Series Completion Questions for MAH – CET 2022 – Download PDF

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Question 1: 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

Question 2: 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

Question 3: 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 4: What approximate value should come in place of the question mark (?) in the following question?
$54.786÷10.121\times 4.454=?$

a) 84

b) 48

c) 118

d) 58

e) 24

Question 5: 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

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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

Question 7: What will come in place of thequestion 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

Question 8: What will come in place of thequestion mark (?) in the following number series?
44  ?  99   148.5  222.75  334.125

a) 44

b) 55

c) 66

d) 33

e) 35

Question 9: What will come in place of thequestion mark (?) in the following number series?
121  238  472  ?  1876  3748

a) 1008

b) 948

c) 944

d) 940

e) 1005

Question 10: 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

Question 11: 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 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

Question 13: 16 8 12 30 ? 472.5

a) 104

b) 103

c) 106

d) 105

e) None of these

Question 14: 2, 5, 12, 27, 58, ?

a) 122

b) 121

c) 123

d) 120

e) None of these

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

Question 16: 68, ?, 77, 104, 168, 293

a) 69

b) 70

c) 68

d) 74

e) None of these

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

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

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

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

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Answers & Solutions:

1) Answer (E)

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

$\because$ 2500 < 2530 < 2601

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

2) 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

3) 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

4) Answer (E)

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

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

= $24.75\approx 24$

5) 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

6) Answer (A)

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

7) Answer (B)

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

8) Answer (C)

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

9) Answer (D)

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

10) Answer (A)

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

11) 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)

12) 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. $2,088.90$

13) 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

14) 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

15) 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

16) 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

17) 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 = $24+24=48$

18) Answer (E)

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

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

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

x = ±52

19) 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

20) Answer (B)

Here

$(46{)}^2$ = 2116

$(11{)}^3$ = 1331

So, 2116 – 1331 = 785

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