SSC CGL 2024 Tier-I · 9 September 2024, Shift 2 · General Intelligence and Reasoning · Question 21 of 100

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 5 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

Three of the following four letter-cluster pairs are alike in a certain way and thus form a group. Which is the letter-cluster pair that does NOT belong to that group? (Note: The odd one out is not based on the number of consonants/vowels or their positions in the letter-cluster.)

(a)EJ - OT
(b)KP - UZ
(c)PF - OI
(d)CH - MR

Answer

Why

Correct — C. Number the alphabet A = 1 … Z = 26, then check two things separately: the gap inside a cluster, and the shift from the first cluster to the second.

Rule: +5 inside the cluster, then +10 to each letter across the pair.

E5 J10 → O15 T20 ✓
K11 P16 → U21 Z26 ✓
C3 H8 → M13 R18 ✓
P16 F6 → O15 I9 ✗

PF – OI breaks both readings: inside the cluster the step is −10 rather than +5, and across the pair the letters shift −1 and +3 rather than +10 each.

Why the others are wrong

(a)
EJ - OT — E to J is +5, and E→O and J→T are both +10. It matches the group on both readings, so it cannot be the odd pair.
(b)
KP - UZ — K to P is +5, and K→U and P→Z are both +10. Identical structure to the majority.
(d)
CH - MR — C to H is +5, and C→M and H→R are both +10. It sits squarely inside the group.

Concept

Letter-cluster items are positional arithmetic. Write A = 1 through Z = 26 and stop reading the letters as letters.

Then check two things, not one: the gap inside each cluster, and the letter-by-letter shift from the first cluster to the second. A pair belongs to the group only if both readings agree with the majority.

The note in the stem rules out counting vowels and consonants. That is SSC telling you the intended rule is positional and nothing else.

Learning the alphabet positions in blocks of five — E is 5, J is 10, O is 15, T is 20, Y is 25 — makes this item almost instant.

Every letter in the three matching pairs lands on or next to a multiple of five.

Key facts

With A = 1 and Z = 26: E5 J10 O15 T20; K11 P16 U21 Z26; C3 H8 M13 R18.
The group rule is +5 inside a cluster and +10 from the first cluster to the second.
In PF – OI the internal gap runs backwards (P16 to F6) and the two clusters shift by −1 and +3.

Study next

Letter-cluster analogies mixing forward and backward shiftsClusters built on the opposite-letter mapping, where A pairs with ZMixed letter-and-number cluster series

Common traps

Checking only the shift between the two clusters and missing that the internal gap in option (c) is reversed.
Counting from A = 0, which throws every position out by one.
Trying a vowel-consonant pattern, which the stem explicitly rules out.

SSC asks for the odd letter-cluster pair in most shifts, and the note about vowels and consonants appears whenever the intended rule is purely positional — treat it as a hint, not boilerplate. The numeric form of the same odd-one-out is Reasoning Q.18 in this shift (9 Sep 2024, 12:30).

Related PYQs

No directly related past PYQ was found.