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Reasoning Questions for SSC CGL Exam 2017

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Dear students only 2 months are left and for SSC CGL(Tier-I) Exam 2017 we have provided 60 Days Plan, Today is the 3rd Day of plan, so according to that plan we are providing miscellaneous of Number & Alphabetical Series of Reasoning quiz in accordance with the syllabus of SSC CGL Tier-I. Daily we will follow this plan which will really help you in getting more marks in the final exam.
Directions(1-6): In the following questions, various terms of a letter series are given with one or two term missing as shown by (?). Choose the missing term out of the given alternatives.

Q1. 1, 3, 3, 6, 7, 9, ?, 12, 21
(a) 13 
(b) 14
(c) 10
(d) 11
Ans.(a)
Sol.
Q2. 3, 10, 29, 66, 127, ?
(a) 189 
(b) 218
(c) 154
(d) 126
Ans.(b)
Sol.
 
Missing term = 127 + (61 + 24 + 6)
= 127 + 91
= 218 

Q3. 0.5, 1.5, 4.5, 13.5, ( )   
(a) 44.5 
(b) 38.5 
(c) 32.5 
(d) 40.5 
Ans.(d)
Sol.   Each term of the series is obtained by multiplying the preceding term by 3. 
Missing number = 13.5 x 3 =  40.5
Q4. W, T, U, Q, S, N, ?, ? 
(a) I, J 
(b) L, J
(c) Q, K
(d) I, G
Ans.(c)
Sol. 
Q5. C, G, K, O, ? 
(a) Q 
(b) R
(c) S
(d) T 
Ans.(c)
Sol.
 

Q6.  a,y,c,w,e,u, (?), (?) 
(a) f q 
(b) f t
(c) g k 
(d) g s
Ans.(d)
Sol.  The series divided into two sequence: 
I- a, c, e II- y, w, u
In I, letters is moved two steps forward.
In II, letters is moved two steps backward.
So, e—-+2–> g
u—- -2—> s
Directions(7-8): Answer the following question according to the given series:
J * R 3 P L 2 # I N 7 O C @ K 5 D = M $ 6 B < A Q 4

Q7. If this series is re-arranged in the reverse order, which will be the eleventh element to the left to the sixteenth element from the left end?
(a) J 
(b) 6 
(c) B
(d) < 
Ans.(c)
Sol.  11th to the left of 16th from left means 5th from the left. But the sequence has been reversed. Therefore, required element will be 5th from right in the original sequence.
5th from right is “B”
Q8. What will come in the place of the question mark (?) in the following series based on the above arrangement?
P R J # L 3 7 I 2 @ O N ?
(a) D K @  
(b) 5 @ O 
(c) D K C
(d) = 5 @ 
Ans.(c)
Sol. 
 

Directions(9-10):The following questions are based upon the alphabetical series given below:
S L U A Y J V E I O N Q G Z B D R H

Q9. What will come in place of question (?) mark in the following series?
LAJEOQ?
(a) ZBH 
(b) IBR 
(c) ZDH
(d) QRH 
Ans.(c)
Sol.  Difference between letter is +2.

Q10. If ‘SU’ is related ‘HD’ and ‘UY’ is related to ‘DZ’ in a certain way, to which of the following is YV related to following the same pattern?
(a) ZQ 
(b) IN 
(c) BG
(d) QO 
Ans.(a)
Sol.  The corresponding letters from the right end.
Directions(11-12): In each of the following questions, various terms of a letter series are given with one or two terms missing as shown by (?). Choose the missing term out of the given alternatives. 
Q11. 1, 6, 15, (?), 45, 66, 91   
(a) 25 
(b) 26 
(c) 27 
(d) 28 
Ans.(d)
Sol.   The pattern is + 5, + 9,+13, +17, + 21, + 25 
Missing number =15 + 13 =28. 

Q12. W, V, T, S, Q, P, N, M, ?, ?   
(a) I, J 
(b) J, I 
(c) J, K 
(d) K, J
Ans.(d)
Sol. The letters are alternately moved one and two steps backward to obtain the successive terms. 

Directions(13-14): In the following questions, one term in the number series is wrong. Find out the wrong term. 

Q13. In this question, one term in the number series is wrong. Find out the wrong term
11, 5, 20, 12, 40, 26, 74, 54   
(a) 5 
(b) 20 
(c) 40
(d) 26
Ans.(c)
Sol.  The given sequence is a combination of two series: 
I. 11, 20, 40, 74  and  II. 5, 12, 26, 54 
The pattern in I is: +9, + 18, + 36, … 
The pattern in II is: +7, +7×2, +14×2
So, 40 is wrong it must be replaced by 38 
Q14. 8, 13, 21, 32, 47, 63, 83 
(a) 21 
(b) 32 
(c) 47 
(d) 83 
Ans.(c)
Sol. The sequence is + 5, + 8, +11, +14, +17, +20 
So. 47 is wrong and must be replaced by (32 + 14) i.e. 46. 

Q15. In this letter series, some of the letters are missing which are given in that order as one of the alternatives below it. Choose the correct alternative. 
 a _ b _ cbb _ aa _ ca _ ba _ bcc.  
(a) ababca 
(b) acccba
(c) accabb
(d) bcacab
Ans.(b)
Sol.   The series is aabcc/bbcaa/ccabb/aabcc. 
The letters move in a cyclic order and in each group, the first and third letters occur twice.
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