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This mock test of Test: Letter Series Type 1 for UPSC helps you for every UPSC entrance exam.
This contains 12 Multiple Choice Questions for UPSC Test: Letter Series Type 1 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Letter Series Type 1 quiz give you a good mix of easy questions and tough questions. UPSC
students definitely take this Test: Letter Series Type 1 exercise for a better result in the exam. You can find other Test: Letter Series Type 1 extra questions,
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QUESTION: 1

A, B, D, H, ____, F

Solution:

A = 1 and 1 × 2 = 2 and the 2nd letter is B.

B = 2 and 2 × 2 = 4 and the 4th letter is D.

Similarly, H = 8 and 8 × 2 = 16, and the 16th letter P is the missing letter.

QUESTION: 2

D, F, G, H, J, K, L, M N, ___

Solution:

D, F, G, H, J, K, L, M, N, ____

The given series consists of consonants in increasing order, starting with D.

The next consonant in the series is P.

QUESTION: 3

G, J, N, Q, U, ___

Solution:

G + 3, J + 4, N + 3, Q + 4, U + 3 , _____

The next letter in the series is U + 3 = X.

QUESTION: 4

BY, CX, EV, GT, KP, ____

Solution:

BY, CX, EV, GT, KP, _____

The given series is a** mixed series**.

- The letters B, C, E, G, and K are the letters in prime value positions. Hence, the next letter is M.
- The second letter in each group forms an opposite pair of the first letter in that group. Hence, the next pair in the series is MN.

QUESTION: 5

AP, BQ, CR, DS, ___

Solution:

- The first letters in all the pairs, i.e., A, B, C, and D, form a series of consecutive letters.
- The second letters P, Q, R, and S form another pair of consecutive letters. Hence, The next pair in the series is ET.

QUESTION: 6

AI, EO, IU, ___, UE

Solution:

- The first letters in all the groups are consecutive vowels starting with A.
- The second letters in all the groups are consecutive vowels starting with I.

Hence, the missing pair **is OA**.

QUESTION: 7

KPD, LOE, MNF, NMG, ____

Solution:

The given series is a mixed series.

__Pattern for the first letter:__

K + 1, L + 1, M + 1, N + 1, O__Pattern for the second letter:__

P – 1, O – 1, N – 1, M – 1, L__Pattern for the third letter:__

D + 1, E + 1, F + 1, G + 1, H- Hence, the next group in the series is OLH.

QUESTION: 8

QLR, JPD, RNU, GNC, SPX, DLB, ____

Solution:

The **alternate groups** are in mixed series.

QLR, RNU, SPX are in one series.

__The Pattern for the first letter:__

Q + 1, R + 1, S + 1, T__The Pattern for the second letter:__

L + 2, N + 2, P + 2, R__The Pattern for the third letter:__

R + 3, U + 3, X + 3, A

Hence, the next group in the series is TRA.

QUESTION: 9

GTB, CYV, YDP, ____, QND

Solution:

The given series is a mixed series.

__Pattern for the first letter:__

G – 4, C – 4, Y – 4, U – 4, Q__Pattern for the second letter:__

T + 5, Y + 5, D + 5, I + 5, N__Pattern for the third letter:__

B – 6, V – 6, P – 6, J – 6, D

Hence, the missing group is UIJ.

QUESTION: 10

2F3, 3O5, 5I7, 7Y11, ____

Solution:

- In each term the numbers are pairs of consecutive prime numbers.
- The letter between them has the place value equal to the product of the two numbers on either side of it.

∵ The next two prime numbers are 11 and 13 whose product is 143. The 143rd letter is (26 × 5 + 13) M.

∴ 11M13 is the next term.

QUESTION: 11

If the first five letters of the alphabet series in the reverse order, then next six are in reverse order, then next seven letters are in reverse order, then remaining letters are in reverse order.

Q. Which letter is 5^{th} to the left of 12^{th} letter from the right hand side of the alphabet series?

Solution:

When the letters are reversed in the group of Five, six, and seven letters.

5th to the left of 12^{th} from the right end (27- 12 = 15^{th} from left) means, 15 – 5 = 10^{th }that lies in the 2^{nd}group(6-11).

Therefore, 6 + 11 = 17 – 10 = 7^{th }letter **i.e.** “G”.

QUESTION: 12

If the first five letters of the alphabet series in the reverse order, then next six are in reverse order, then next seven letters are in reverse order, then remaining letters are in reverse order.

Q. Which letter is 5^{th} to the right of 12^{th} letter from the right hand side of the alphabet series?

Solution:

When the letters are reversed in the group of Five, six, seven and letters.

5th to the right of 12th from right end (27 – 12 = 15th from left) means, 15 + 5 = 20th

Which lies in 4th group(19-26). Therefore,

19 + 26 = 45 ; 45 – 20 = 25th = “Y”.

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