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# LIC AAO Number Series Questions PDF

Question 1: Find the missing number in series:
2, 5, 9, ?, 20, 27

a) 14

b) 16

c) 18

d) 24

Question 2: How many 7’s are there in the following series which are preceded by 6, which is not preceded by 8 ?
8 7 6 7 8 6 7 5 6 7 9 7 6 1 6 7 7 6 8 8 6 9 7 6 8 7

a) Two

b) Nil

c) One

d) Three

Instructions

Study the following arrangement of letters/numbers/symbols carefully and answer the questions given below :

J & B L E H $3 C % 8 I F M 4 1 A R 6 V$\uparrow$9 D 5 K U 2 T ? Z Question 3: How many such symbols are there in the above arrangement each of which is immediately preceded by a digit and also immediately followed by a consonant? a) None b) One c) Two d) Three e) More than three Question 4: On the basis of the positions in the above arrangement if H is related to I, then F is related to a)$\uparrow$b) 9 c) D d) 6 e) R Question 5: If all the numbers are deleted from the above arrangement, which of the following would be the seventh from the left end? a) C b)$

c) H

d) %

e) I

Question 6: If all the alphabets are deleted from the above arrangement, which of the following will be the third from the right end?

a) 4

b) $\uparrow$

c) 1

d) 5

e) 9

Question 7: The position of how many digits will remain the same if the digits in the number 7029351648 are rearranged in the ascending order from the left to right?

a) None

b) One

c) Two

d) Three

e) More than three

Question 8: What should come in the place of question mark (?) in the following number series?
3 4 4 3 4 4 5 3 4 4 5 6 ? 4 4 5 6 7 3 4 4 5 6 7 8

a) 4

b) 3

c) 6

d) 5

e) None of these

Instructions

What will come in the place of question mark (?) in the following number series ?

Question 9: 13 14 22 31 95 120 ?

a) 336

b) 332

c) 325

d) 225

e) 425

Question 10: 9 ? 161 504 1233 2564

a) 23

b) 34

c) 32

d) 36

e) None of these

Question 11: 11 ? 87 258 771 2310

a) 24

b) 30

c) 36

d) 45

e) 54

Question 12: 14 9 11 18.5 ? 99.5 300.5

a) 37.5

b) 39

c) 36

d) 38.5

e) None of these

Question 13: 36.2 37.6 39.4 41.6 ? 47.2 50.6

a) 44.2

b) 44.4

c) 44.6

d) 42.2

e) None of these

Instructions

Find the missing number (?) in the given series from the options provided.

Question 14: 2, 4, 7, 11, 16, ?

a) 22

b) 21

c) 20

d) 23

e) 24

Question 15: 9, 27, 45, 63, ?

a) 79

b) 80

c) 81

d) 82

e) 83

Question 16: 3, 8, 15, 24, 35, ?

a) 63

b) 49

c) 80

d) 48

e) 56

Question 17: -3, -1, 11, 39, 89, ?

a) 198

b) 189

c) 210

d) 174

e) 167

Question 18: 9, 12, 19, 34, 65, ?

a) 127

b) 128

c) 129

d) 130

e) 134

Instructions

Find the missing number (?) in the given series from the options provided.

Question 19: 3, 12, 25, 42, 63,?

a) 88

b) 77

c) 78

d) 83

e) 87

Question 20: 6, 24, 60, 120, 210, ?

a) 345

b) 336

c) 326

d) 316

e) 320

There are 3 such 7 preceeded by  which is not preceeded by 8.
8 7 6 7 8 6 7 5 6 7 9 7 6 1 6 7 7 6 8 8 6 9 7 6 8 7
Therefore, the correct option is option D.

$\Rightarrow$ $T_{2} – T _{1} = 4 – 2 = 2$
$\Rightarrow$ $T_{3} – T _{2} = 7 – 4 = 3$
$\Rightarrow$ $T_{4} – T _{3} = 11 – 7 = 4$
$\Rightarrow$ $T_{5} – T _{4} = 16 – 11 = 5$
Here we can see the pattern that difference between consecutive terms are in A.P. and it is increasing by 1. Therefore,
$\Rightarrow$ $T_{6} = 6 + T _{5} = 6+16 = 22$
Hence, option A is the correct answer.

$\Rightarrow$ $T_{1} = 9 = 5^2 – 4^2$
$\Rightarrow$ $T_{2} = 27 = 6^2 – 3^2$
$\Rightarrow$ $T_{3} = 45 = 7^2 – 2^2$
$\Rightarrow$ $T_{4} = 63 = 8^2 – 1^2$
Hence, we can say that $\Rightarrow$ $T_{n} = (n+4)^2 – (5-n)^2$
$\Rightarrow$ $T_{5} = (5+4)^2 – (5-4)^2 = 81$
Hence, option C is the correct answer.

$\Rightarrow$ $T_{1} = 3 = 2^2 – 1$
$\Rightarrow$ $T_{2} = 8 = 3^2 – 1$
$\Rightarrow$ $T_{3} = 15 = 4^2 – 1$
$\Rightarrow$ $T_{4} = 24 = 5^2 – 1$
$\Rightarrow$ $T_{5} = 35 = 6^2 – 1$
Hence, we can say that $\Rightarrow$ $T_{n} = (n+1)^2 – 1$
$\Rightarrow$ $T_{6} = (6+1)^2 – 1 = 48$
Hence, option D is the correct answer.

$\Rightarrow$ $T_{1} = -3 = 1^3 – 2^2$
$\Rightarrow$ $T_{2} = -1 = 2^3 – 3^2$
$\Rightarrow$ $T_{3} = 11 = 3^3 – 4^2$
$\Rightarrow$ $T_{4} = 39 = 4^3 – 5^2$
$\Rightarrow$ $T_{5} = 89 = 5^3 – 6^2$
Hence, we can say that $\Rightarrow$ $T_{n} = n^3 – (n+1)^2$
$\Rightarrow$ $T_{6} = 6^3 – 7^2 = 167$
Hence, option E is the correct answer.

$\Rightarrow$ $T_{1} = 9 = 2^2 + 5$
$\Rightarrow$ $T_{2} = 12 = 2^3 + 4$
$\Rightarrow$ $T_{3} = 19 = 2^4 + 3$
$\Rightarrow$ $T_{4} = 34 = 2^5 + 2$
$\Rightarrow$ $T_{5} = 65 = 2^6 + 1$
Hence, we can say that $\Rightarrow$ $T_{n} = 2^{n+1} + (6 -n)$
$\Rightarrow$ $T_{6} = 2^{6+1} + (6 – 6) = 128$
Hence, option B is the correct answer.

$\Rightarrow$ $T_{1} = 3 = 3*1$
$\Rightarrow$ $T_{2} = 12 = 4*3$
$\Rightarrow$ $T_{3} = 25 = 5*5$
$\Rightarrow$ $T_{4} = 42 = 6*7$
$\Rightarrow$ $T_{5} = 63 = 7*9$
Hence, we can say that $\Rightarrow$ $T_{n} = (n+2)*(2n -1)$
$\Rightarrow$ $T_{6} = 8*11 = 88$
Hence, option A is the correct answer.

$\Rightarrow$ $T_{1} = 6 = 1*2*3$
$\Rightarrow$ $T_{2} = 24 = 2*3*4$
$\Rightarrow$ $T_{3} = 60 = 3*4*5$
$\Rightarrow$ $T_{4} = 120 = 4*5*6$
$\Rightarrow$ $T_{5} = 120 = 5*6*7$
Hence we can generalise that $T_{n} = (n)*(n+1)*(n+2)$
$\Rightarrow$ $T_{6} = 6*7*8 = 336$