In this letter analogy, “ACE is to HJL as LNP is to ______”. Select the group of letters that completes the analogy by applying the same constant forward shift in the alphabet.

Introduction / Context:
This question is a classic alphabet coding analogy from verbal reasoning. The pair ACE : HJL shows how one group of letters is transformed into another group using a consistent shift in alphabet positions. The objective is to discover that pattern and apply it to the second group LNP so that the relationship is preserved. Such questions test the candidate's ability to work with alphabet positions and recognise hidden patterns in sequences of letters.

Given Data / Assumptions:

• First letter group: ACE. • Second letter group: HJL. • Third group: LNP with an unknown partner. • Options: SUW, MOQ, IJK, BDF. • Alphabet positions: A = 1, B = 2, ..., Z = 26 are used for analysis.

Concept / Approach:
We compare ACE with HJL letter by letter and calculate how many positions each letter moves forward in the alphabet. If the shift is constant for all positions, we have a straightforward rule. Once this shift is determined, we apply it to each letter of LNP. The correct answer will be the option that exactly matches the result, both in character and order. No reversal or mixing of letters is used in the example, so we expect a direct positional shift.

Step-by-Step Solution:
Step 1: Convert ACE to alphabet positions. A = 1, C = 3, E = 5. Step 2: Convert HJL to alphabet positions. H = 8, J = 10, L = 12. Step 3: Compute the shift for each letter. 1 → 8 is a shift of +7. 3 → 10 is a shift of +7. 5 → 12 is a shift of +7. So the rule is: add 7 to the position of each letter. Step 4: Apply the same rule to LNP. L = 12, N = 14, P = 16. Add 7: 12 + 7 = 19 → S, 14 + 7 = 21 → U, 16 + 7 = 23 → W. The resulting group is SUW.

Verification / Alternative check:
To verify, you can reverse the shift. Subtract 7 from HJL: H (8) − 7 = A (1), J (10) − 7 = C (3), L (12) − 7 = E (5). This returns ACE, confirming that the shift is correct and consistent. Doing the same with SUW returns LNP: S (19) − 7 = L (12), U (21) − 7 = N (14), W (23) − 7 = P (16). Thus, SUW is the only option that maintains the exact same pattern used in the example.

Why Other Options Are Wrong:
• MOQ: The letters do not correspond to a uniform +7 shift from LNP in all positions. • IJK: These letters are earlier in the alphabet and cannot be obtained from LNP by adding 7. • BDF: This group is unrelated to LNP by a constant forward shift of 7 and looks more like the pattern of ACE itself.

Common Pitfalls:
Candidates sometimes focus on visual similarity of letter shapes or guess based on partial matching. Another frequent error is to forget that each position must follow the same rule that works for all letters of the example pair. Always convert letters to numbers, compute the exact shifts, and then apply the same arithmetic carefully to the new group. This systematic approach avoids guesswork and ensures the correct choice.

Final Answer:
The group of letters that correctly completes the analogy is SUW.