Classification – Alphanumeric codes: identify the odd one out by checking whether the middle number equals the sum of the alphabetical positions of the first and last letters (A=1 … Z=26). Exactly three obey this rule; one does not. Which code is different? Options: C9F, H20L, N31Q, E29Y.

Introduction / Context:Many verbal-reasoning “classification” items hide a deterministic rule. Here, each code has the pattern Letter–Number–Letter. A frequent device is to relate the number to the alphabetical positions of the two letters. We must find which code violates the common rule.

Given Data / Assumptions:

• Alphabet index uses A=1 to Z=26.
• Codes given: C9F, H20L, N31Q, E29Y.
• Task: test if the middle number equals position(first letter) + position(last letter).

Concept / Approach:Compute the position of each letter and compare the sum with the embedded number. The odd one will be the single code where the equality fails.

Step-by-Step Solution:C9F: C=3, F=6 → 3+6=9 → matches.H20L: H=8, L=12 → 8+12=20 → matches.N31Q: N=14, Q=17 → 14+17=31 → matches.E29Y: E=5, Y=25 → 5+25=30 ≠ 29 → mismatch.

Verification / Alternative check:Re-run the additions or cross-verify with a quick alphabet index list; only E29Y fails the equality, so it must be the outlier.

Why Other Options Are Wrong:C9F, H20L, N31Q: In each case, the number equals the sum of the letter positions, so they belong to the majority pattern.

Common Pitfalls:Confusing zero-based indexing with A=1, or attempting multiplication rather than addition. Also avoid off-by-one errors near the end of the alphabet.

Final Answer:E29Y