Classification – Odd one out (letter quartets – consecutive endings): Which quartet does not end with two consecutive letters: ABYZ, CDWX, EFUV, GHTV?

Introduction / Context:
In these four-letter strings, the last two letters can be checked for being consecutive in the alphabet (e.g., YZ, WX, UV). Three options satisfy this “consecutive tail” property; one does not. Identify the string whose final pair is not consecutive.

Given Data / Assumptions:

• ABYZ → Y(25), Z(26) consecutive.
• CDWX → W(23), X(24) consecutive.
• EFUV → U(21), V(22) consecutive.
• GHTV → T(20), V(22) not consecutive (gap of +2).

Concept / Approach:
Because the first two letters vary across options, focus on the tail property. ABYZ, CDWX, and EFUV each ends with two immediately successive letters. GHTV ends with T and V, which are separated by one letter (U) and are not consecutive. Therefore, GHTV is the odd one out.

Step-by-Step Solution:
1) Extract the last two letters in each quartet.2) Check if they differ by exactly +1 in index.3) Only GHTV fails; select it as the answer.

Verification / Alternative check:
Even if you inspect the first two letters (AB, CD, EF, GH), they are consecutive pairs, but the discriminator intentionally lives at the end to create a single outlier.

Why Other Options Are Wrong:
They maintain the “consecutive pair” pattern at the tail.

Common Pitfalls:
Overanalyzing step sizes across all four letters; the simpler tail check is sufficient and robust.

Final Answer:
GHTV