frac{30}{2/5}-1$ = 74

27) Answer (B)

The difference between consecutive terms is doubling.

For example, 9.4 – 8 = 1.4
12.2 – 9.4 = 2.8
17.8 – 12.2 = 5.6
29 – 17.8 = 11.2

Hence, the term to come next is 29 + 11.2*2 = 51.4

28) Answer (E)

16 = 17*1 – 1

30 = 16*2 – 2

87 = 30*3 – 3

344 = 87*4 – 4

So, the next number is 344*5 – 5 = 1715

29) Answer (B)

$13 \times 0.5 + 6.5 = 13$

$13 \times 1 + 7 = 20$

$20 \times 1.5 + 7.5 = 37.5$

$37.5 \times 2 + 8 = 83$

$83 \times 2.5 + 8.5 = 216$

30) Answer (B)

+The difference between the consecutive terms is:
31-29 = 2
37 -31 = 6
49- 37 = 12
69-49 = 20
Here, the difference between the consecutive difference terms is
6-2 = 4
12-6 = 6
20-12 = 8
Hence, the next term in difference series = 10
Therefore the next term in series = 69+30 = 99

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