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# Alphabet and Numeric Puzzles Questions for IBPS RRB and PO Prelims

Question 1: Complete the series; FAQ, AQF, ___, FAQ, AQF

a) QAF

b) QFA

c) FQA

d) AFQ

Solution:

From the 2nd term onwards,

1st letter of every term is 2nd letter of the previous term

2nd letter of every term is 3rd letter of the previous term

3rd letter of every term is 1st letter of the previous term

2nd term is AQF so as per the above logic third term will be QFA.

Question 2: Complete the series; DIT, GMQ, KQM, PUH,____

a) VYB

b) UYB

c) VXC

d) TUJ

Solution:

Considering the first letters of each term.

There are 2 alphabets in between D and G (E and F).

There are 3 alphabets between G and K (H, I and J).

There are 4 alphabets between K and P (L, M, N, and O)

Hence, there will be 5 alphabets between P and the required letter. Thus, the required letter will be V (Q, R, S, T and U)

Considering the second letter of each term.

There are three alphabets between each of the second terms, I and M (J, K, L), M and Q (N, O, P), and Q and U (R, S, T).

Hence, there will be three terms between the required term and U, i.e., V, W, X. Thus, the required second term is Y.

Hence, the answer is option A.

Question 3: The missing character in the series A D I _ Y is

a) Q

b) P

c) R

d) O

Solution:

There are two alphabets between A and D.
There are four alphabets between D and I.
There must be six alphabets between I and next letter.
I, J, K, L, M, N, O, P
There are eight alphabets between P and Y
Therefore, missing character is ‘P’.

Question 4: The seventh term in the series 5, 13, 29, 61, ……. is ________________.

a) 97

b) 191

c) 253

d) 509

Solution:

5 x 2 + 3 = 13

13 x 2 + 3 = 29

29 x 2 + 3 = 61

61 x 2 + 3 = 125

125 x 2 + 3 = 253

253 x 2 + 3 = 509

Question 5: Complete the following series of numbers: 6, 26, 126, 626, …….

a) 3126

b) 3125

c) 3216

d) 3215

Solution:

6 x 5 – 4 = 26

26 x 5 – 4 = 126

126 x 5 – 4 = 626

626 x 5 – 4 = 3126

Question 6: Select the combination of letters that when sequentially placed in the blanks of the given letter series will complete the series.
dabx_dab_ _d_b_xdab_x_

a) axdaxxd

b) xxxaxxd

Solution:

By Trial and Error method,

Option A

dabxa | dabxd | dabxx | dabxx | d

The above letters do not form a series. Hence option A is incorrect.

Option B

dabxx | dabxx |dabxx | dabxx | d

The above letters form a series.

Hence, the correct answer is Option B

Question 7: Select the combination of letters that when sequentially placed in the gaps of the given letter series will complete the series.
K _ _ MKXZ _ KX _ MK _ ZM

a) XZMZK

b) MZMZX

c) XXMKX

d) XZMZX

Solution:

By Trial and Error method,

Option A

K X Z M | K X Z M | K X Z M | K K Z M

The above letters do not form a series. Hence option A is incorrect.

Option B

K M Z M | K X Z M | K X Z M | K X Z M

The above letters do not form a series. Hence option B is incorrect.

Option C

K X X M | K X Z M | K X K M | K X Z M

The above letters do not form a series. Hence option C is incorrect.

Option D

K X Z M | K X Z M | K X Z M | K X Z M

The above letters form a series.

Hence, the correct answer is Option D

Question 8: Select the letter-cluster that can replace the question mark (?) in the following series.
BDF, DHL, FLR, ?

a) HPX

b) IQY

c) HPY

d) HOX

Solution:

The logic here is

B + 2 = D  $\longrightarrow$  D + 2 = F  $\longrightarrow$  F + 2 = H

D + 4 = H  $\longrightarrow$  H + 4 = L  $\longrightarrow$  L + 4 = P

F + 6 = L  $\longrightarrow$  L + 6 = R  $\longrightarrow$  R + 6 = X

Similarly, the next letter-cluster in the series is HPX

Hence, the correct answer is Option A

Question 9: Select the combination of letters that when sequentially placed in the gaps of the given letter series will complete the series.
U_D_UP_M_PDMU_DM

a) UMDUD

b) MPUDU

c) PMUPD

d) PMDUP

Solution:

By Trial and Error method,

Option A

UUDM | UPDM | UPDM | UDDM

The above letters do not form a series. Hence option A is incorrect.

Option B

UMDP | UPUM | PPDM | UUDM

The above letters do not form a series. Hence option B is incorrect.

Option C

UPDM | UPUM | PPDM | UDDM

The above letters do not form a series. Hence option C is incorrect.

Option D

UPDM | UPDM | UPDM | UPDM

The above letters form a series.

Hence, the correct answer is Option D

Question 10: Select the combination of letters that when sequentially placed in the blanks of the given series will complete the series.
a b b _ b_ _ b a b _ _ b a b b b b_ _

a) b a b b a a b

b) a b b a a b b

c) a b b a b a b

d) a b b a a a b

Solution:

By Trial and Error method,

Option A

a b b b b | a b b a b | b a b a b | b b b a b

The above letters do not form a series. Hence option A is incorrect.

Option B

a b b a b | b b b a b | a a b a b | b b b b b

The above letters do not form a series. Hence option B is incorrect.

Option C

a b b a b | b b b a b | a b b a b | b b b a b

The above letters form a series.

Hence, the correct answer is Option C