Coding — What is the code for 'walks'? Statements: I. In a code language, 'she walks fast' → 'he ka to'. II. In the same language, 'she learn fast' → 'jo ka to'.

Difficulty: Medium

Correct Answer: Both Statements I and II together are sufficient, but neither alone is sufficient.

Explanation:


Introduction / Context:
We need the unique codeword for the English word 'walks' by comparing two coded sentences with overlapping vocabulary.


Given Data / Assumptions:

  • Sentence A: 'she walks fast' ↔ 'he ka to'.
  • Sentence B: 'she learn fast' ↔ 'jo ka to'.
  • Same language, one-to-one mapping within each sentence.


Concept / Approach:
Use set intersection: common English words across sentences share common code tokens; the remaining token in A must map to 'walks'.


Step-by-Step Solution:

Common English words in A and B: {she, fast}. Common codes in their translations: {ka, to} (order unknown). Therefore, 'she' and 'fast' map to 'ka' and 'to' in some order.The remaining English word in A is 'walks' and the remaining code token is 'he'. Hence, 'walks' ↔ 'he'.


Verification / Alternative check:
Confirm that no alternative mapping exists without violating bijection per sentence. If we reassigned 'he' to either 'she' or 'fast', we would have to reuse one of {ka,to} for 'walks', which contradicts the one-to-one mapping in sentence A.


Why Other Options Are Wrong:

  • I alone: Without B, we cannot separate which two of {he,ka,to} belong to {she,fast}; however, the cross-sentence intersection provided by II enables elimination.
  • II alone: Does not contain 'walks'; cannot identify its code directly.
  • Either alone: false; each alone is insufficient.
  • Both insufficient: false; together they identify 'walks' uniquely.


Common Pitfalls:
Assuming a fixed order or mapping 'fast'→'fast-sounding' tokens; ignoring that only the overlap pins two tokens, leaving the third for 'walks'.


Final Answer:
Both together are sufficient ('walks' is coded as 'he').

More Questions from Data Sufficiency

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion