Based on the English alphabetical order, three of the following four…
2026
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 letter-cluster pair 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.)
Answer: C. TM-PS — Concept. A letter-cluster pair question defines its group by a RELATION, never by the letters themselves. Place the letters on the alphabet scale (A = 1, B =…
- A.
QJ-MO
- B.
IB-EG
- C.
TM-PS
- D.
NG-JL
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Show answer & explanation
Correct answer: C
Concept. A letter-cluster pair question defines its group by a RELATION, never by the letters themselves. Place the letters on the alphabet scale (A = 1, B = 2, ..., Z = 26) and two measurements describe a pair of two-letter clusters: the internal gap, i.e. how many places the second letter of a cluster lies from its first letter, and the slot-wise transfer shift, i.e. how many places each letter of the first cluster must move to reach the letter occupying the same slot in the second cluster. Other relations, such as the cross-diagonal shifts, can be measured as well, but they are algebraically determined by these two, so measuring these two is enough. Whichever relation three pairs share is the rule of the group, and the pair whose measurement differs on that relation is the odd one out.
Application. Measuring all four pairs on both relations:
Pair | Internal gap, 1st cluster | Internal gap, 2nd cluster | Slot-1 shift | Slot-2 shift |
|---|---|---|---|---|
QJ-MO | Q (17) to J (10) = 7 back | M (13) to O (15) = 2 forward | Q to M = 4 back | J to O = 5 forward |
IB-EG | I (9) to B (2) = 7 back | E (5) to G (7) = 2 forward | I to E = 4 back | B to G = 5 forward |
TM-PS | T (20) to M (13) = 7 back | P (16) to S (19) = 3 forward | T to P = 4 back | M to S = 6 forward |
NG-JL | N (14) to G (7) = 7 back | J (10) to L (12) = 2 forward | N to J = 4 back | G to L = 5 forward |
In every pair the first cluster has an internal gap of 7 places backwards, so that measurement is common to the whole set and cannot separate anything.
The slot-1 shift is 4 places backwards in every pair as well, so it is also common to the whole set.
The internal gap of the second cluster is 2 places forward for QJ-MO, IB-EG and NG-JL, but 3 places forward for TM-PS.
The slot-2 shift shows the very same single deviation: 5 places forward for QJ-MO, IB-EG and NG-JL, and 6 places forward for TM-PS. These are not two separate findings, because the four quantities always satisfy: internal gap of the second cluster = internal gap of the first cluster + slot-2 shift - slot-1 shift. Since the first-cluster gap and the slot-1 shift are identical in all four pairs, a deviation in one of the two remaining quantities is the same deviation seen from the other side.
So the rule shared by three pairs is: second cluster = (first letter moved 4 places back, second letter moved 5 places forward), which makes the second-cluster gap 2 places forward. The pair that does not follow this rule is TM-PS.
Cross-check. Two checks make this safe. First, rebuild each second cluster from its own first cluster using the shared rule (4 places back, 5 places forward) and compare it with what is printed: QJ gives M and O, matching MO; IB gives E and G, matching EG; NG gives J and L, matching JL; but TM gives P and R, whereas PS is printed. Second, exclusivity: the internal gap of the first cluster is 7 places back in all four pairs and the slot-1 shift is 4 places back in all four, so neither of them can discriminate. That leaves only one independent variation, which may be read either as the second-cluster internal gap or as the slot-2 shift, and it deviates for exactly one pair - so exactly one pair falls outside the group. This deviation also survives the note in the question, because nothing in this reading counts vowels or consonants. Answer: TM-PS.