In this letter-series odd-man-out question, select the odd group of letters from the given alternatives.

Introduction / Context:
This question is another alphabet-based odd-man-out problem. Each option consists of three letters. In three of the groups, the letters are equally spaced in the alphabet by a constant number of positions. In one group, the last step differs, so it does not maintain equal spacing. Your goal is to convert letters to their numerical positions and check how many positions separate consecutive letters.

Given Data / Assumptions:

• Options: KQW, RXD, BHN and AGL.
• We use the standard order A = 1, B = 2, ..., Z = 26.
• We assume circular counting from Z to A is allowed when needed.

Concept / Approach:
The intended pattern is that the difference between the first and second letters, and between the second and third letters, is the same in three options. We will compute these differences for each group. In KQW, RXD and BHN, both steps are of size 6 (moving forward), while in AGL the first step is 6 but the second step is only 5. This mismatch in step size makes AGL the odd group.

Step-by-Step Solution:
Step 1: Convert letters in KQW to positions. K = 11, Q = 17 and W = 23. The differences are Q - K = 17 - 11 = 6 and W - Q = 23 - 17 = 6. So KQW has a pattern of +6, +6. Step 2: Convert letters in RXD. R = 18, X = 24 and D = 4. The difference from R to X is 24 - 18 = 6. To go from X to D using circular counting, we move forward from 24 by 6 positions: 24 + 6 = 30, and 30 - 26 = 4, which corresponds to D. Thus, the second step is effectively +6 as well, giving RXD a pattern of +6, +6. Step 3: Convert letters in BHN. B = 2, H = 8 and N = 14. The differences are H - B = 8 - 2 = 6 and N - H = 14 - 8 = 6. So BHN also follows +6, +6. Step 4: Convert letters in AGL. A = 1, G = 7 and L = 12. The difference from A to G is 7 - 1 = 6. The difference from G to L is 12 - 7 = 5. So the pattern is +6, +5, not +6, +6. Step 5: Summarise. Three groups have equal jumps of 6 positions each between consecutive letters, while one group has unequal jumps of 6 and 5. Step 6: Conclude that AGL is the odd group, because it breaks the equal-step pattern.

Verification / Alternative check:
We can visualise the letters on a number line or list the alphabet to verify the steps manually. Starting from K, count 6 steps forward to Q and another 6 to W. Similarly, from B to H to N and from R to X to D (using circular counting), we always move 6 steps. For AGL, after moving 6 steps from A to G, moving 6 steps again would land at M, not L. The fact that L is only 5 steps from G confirms that AGL deviates from the common pattern.

Why Other Options Are Wrong:
KQW is not odd because it uses a consistent +6 step between its letters. RXD also respects the +6 step size in a circular manner. BHN follows the same pattern. These three groups are therefore structurally identical under the equal-step rule and cannot be the odd-man-out. Only AGL has different step sizes and so does not belong with the others.

Common Pitfalls:
Some learners ignore circular counting from Z to A, which can make RXD look irregular at first sight. However, reasoning questions often assume that the alphabet can be treated cyclically. Another mistake is to test only one of the differences and not both. To be thorough, you must always check the step between first and second letters and between second and third letters. This reveals that AGL fails the equal-step condition and is the intended odd option.

Final Answer:
The odd group of letters is AGL, because in this group the letters are separated by +6 and +5 positions, whereas in the other groups they are separated consistently by +6 and +6 positions.