In this odd-man-out reasoning question, select the odd letter-number group from the given alternatives.

Introduction / Context:
This question belongs to the common aptitude topic called odd-man-out or classification. You are given four alphanumeric groups, and exactly one of them follows a different letter-number pattern compared with the others. Such questions test your ability to map letters to their positions in the English alphabet and to notice when a numeric value is deliberately linked to one of those positions.

Given Data / Assumptions:

• Given groups: E5T, O9S, U3A and I7M.
• We assume standard English alphabet positions: A = 1, B = 2, ..., Z = 26.
• We look for a consistent pattern that three groups share and one group breaks.

Concept / Approach:
The most natural approach is to convert each letter to its numeric position in the alphabet and then compare that with the middle digit. For patterns of this kind, the digit may equal the position of the first letter, the last letter, the difference between them, or some other simple function. Once we compute and compare these values, the odd group will be the one that does not follow the shared rule.

Step-by-Step Solution:
Step 1: Write the alphabet positions of each letter. E = 5, T = 20, O = 15, S = 19, U = 21, A = 1, I = 9 and M = 13. Step 2: Analyse E5T. Here the first letter E has position 5 in the alphabet. The middle symbol is the digit 5, which exactly matches the alphabetical position of the first letter. Step 3: Analyse O9S. O has position 15, S has position 19, while the middle digit is 9. The digit 9 does not equal 15 or 19, nor is it a simple difference such as 19 - 15 = 4. Step 4: Analyse U3A. U has position 21 and A has position 1, while the digit is 3. Again, 3 does not match 21, 1 or an obvious difference such as 21 - 1 = 20. Step 5: Analyse I7M. I has position 9 and M has position 13. The digit 7 does not match 9 or 13, and the difference 13 - 9 = 4 is also different from 7. Step 6: Conclude that in O9S, U3A and I7M the middle digit has no direct equality link with the alphabetical position of the first letter, but in E5T the middle digit is exactly the same as the position of the first letter.

Verification / Alternative check:
As an additional check, we can attempt other simple rules such as the digit representing the position of the last letter, the average of positions of first and last letters, or their difference. None of these consistently fit three groups while excluding one. The only clean, simple and unique rule is that in E5T the digit equals the alphabetical position of the first letter. Since a good odd-man-out rule must single out exactly one option, this confirms E5T as the correct answer.

Why Other Options Are Wrong:
O9S is incorrect as the digit 9 does not match the position of O or S, nor does it represent a simple function of those positions that is shared by other groups. U3A is incorrect for the same reason: the digit 3 is unrelated to positions 21 and 1 in a way that matches other groups. I7M is also incorrect because 7 does not stand in a similar relationship to 9 and 13, so it cannot be the unique special group.

Common Pitfalls:
Many learners try to invent very complicated relationships like sums of positions or multi-step arithmetic. In standard exam reasoning questions, the intended pattern is usually simple and elegant. Another common mistake is to accept a pattern that fits only one or two groups instead of three. The correct approach is to identify a rule that holds for three options, leaving exactly one as the odd-man-out.

Final Answer:
The odd letter-number group is E5T because only in this group does the middle digit equal the alphabetical position of the first letter, while no such relationship holds for the other three groups.