Which of the following letter-number clusters will replace the question mark…
2026
Which of the following letter-number clusters will replace the question mark (?) in the given series to make it logically complete? IQX33, LTA49, OWD65, RZG81, ?
Answer: D. UCJ97 — Concept: an alphanumeric series is not one rule but several parallel tracks. Each letter slot is its own sequence — convert the letter to its alphabet…
- A.
VCJ97
- B.
UCK97
- C.
UDJ97
- D.
UCJ97
Attempted by 55 students.
Show answer & explanation
Correct answer: D
Concept: an alphanumeric series is not one rule but several parallel tracks. Each letter slot is its own sequence — convert the letter to its alphabet position (A = 1 … Z = 26), find that slot's constant step, and wrap back to the start of the alphabet whenever a position passes 26. The numeric slot is a separate arithmetic sequence with its own common difference. Extend every track by one step and reassemble the cluster.
Application to this series, track by track:
First letters: I = 9, L = 12, O = 15, R = 18. The step is a constant +3, so the next position is 18 + 3 = 21, which is U.
Second letters: Q = 17, T = 20, W = 23, Z = 26. The step is again +3, so the next position is 26 + 3 = 29; that passes 26, so subtract 26 to wrap, giving 3, which is C.
Third letters: X = 24, A = 1, D = 4, G = 7. The same +3 step applies and the wrap has already been used once (24 + 3 = 27, and 27 − 26 = 1), so the next position is 7 + 3 = 10, which is J.
Numbers: 33, 49, 65, 81. The common difference is +16, so the next number is 81 + 16 = 97.
Summary of the four tracks:
Track | Positions in the given terms | Step | Next value |
|---|---|---|---|
First letter | I(9), L(12), O(15), R(18) | +3 | 21 → U |
Second letter | Q(17), T(20), W(23), Z(26) | +3 | 29 − 26 = 3 → C |
Third letter | X(24), A(1), D(4), G(7) | +3 (wrap: 27 → 1) | 10 → J |
Number | 33, 49, 65, 81 | +16 | 97 |
Cross-check by a second route: all three letter tracks move by the same step, so the positions 18, 26, 7 of RZG advance together to 21, 3, 10 — that is U, C, J. The numeric track also fits the form 16n + 1 (33 = 16×2 + 1, 49 = 16×3 + 1, 65 = 16×4 + 1, 81 = 16×5 + 1), so the sixth term is 16×6 + 1 = 97. Both routes give the same cluster.
The cluster that completes the series is UCJ97.