# Number Series Questions for MBA-CET PDF

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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.

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

a)Â 96

b)Â 106

c)Â 126

d)Â 130

e)Â None of these

Solution:

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

$1^3+1$ = 2

$2^3+1$ =Â 9

$3^3+1$ =Â 28

$4^3+1$ =Â 65

$5^3+1$ =Â 126

=> Ans – (C)

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

a)Â 1589

b)Â 1598

c)Â 1958

d)Â 1985

e)Â 1835

Solution:

The pattern followed isÂ :

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

10Â $\times 3 + 3^2$ =Â 39

39Â $\times 5 + 5^2$ =Â 220

220Â $\times 7 + 7^2$ =Â 1589

1589Â $\times 9 + 9^2$ =Â 14382

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

a)Â 1008

b)Â 948

c)Â 944

d)Â 940

e)Â 1005

Solution:

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

121 $\times 2 – 4$ = 238

238Â $\times 2 – 4$ =Â 472

472Â $\times 2 – 4$ =Â 940

940Â $\times 2 – 4$ =Â 1876

1876Â $\times 2 – 4$ =Â 3748

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

a)Â 44

b)Â 55

c)Â 66

d)Â 33

e)Â 35

Solution:

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

44 $\times \frac{3}{2}$ =Â 66

66Â $\times \frac{3}{2}$ =Â 99

99Â $\times \frac{3}{2}$ =Â 148.5

148.5Â $\times \frac{3}{2}$ =Â 222.75

222.75Â $\times \frac{3}{2}$ =Â 334.125

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

a)Â 18.5

b)Â 16.5

c)Â 8.5

d)Â 8.25

e)Â None of these

Solution:

The pattern followed isÂ :

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

16.5Â $\times \frac{2}{2}$ =Â 16.5

16.5Â $\times \frac{3}{2}$ =Â 24.75

24.75Â $\times \frac{4}{2}$ =Â 49.5

49.5Â $\times \frac{5}{2}$ =Â 123.75

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

a)Â 95

b)Â 90

c)Â 100

d)Â 105

e)Â 96

Solution:

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

23 $+ 2^2 – 1$ = 23

23Â $+ 3^2 – 2$ =Â 30

30Â $+ 4^2 – 3$ =Â 43

43Â $+ 5^2 – 4$ =Â 64

64Â $+ 6^2 – 5$ =Â 95

Question 7:Â What should come in place of the question mark (?) in the following number series?
1, 5, 17, 53, 161, 485, ?

a)Â 1168

b)Â 1254

c)Â 1457

d)Â 1372

e)Â None of these

Solution:

The pattern here followed isÂ :

1 * 3 + 2 = 5

5Â * 3 + 2 =Â 17

17Â * 3 + 2 =Â 53

53Â * 3 + 2 =Â 161

161Â * 3 + 2 =Â 485

485Â * 3 + 2 =Â 1457

Question 8:Â What approximate value should come in place of the question mark (?) in the following question?
$54.786 \div 10.121 \times 4.454 = ?$

a)Â 84

b)Â 48

c)Â 118

d)Â 58

e)Â 24

Solution:

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

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

= $24.75 \approx 24$

Question 9:Â What should come in place of the question mark (?) in the following number series?
2 5 11 23 47 95 ?

a)Â 168

b)Â 154

c)Â 191

d)Â 172

e)Â None of these

Solution:

The pattern here followed isÂ :

2 * 2 + 1 = 5

5 *Â 2 + 1 =Â Â 11

11 *Â 2 + 1 =Â 23

23 *Â 2 + 1 =Â 47

47 *Â 2 + 1 =Â 95

95 *Â 2 + 1 =Â 191

Question 10:Â What should come in place of the question mark (?) in the following number series?
1 4 14 45 139 422 ?

a)Â 1268

b)Â 1234

c)Â 1272

d)Â 1216

e)Â None of these

Solution:

The pattern here followed isÂ :

1Â * 3 +Â 1 = 4

4Â * 3 +Â 2 = 14

14Â * 3 + 3 = 45

45Â * 3 +Â 4 = 139

139Â * 3 + 5 = 422

422 * 3 + 6 =Â 1272

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

a)Â 50

b)Â 65

c)Â 75

d)Â 80

e)Â None of these

Solution:

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

$\because$ 2500Â < 2530Â < 2601

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

Question 12:Â 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)Â Rs. 2089.90

b)Â Rs. 2140.90

c)Â Rs. 2068.50

d)Â Rs. 2085.50

e)Â None of these

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:Â 16 8 12 30 ? 472.5

a)Â 104

b)Â 103

c)Â 106

d)Â 105

e)Â None of these

Solution:

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

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

8Â $\times \frac{3}{2}$ =Â 12

12Â $\times \frac{5}{2}$ =Â 30

30Â $\times \frac{7}{2}$ =Â 105

105Â $\times \frac{9}{2}$ =Â 472.5

Question 14:Â 2, 5, 12, 27, 58, ?

a)Â 122

b)Â 121

c)Â 123

d)Â 120

e)Â None of these

Solution:

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

2 $\times 2 + 1$ = 5

5Â $\times 2 + 2$ =Â 12

12Â $\times 2 + 3$ =Â 27

27Â $\times 2 + 4$ =Â 58

58Â $\times 2 + 5$ =Â 121

Question 15:Â 18 19.7 16.3 23.1 9.5 ?

a)Â 36.5

b)Â 36.8

c)Â 36.7

d)Â 36.9

e)Â None of these

Solution:

The pattern isÂ :

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

19.7Â $- 1.7 \times 2^1$ =Â 16.3

16.3Â $+ 1.7 \times 2^2$ =Â 23.1

23.1Â $- 1.7 \times 2^3$ =Â 9.5

9.5Â $+ 1.7 \times 2^4$ =Â 36.7

Question 16:Â 68, ?, 77, 104, 168, 293

a)Â 69

b)Â 70

c)Â 68

d)Â 74

e)Â None of these

Solution:

Cubes of consecutive natural numbers are added

68 $+ 1^3$ =Â 69

69Â $+ 2^3$ =Â 77

77Â $+ 3^3$ =Â 104

104Â $+ 4^3$ =Â 168

168Â $+ 5^3$ =Â 293

Question 17:Â In how many different ways can the numbers â€˜256974â€™ 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)Â 48

b)Â 720

c)Â 36

d)Â 360

e)Â None of these

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

a)Â 58

b)Â -48

c)Â -56

d)Â 42

e)Â -52

Solution:

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

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

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

x = Â±52

Question 19:Â 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)Â $\frac{7}{13}$

b)Â $\frac{3}{8}$

c)Â $\frac{4}{7}$

d)Â $\frac{1}{6}$

e)Â None of these

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

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

Question 20:Â If $(11)^{3}$ is subtracted from $(46)^{2}$ . what will be the remainder?

a)Â 787

b)Â 785

c)Â 781

d)Â 783

e)Â None of these

$(46)^2$ = 2116
$(11)^3$ = 1331