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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

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

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 = $\sqrt{?}$

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

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)

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

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

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

= $24.75 \approx 24$

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

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

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

= $9000 [(1 + \frac{11}{100})^2 – 1]$

= $9000 [(1.11)^2 – 1]$

= $9000 \times (1.2321 – 1)$

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

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

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

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

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

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

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

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 = ±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

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

Here

$(46)^2$ = 2116

$(11)^3$ = 1331

So, 2116 – 1331 = 785

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

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.

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.

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.

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.

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.

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

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

= 30.

Option A is the right answer.

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

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

= $324 + 4*18$

= $396$

Option A is the right answer.

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

$9 =\sqrt{?}$

=> ? = 729

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

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

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

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

$13 \times 0.5 + 6.5 = 13$

$13 \times 1 + 7 = 20$

$20 \times 1.5 + 7.5 = 37.5$

$37.5 \times 2 + 8 = 83$

$83 \times 2.5 + 8.5 = 216$