In a certain code language, each letter in a word is replaced by the letter two places before it in the English alphabet (with wrap around from A to Z). Under this rule, BOULDER is coded as ZMSJBCP. Using the same pattern, how is ELK written in this code language?

Introduction / Context:
This letter coding question uses a fixed backward shift in the alphabet. You are told that BOULDER becomes ZMSJBCP under the code, and you must infer the pattern and apply it to the word ELK. Such questions check whether you can detect a consistent shift and handle wrap around correctly for letters at the beginning of the alphabet.

Given Data / Assumptions:

• BOULDER is coded as ZMSJBCP.
• The code is letter by letter and deterministic.
• The pattern described is that each letter is replaced by the letter two positions earlier in the alphabet, with A mapped back to Y and B to Z.
• We must code ELK using this same rule.

Concept / Approach:
To confirm the rule, we compare each letter in BOULDER with its coded partner in ZMSJBCP and compute the shift in terms of alphabetical positions. If each coded letter is exactly two positions before the original letter, with B mapping to Z to account for wrapping, then the pattern is simply a uniform backward shift of two. Once verified, we apply this shift to E, L and K to obtain the code for ELK.

Step-by-Step Solution:
Step 1: Write the mapping from BOULDER to ZMSJBCP. B → Z, O → M, U → S, L → J, D → B, E → C, R → P. Step 2: Convert letters to their alphabetical positions. B(2) → Z(26). Treating the alphabet cyclically, 2 - 2 = 0 and 0 corresponds to 26, so this is effectively a shift of -2. O(15) → M(13): 15 - 2 = 13. U(21) → S(19): 21 - 2 = 19. L(12) → J(10): 12 - 2 = 10. D(4) → B(2): 4 - 2 = 2. E(5) → C(3): 5 - 2 = 3. R(18) → P(16): 18 - 2 = 16. Step 3: All mappings are consistent with a shift of two positions backward, wrapping as needed. Step 4: Now apply this rule to ELK. E is the 5th letter; 5 - 2 = 3, which is C. L is the 12th letter; 12 - 2 = 10, which is J. K is the 11th letter; 11 - 2 = 9, which is I. Step 5: Combine the coded letters in order: C J I → CJI.

Verification / Alternative check:
As a quick consistency check, you can decode CJI back to ELK by shifting forward by two positions: C to E, J to L and I to K. This recovers ELK exactly, which confirms that the coding process is reversible and the shift amount has been chosen correctly. Rechecking the BOULDER example with the same backward shift of two for each letter further confirms the general rule.

Why Other Options Are Wrong:
Options XIG, EOC, BXM and ZIG require different shifts or mixtures of forward and backward movement that do not match the -2 pattern verified with BOULDER. For example, mapping E to X would involve a large backward shift, not just two positions. Because the coding rule must be uniform and consistent across all words, only CJI respects the shift by two positions backward for every letter of ELK.

Common Pitfalls:
A common mistake is to assume a forward shift because many code questions use that pattern. Another is to ignore wrapping and think that letters at the start of the alphabet cannot map to letters near the end. Remember that with wrap around, B shifted back by two becomes Z. Writing down the alphabet in a circle or noting pairs explicitly helps visualise this. Once the shift is confirmed, applying it to any given word is straightforward.

Final Answer:
Therefore, using the same shift of two letters backward, ELK is written as CJI in the code language.