Based on the English alphabetical order, three of the following four letter-cluster pairs are alike in a certain way and thus for…

Question

Based on the English alphabetical order, three of the following four letter-cluster pairs 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 one out is not based on the number of consonants/vowels or their position in the letter cluster.)

1. A. RP−NL
2. B. LJ−HF
3. C. OM−KI
4. D. EC−ZA (Correct answer)

Correct Answer

Option D — EC−ZA

Detailed Solution & Explanation

The correct answer is EC−ZA.

Key Points

• Test the shift applied to each letter across the dash:
  • RP−NL: R−4=N, P−4=L ✓
  • LJ−HF: L−4=H, J−4=F ✓
  • OM−KI: O−4=K, M−4=I ✓
  • EC−ZA: E−4 would be A, not Z; and C−2=A, not C−4
• Three pairs apply a uniform −4 to both letters. EC−ZA does not, so it is the odd one out.

Additional Information

• The trap is that EC−ZA looks plausible because E and Z sit at opposite ends of the alphabet and A follows Z in a wrap-around. But a wrap from E backwards by 4 lands on A, not Z — the arithmetic simply does not hold.
• Convert to a numeric signature for each pair: (−4,−4), (−4,−4), (−4,−4), and for EC−ZA a mismatch. The outlier then needs no memory of letters at all.
• Where a wrap-around is genuinely intended, the alphabet is treated as circular so that A−1 = Z. Check whether a consistent wrap makes the pattern work before rejecting a pair.
• The question's note excluding vowel/consonant counts removes a plausible false trail — EC−ZA is also the only pair with two vowels, which is exactly the reasoning the note forbids.