# Alphanumeric Questions for SSC CGL Exam

Directions (1-6): In each of the following questions, various terms of an alphabet series are given with one or more terms missing as shown by (?). Choose the missing terms out of the given alternatives.

Q1. T, R, P, N, L, ?, ?
(a) J, G
(b) J, H
(c) K, H
(d) I, K
Ans.(b)
Sol.

Q2. Z, W, S, P, L, I, E, ?
(a) B
(b) D
(c) F
(d) K
Ans.(a)
Sol.

Q3. Z, X, S, I, R, R, ?, ?
(a) G, I
(b) J, I
(c) J, K
(d) K, M
Ans.(a)
Sol.

Note that the numbers representing the difference between the consecutive terms of the series again form a series – 2, 5, 10, 17, 26, 37, 50 – in which the patterns is +3, +5, +7, +9, +11, +13.

Q4. Y, B, T, G, O, ?
(a) N
(b) M
(c) L
(d) K
Ans.(c)
Sol. The given sequence is a combination of two series:
I. Y, T O and  II. B, G, ?
I consists of 2nd, 7th and 12th letters from the end of the English alphabet, while
II consists of 2nd, 7th, and 12th letters from the beginning of the English alphabet.
So, the missing letter in II is the 12th letter from the beginning of the English alphabet, which is L.

Q5. BZA, DYC, FXE, ?, JVI
(a) HUG
(b) HWG
(c) UHG
(d) WHG
Ans.(b)
Sol.

Q6. A, CD, GHI, ?, UVWXY
(a) LMNO
(b) MNPQ
(c) MNOP
(d) NOPQ
Ans.(c)
Sol.

Each term consists of consecutive letters in order. The number of letters in the terms goes on increasing by one at each step.

Directions (7-15): In each of the following questions, a number series is given with one term missing. Choose the correct alternative that will continue the same pattern and replace the question mark in the given series:

Q7. 0, 2, 8, 14, ?, 34
(a) 20
(b) 23
(c) 24
(d) 25
Ans.(c)
Sol. The pattern is +2, +6, +6, +10, +10 …..
So, missing term = 14 + 10 = 24

Q8. 4, 6, 9, 13 1/2, ?
(a) 17 1/2
(b) 19
(c) 20 1/4
(d) 22 3/4
Ans.(c)

Q9. 48, 24, 96, 48, 192, ?
(a) 76
(b) 90
(c) 96
(d) 98
Ans.(c)
Sol. The pattern is ÷2 × 4, ÷ 2, × 4, ….
So, missing term = 192 ÷ 2 = 96

Q10. 4, 10, ?, 82, 244, 730
(a) 24
(b) 28
(c) 77
(d) 218
Ans.(b)
Sol. Each number in the series is 2 less than thrice the preceding number
So, missing number = (10 × 3) – 2 = 28

Q11. 6, 13, 28, 59, ?
(a) 111
(b) 113
(c) 114
(d) 122
Ans.(d)
Sol. The pattern is × 2 + 1, × 2 + 2, × 2 + 3, …..
So, missing term = 59 × 2 + 4 = 122

Q12. 1, 5, 14, 30, 55, 91, ?
(a) 130
(b) 140
(c) 150
(d) 160
Ans.(b)

Q13. 0, 1, 3, 4, 8, 15, 27, ?
(a) 37
(b) 44
(c) 50
(d) 55
Ans.(c)
Sol. The sum of any three consecutive terms of the series gives the next term
So, missing number = 8 + 15 + 27 = 50

Q14. 3, 8, 13, 24, 41, ?
(a) 70
(b) 75
(c) 80
(d) 85
Ans.(a)
Sol. The pattern followed is:
nth term + (n + 1)th term + (n + 1) = (n + 2)th term
Thus, 1st term + 2nd term + 2 = 3rd term;
2nd term + 3rd term + 3 = 4th term and so on
So, missing term = 6th term = 4th term + 5th term + 5 = 24 + 41 + 5 = 70

Q15. 13, 32, 24, 43, 35, ?, 46, 65, 57, 76
(a) 45
(b) 52
(c) 54
(d) 55
Ans.(c)
Sol. The given sequence is a combination of two series:
I. 13, 24, 35, 46, 57 and  II. 32, 43, ?, 65, 76
The pattern in both I and II is + 11
So, missing term = 43 + 11 = 54

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