Four letter-clusters have been given, out of which three are alike in some manner and one is different. Select the one that is different. Note :The odd one out is not based on the number of consonants/vowels or their position in the letter cluster

Correct — D.

Convert each cluster to alphabet positions.

Rule: a letter, then one back, then one forward — n, n−1, n+1.

LKM = 12, 11, 13 ✓
HGI = 8, 7, 9 ✓
TSU = 20, 19, 21 ✓
XYW = 24, 25, 23 ✗

Three clusters read n, n−1, n+1. XYW reads n, n+1, n−1 — the two neighbours are written in the reverse order, so it is the odd one out → option (d).

Why the others are wrong

• (a)
  LKM — LKM is 12, 11, 13 — one back, then one forward, which is the shared shape. It belongs with HGI and TSU.
• (b)
  HGI — HGI is 8, 7, 9, the same n, n−1, n+1 shape. Nothing else about it separates it from the group.
• (c)
  TSU — TSU is 20, 19, 21, again n, n−1, n+1. Its letters sit late in the alphabet, but where a cluster starts is not what divides these four.

Odd-one-out on letter clusters is a two-step routine: convert to alphabet positions, then look at the shape the numbers make relative to the first letter.

Every cluster here is built round its own first letter. L(12) is given 11 and 13, H(8) is given 7 and 9, T(20) is given 19 and 21, X(24) is given 25 and 23.

So the neighbours used are identical in all four. What differs is the order in which the two neighbours are written, and the printed note ruling out vowel and consonant counts means the difference has to be arithmetic.

Key facts

• LKM, HGI and TSU each read n, n−1, n+1 in alphabet positions.
• XYW reads n, n+1, n−1, with the two neighbours in reverse order.
• All four clusters pair a letter with its two immediate alphabet neighbours.

Common traps

• Stopping at 'each cluster is a letter with its neighbours', which is true of all four
• Hunting for a vowel or consonant count that the printed note has already excluded

No directly related past PYQ was found.