Number Series Questions for MBA-CET PDF

Here you can download the important Number series questions PDF with solutions for MAH MBA CET by Cracku. These questions will help you to make practice and solve the Number series aptitude section questions in the MAH MBA CET exam. Utilize this best PDF practice set which is included answers in detail. Click on the below link to download the Number series questions PDF for MBA-CET 2022.

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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

1) Answer (C)

Solution:

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)

Question 2: What will come in place of the question mark (?) in the following number series?
9  10  39  220  ?  14382

a) 1589

b) 1598

c) 1958

d) 1985

e) 1835

2) Answer (A)

Solution:

The pattern followed is :

9 $\times 1 + 1^2$ = 10

10 $\times 3 + 3^2$ = 39

39 $\times 5 + 5^2$ = 220

220 $\times 7 + 7^2$ = 1589

1589 $\times 9 + 9^2$ = 14382

Question 3: What will come in place of the question mark (?) in the following number series?
121  238  472  ?  1876  3748

a) 1008

b) 948

c) 944

d) 940

e) 1005

3) Answer (D)

Solution:

Each number is multiplied by 2 and then 4 is subtracted from it.

121 $\times 2 – 4$ = 238

238 $\times 2 – 4$ = 472

472 $\times 2 – 4$ = 940

940 $\times 2 – 4$ = 1876

1876 $\times 2 – 4$ = 3748

Question 4: What will come in place of the question mark (?) in the following number series?
44  ?  99   148.5  222.75  334.125

a) 44

b) 55

c) 66

d) 33

e) 35

4) Answer (C)

Solution:

Each number is multiplied by $\frac{3}{2}$

44 $\times \frac{3}{2}$ = 66

66 $\times \frac{3}{2}$ = 99

99 $\times \frac{3}{2}$ = 148.5

148.5 $\times \frac{3}{2}$ = 222.75

222.75 $\times \frac{3}{2}$ = 334.125

Question 5: What will come in place of the question mark (?) in the following number series?
33  16.5?  24.75    49.5  123.75

a) 18.5

b) 16.5

c) 8.5

d) 8.25

e) None of these

5) Answer (B)

Solution:

The pattern followed is :

33 $\times \frac{1}{2}$ = 16.5

16.5 $\times \frac{2}{2}$ = 16.5

16.5 $\times \frac{3}{2}$ = 24.75

24.75 $\times \frac{4}{2}$ = 49.5

49.5 $\times \frac{5}{2}$ = 123.75

Question 6: What will come in place of the question mark (?) in the following number series?
20  23  30  43  64  ?

a) 95

b) 90

c) 100

d) 105

e) 96

6) Answer (A)

Solution:

Numbers of the form $n^2 – (n-1)$ are added, where $n$ is an integer starting from 2

23 $+ 2^2 – 1$ = 23

23 $+ 3^2 – 2$ = 30

30 $+ 4^2 – 3$ = 43

43 $+ 5^2 – 4$ = 64

64 $+ 6^2 – 5$ = 95

Question 7: 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

7) Answer (C)

Solution:

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

Question 8: 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

8) Answer (E)

Solution:

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

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

= $24.75 \approx 24$

Question 9: 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

9) Answer (C)

Solution:

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

Question 10: 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

10) Answer (C)

Solution:

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

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Question 11: What is the least number to be added to 2530 to make it a perfect square?

a) 50

b) 65

c) 75

d) 80

e) None of these

11) Answer (E)

Solution:

We know that $50^2 = 2500$ and $51^2 = 2601$

$\because$ 2500 < 2530 < 2601

$\therefore$ Required number = 2601 – 2530 = 71

Question 12: 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

12) Answer (E)

Solution:

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

= $9000 [(1 + \frac{11}{100})^2 – 1]$

= $9000 [(1.11)^2 – 1]$

= $9000 \times (1.2321 – 1)$

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

Question 13: 16 8 12 30 ? 472.5

a) 104

b) 103

c) 106

d) 105

e) None of these

13) Answer (D)

Solution:

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

Question 14: 2, 5, 12, 27, 58, ?

a) 122

b) 121

c) 123

d) 120

e) None of these

14) Answer (B)

Solution:

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

Question 15: 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

15) Answer (C)

Solution:

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

Question 16: 68, ?, 77, 104, 168, 293

a) 69

b) 70

c) 68

d) 74

e) None of these

16) Answer (A)

Solution:

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

Question 17: 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

17) Answer (A)

Solution:

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 = $24 + 24 = 48$

Question 18: 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

18) Answer (E)

Solution:

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

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

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

x = ±52

Question 19: 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

19) Answer (B)

Solution:

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

Question 20: 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

20) Answer (B)

Solution:

Here

$(46)^2$ = 2116

$(11)^3$ = 1331

So, 2116 – 1331 = 785

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