Difficulty: Medium
Correct Answer: TOJCFQNZO
Explanation:
Introduction / Context:
This question tests pattern recognition in an alphabet-based coding system. You are told how the word SUPERMAN is transformed to TTQDSLBM, and then asked to encode SPIDERMAN in the same way. The pattern turns out to involve shifting letters forward or backward in the alphabet in an alternating manner. Such alternating shift patterns are very popular in exam coding–decoding questions because they are simple once spotted but still trap careless candidates.
Given Data / Assumptions:
Concept / Approach:
The idea is to compare each letter of SUPERMAN with the corresponding letter of TTQDSLBM and measure how many positions forward or backward it moves in the alphabet. When we write down these shifts, we notice a clear alternating pattern: the first letter moves one step forward, the second moves one step backward, the third again moves forward, and so on. Once this alternating +1, -1 pattern is confirmed, we apply it to each corresponding position of the new word SPIDERMAN to generate its code. The core concept is positional letter shifting with alternating directions.
Step-by-Step Solution:
Step 1: Write positions for SUPERMAN: S, U, P, E, R, M, A, N.
Step 2: Compare with TTQDSLBM: T, T, Q, D, S, L, B, M.
Step 3: Check shifts: S → T (+1), U → T (-1), P → Q (+1), E → D (-1), R → S (+1), M → L (-1), A → B (+1), N → M (-1).
Step 4: Conclude: for odd positions (1st, 3rd, 5th, 7th) we add 1 to the letter; for even positions (2nd, 4th, 6th, 8th) we subtract 1 from the letter.
Step 5: Now take SPIDERMAN: S, P, I, D, E, R, M, A, N.
Step 6: Apply +1 shift to positions 1, 3, 5, 7, 9 and -1 shift to positions 2, 4, 6, 8.
Step 7: Position 1: S (+1) → T.
Step 8: Position 2: P (-1) → O.
Step 9: Position 3: I (+1) → J.
Step 10: Position 4: D (-1) → C.
Step 11: Position 5: E (+1) → F.
Step 12: Position 6: R (-1) → Q.
Step 13: Position 7: M (+1) → N.
Step 14: Position 8: A (-1) → Z (before A comes Z in cyclic alphabet).
Step 15: Position 9: N (+1) → O.
Step 16: Combine all letters: T O J C F Q N Z O = TOJCFQNZO.
Verification / Alternative check:
To verify, you can reverse the coding on TOJCFQNZO back to SPIDERMAN using the opposite shifts: subtract 1 from odd positions and add 1 to even positions. Doing this recovers S, P, I, D, E, R, M, A, N exactly. Similarly, reversing the shifts from TTQDSLBM yields SUPERMAN again. The successful reverse check shows that our alternating pattern is correct and consistently applied. Ensuring reversibility is a strong way to validate any coding-decoding logic in such questions.
Why Other Options Are Wrong:
Option B (TOJCFSLBM) breaks the pattern after the sixth letter and introduces letters that would correspond to a different shifting rule. Option C (TOJCFONZO) alters the mapping in the middle segment and does not follow the strict +1, -1 alternation. Option D (TOJCOSLBM) changes multiple letters inconsistently. Option E (TPKCFQNZO) breaks the pattern right from the second position. Only Option A matches the precise alternating shift (+1 at odd positions, -1 at even positions) when tested letter by letter.
Common Pitfalls:
Exam takers sometimes miscount positions (mixing up which positions are odd or even) or forget that the alphabet wraps around (for example, moving backward from A gives Z). Another common error is to use a consistent forward shift for all letters without verifying whether odd and even positions behave differently. Always compute a few letter shifts explicitly and note the direction for each position before committing to a rule. Writing down alphabetical positions (1–26) can also reduce mistakes.
Final Answer:
Therefore, using the same coding rule as for SUPERMAN, the word SPIDERMAN is written as TOJCFQNZO in that code language.
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