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Test: Letter Series Type 1


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12 Questions MCQ Test CSAT Preparation for UPSC CSE | Test: Letter Series Type 1

Test: Letter Series Type 1 for SSC CGL 2022 is part of CSAT Preparation for UPSC CSE preparation. The Test: Letter Series Type 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The Test: Letter Series Type 1 MCQs are made for SSC CGL 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Letter Series Type 1 below.
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Test: Letter Series Type 1 - Question 1

A, B, D, H, ____, F

Detailed Solution for Test: Letter Series Type 1 - Question 1

Each letter in the given series is multiplied with 2 to get the next letter in the series.

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.

So, Correct Option is Option A

Test: Letter Series Type 1 - Question 2

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

Detailed Solution for Test: Letter Series Type 1 - Question 2

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.

So, Correct Option is Option B

Test: Letter Series Type 1 - Question 3

G, J, N, Q, U, ___

Detailed Solution for Test: Letter Series Type 1 - Question 3

G + 3, J + 4, N + 3, Q + 4, U + 3 , _____
The next letter in the series is U + 3 = X.

So, Correct Option is Option C.

Test: Letter Series Type 1 - Question 4

BY, CX, EV, GT, KP, ____

Detailed Solution for Test: Letter Series Type 1 - Question 4

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.
Test: Letter Series Type 1 - Question 5

AP, BQ, CR, DS, ___

Detailed Solution for Test: Letter Series Type 1 - Question 5
  • 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.
Test: Letter Series Type 1 - Question 6

AI, EO, IU, ___, UE

Detailed Solution for Test: Letter Series Type 1 - Question 6
  • 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.

Test: Letter Series Type 1 - Question 7

KPD, LOE, MNF, NMG, ____

Detailed Solution for Test: Letter Series Type 1 - Question 7

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.
Test: Letter Series Type 1 - Question 8

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

Detailed Solution for Test: Letter Series Type 1 - Question 8

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.

Test: Letter Series Type 1 - Question 9

GTB, CYV, YDP, ____, QND

Detailed Solution for Test: Letter Series Type 1 - Question 9

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.

Test: Letter Series Type 1 - Question 10

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

Detailed Solution for Test: Letter Series Type 1 - Question 10
  • 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.

Test: Letter Series Type 1 - 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 5th to the left of 12th letter from the right hand side of the alphabet series?

Detailed Solution for Test: Letter Series Type 1 - Question 11

When the letters are reversed in the group of Five, six, and seven letters.
5th to the left of 12th from the right end (27- 12 = 15th from left) means, 15 – 5 = 10th that lies in the 2ndgroup(6-11).
Therefore, 6 + 11 = 17 – 10 = 7th letter i.e. “G”.

Test: Letter Series Type 1 - 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 5th to the right of 12th letter from the right hand side of the alphabet series?

Detailed Solution for Test: Letter Series Type 1 - Question 12

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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