# Tricky questions of Reasoning(Series) for SSC CGL Exam 2017

Directions (1-4): In each of the following question, 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.
Q1. 1, 3, 3, 8, 7, 13, ?, 18, 21
(a) 10
(b) 11
(c) 12
(d) 13

Q2. 3, 7, 15, ?, 63, 127
(a) 30
(b) 31
(c) 47
(d) 52

Q3. 1, 5, 13, 25, 41, ?
(a) 51
(b) 57
(c) 61
(d) 63

Q4. 1, 1, 8, 4, 27, 9, 64, ?
(a) 12
(b) 16
(c) 81
(d) 49

Q5. Which fraction comes next in the sequence 1/2,3/4,5/8,7/16,  ?
(a) 9/32
(b) 10/17
(c) 11/34
(d) 12/35

Directions (6-9): In each of the following questions, one term in the number series is wrong. Find out the wrong term.
Q6. 125, 126, 124, 127, 123, 129
(a) 124
(b) 127
(c) 123
(d) 129

Q7. 5, 10, 40, 80, 320, 550, 2560
(a) 10
(b) 80
(c) 320
(d) 550

Q8. 8, 13, 23, 43, 86, 163, 323
(a) 23
(b) 163
(c) 86
(d) 323

Q9. 1, 5, 5, 9, 7, 11, 11, 15, 12, 17
(a) 11
(b) 12
(c) 17
(d) 15

Directions (10-15): 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.
Q10. A, I, P, V, A, E, ?
(a) G
(b) F
(c) J
(d) H

Q11. b e d f ? h j ? l
(a) i m
(b) m i
(c) i n
(d) j m

Q12. ajs, gpy, ? , sbk, yhq
(a) emv
(b) mve
(c) oua
(d) qzi

Q13. DEF, HIJ, MNO, ?
(a) RST
(b) STU
(c) RTV
(d) SRQ

Q14. C4X, F9U, I16R, ?
(a) L25Q
(b) L25P
(c) L25O
(d) L27P

Q15. Q1F, S2E, U6D, W2IC, ?
(a) Y99B
(b) Y77B
(c) Y88B
(d) Z88B

Solutions

S1. Ans.(d)
Sol. Clearly, the given sequence is a combination of two series:
I. 1, 3, 7, ?, 21 and  II. 3, 8, 13, 18
The pattern followed in I is +2, +4, ……. and the pattern followed in II is +5.
So, missing number = 7 + 6 = 13

S2. Ans.(b)
Sol. Each number in the series is one more than twice the preceding number.
So, missing term = (15 × 2) + 1 = 31.

S3. Ans.(c)
Sol. The pattern is +4, +8, +12, +16, …..
So, missing term = 41 + 20 = 61

S4. Ans.(b)
Sol. The series consists of squares and cubes of consecutive natural numbers i.e.
13, 12, 23, 22,, 33, 32, 43,42, ……
So, missing term = 42 = 16.
S5. Ans.(a)
Sol. Clearly, the numerators of the fractions in the given sequence from the series 1, 3, 5, 7, in which each term is obtained by adding 2 to the previous term.
The denominators of the fractions form the series 2, 4, 8, 16 i.e. 21, 22, 23, 24.
So, the numerator of the next fraction will be (7 + 2) i.e. 9 and the denominator will be 25 i.e. 32.
Thus, the next term is . Hence, the answer is (a).
S6. Ans.(d)
Sol. The correct pattern is +1, -2, +3, -4, +5
So, 129 is wrong and must be replaced by (123 + 5) i.e. 128.

S7. Ans.(d)
Sol. The correct pattern is ×2, ×4, ×2, ×4, …..
So, 550 is wrong and must be replaced by (320 × 2) i.e. 640.

S8. Ans.(c)
Sol. The correct pattern is ×2 – 3
So, 86 is wrong and must be replaced by (43 × 2 – 3) i.e. 83.

S9. Ans.(b)
Sol. The given sequence is a combination of two series:
I. 1, 5, 7, 11, 12 and  II. 5, 9, 11, 15, 17
The pattern in both I and II is +4, +2, +4, +2
So, 12  is wrong and must be replaced by (11 + 2 ) i.e. 13.

S10. Ans.(d)
Sol.

S11. Ans.(a)
Sol. The series may be divided into groups as shown:
b e d / f ? h / j ? l
Clearly in the first group, the second and third letters are respectively three and two steps ahead of the first letter. A similar pattern would follow in the second and third groups.

S12. Ans.(b)
Sol.

S13. Ans.(b)
Sol.

S14. Ans.(c)
Sol.

S15. Ans.(c)
Sol.

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