Based on the English alphabetical order, three of the following four letter-clusters are alike in a certain way and thus form a group. Which is the one that DOES NOT belong to that group? (Note: The odd man out is not based on the number of consonants/vowels or their position in the letter-cluster.)

Correct — B. Rule: +2 from the first letter to the second, then +1 to the third.

QST: Q(17) to S(19) is +2, S to T(20) is +1.
IKL: I(9) to K(11) is +2, K to L(12) is +1.
ACD: A(1) to C(3) is +2, C to D(4) is +1.

TWW: T(20) to W(23) is +3, and W to W is +0.

TWW breaks both steps -> option (b).

Why the others are wrong

• (a)
  QST — QST is 17, 19, 20 - +2 then +1, the group pattern exactly.
• (c)
  IKL — IKL is 9, 11, 12 - +2 then +1, so it sits inside the group.
• (d)
  ACD — ACD is 1, 3, 4 - +2 then +1, so it sits inside the group.

A letter-cluster group is defined by its gaps, not by its letters. Convert to positions first, A=1 through Z=26.

Write each cluster as a pair of differences and the odd one shows itself with no further thought. Three read (+2, +1) here and one reads (+3, 0).

The bracketed note removes the shortcut candidates reach for - counting vowels and consonants - which is SSC saying the answer is in the spacing.

Key facts

• QST, IKL and ACD each run +2 then +1 in alphabet positions.
• TWW runs +3 then 0, failing on both steps.
• T is the 20th letter of the alphabet and W the 23rd.

Common traps

• Comparing the letters themselves instead of the gaps between them
• Chasing the vowel and consonant count the stem has already ruled out
• Checking the first gap only and never the second

No directly related past PYQ was found.