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# Series Completion Questions for MAH – CET 2022 – Download PDF

Here you can download CMAT 2022 – important MAH – CET Series Completion Questions PDF by Cracku. Very Important MAH – CET 2022 and These questions will help your MAH – CET preparation. So kindly download the PDF for reference and do more practice.

Question 1: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

Question 2: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

Question 3: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

Question 4: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

Question 5: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

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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

Question 7:Â What will come in place of thequestion mark (?) in the following number series?
33Â  16.5Â  ?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

Question 8:Â What will come in place of thequestion 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

Question 9:Â What will come in place of thequestion 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

Question 10: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

Question 11: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

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

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

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

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

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

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

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

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

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

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

$\because$ 2500Â < 2530Â < 2601

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

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

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

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

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

= $24.75 \approx 24$

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

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

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

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

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

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

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)

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

= $9000 [(1 + \frac{11}{100})^2 – 1]$

= $9000 [(1.11)^2 – 1]$

= $9000 \times (1.2321 – 1)$

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

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

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

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

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

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 = $24 + 24 = 48$

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

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

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

x = Â±52

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

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