Logical Reasoning in English ALPHABET Series #MCQ_CetJob

Chapter Wise Logical Reasoning in English ALPHABET Series

"1 >>AZ, GT, MN, ?, YB ?

• (A) KF
• (B) RX
• (C) SH
• (D) TS

Solution First letter moves 6 steps forward. Second letter moves 6 steps backward.

2 >>AZ, CX, FU, ? ?

• (A) IR
• (B) IV
• (C) JQ
• (D) KP

Solution The first letter moves 2, 3, 4 ... steps forward. The second letter moves 2, 3, 4 ... steps backward.

3 >>AZ, BY, CX, ? ?

• (A) EF
• (B) GH
• (C) IJ
• (D) DW

Solution The first letter moves one step forward. The second letter moves one step backward.

4 >>DKY, FJW, HIU, JHS, ? ?

• (A) KGR
• (B) LFQ
• (C) KFR
• (D) LGQ

Solution The first letters are alternate. The second letter of moves one step backward to obtain the second letter of the successive term. The third letter moves two steps backward to obtain the third letter of the subsequent term.

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5 >>A, B, N, C, D, O, E, F, P, ?, ?, ? ?

• (A) G, H, I
• (B) G, H, J
• (C) G, H, Q
• (D) J, K, L

Solution The series consists of 2 groups (A, B, C, D, E, F, ?, ?) and (N, O, P, ?). The series is made of two terms of first group followed by one term of second group.

6 >>Z, W, S, P, L, I, E, ? ?

• (A) B
• (B) D
• (C) F
• (D) K

Solution The letters move 3, 4, 3, 4, ... steps backward.

7 >>A, I, P, V, A, E, ? ?

• (A) E
• (B) F
• (C) G
• (D) H

Solution The letters move 8, 7, 6, 5... steps forward.

Logical Reasoning in English

8 >>C4X, F9U, I16R, ? ?

• (A) K25P
• (B) L25P
• (C) L25O
• (D) L27P

Solution The first letter moves three steps forward. The last letter moves three steps backward. The nubers are square in 22, 32, 42, 52, etc.,

9 >>ba_ba_bac_acb_cbac ?

• (A) aacb
• (B) bbca
• (C) ccba
• (D) cbac

Solution The series is bac/bac/bac/bac/bac/bac. The pattern bac is repeated.

10 >>m _ nm _ n _ an _ a _ ma _ ?

• (A) aamnan
• (B) ammanm
• (C) aammnn
• (D) amammn

Solution The series is man/man/man/man/man. The pattern 'man' is repeated.

11 >>adb_ac_da_cddcb_dbc_cbda ?

• (A) bccba
• (B) cbbaa
• (C) ccbba
• (D) bbcad

Solution The series is adbcacbdabcddcbadbcacbda.

12 >>EV, JQ, OL, ? ?

• (A) TH
• (B) TG
• (C) FT
• (D) US

Solution E is fifth from beginning of the alphabet and V is fifth from the end.Similarly J is tenth from the beginning and Q is 10th from the end and the same rule is followed in the other pairs of the series.

13 >>prt ......... bdf , hjl , npr ?

• (A) uwy
• (B) vya
• (C) vxz
• (D) axy

Solution There is a gap of one letter in every letter of the group. Moreover, every second group of the series starts after a gap of one letter, i.e., after prt, we see vxz.

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14 >>KPA, LQB, MRC, NSD, ...... ?

• (A) IJOT
• (B) OTE
• (C) EOT
• (D) TOE

Solution The first letter of every group is in alphabetical order, i.e., KLMNO. Same is the case with II and III letters of the group. Such as PQRST and ABCDE

15 >>cc _ a _ca_ ccb _cca_c_ba ?

• (A) cbaba
• (B) bcabc
• (C) bacbc
• (D) None of these

Solution The arrangement is ccba, ccab, ccba, ccab,ccba.

16 >>In following alphabet series , one term missing as shown by question mark (?).

Choose missing term from options.
U, O, I, ?, A ?

• (A) E
• (B) C
• (C) S
• (D) G

Solution The series consists of vowels A, E, I, O, U written in a reverse order.

17 >>In following alphabet series , one term missing as shown by question mark (?).

Choose missing term from options.
Y, W, U, S, Q, ?, ? ?

• (A) N, J
• (B) M, L
• (C) J, R
• (D) O, M

Solution The series consits of alternate letters in reverse order.

18 >>In following alphabet series , one term missing as shown by question mark (?).

Choose missing term from options.
A, B, D, G, ? ?

• (A) M
• (B) L
• (C) K
• (D) H

Solution The first, second, third,.... letters of the series are respectively moved one, two, three,... steps forward to obtain the seccessive terms.

19 >>Find out the wrong number from the given series125, 126, 124, 127, 123, 129 ?

• (A) 126
• (B) 124
• (C) 123
• (D) 129

Solution The sequence is + 1, - 2, + 3, - 4, + 5.So, 129 is wrong and and must be replaced by (123 + 5) i.e. 128.

20 >>Find out the wrong number from the given series 3, 4, 10, 32, 136, 685, 4116 ?

• (A) 10
• (B) 32
• (C) 136
• (D) 685

Solution The sequence is as follows :2nd term = (1st term + 1) * 13rd term = (2nd term + 1) * 24th term = (3rd term + 1) * 3 and so on.So, 32 is wrong and must be replaced by (10 + 1) * 3 i.e. 33.

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