Among M, P, K, J, T, and W, who is lighter than only the heaviest (i.e., second heaviest)? I. P is heavier than M and T. II. W is heavier than P but lighter than J, and J is not the heaviest.

Difficulty: Medium

Statement I alone is sufficient; Statement II alone is not.
Statement II alone is sufficient; Statement I alone is not.
Either Statement I alone or Statement II alone is sufficient.
Both statements together are NOT sufficient.
Both statements together are sufficient, but NEITHER alone is sufficient.

Correct Answer: Both statements together are NOT sufficient.

Explanation:

Introduction / Context:
This is a relative ordering (inequalities) Data Sufficiency question over six individuals: M, P, K, J, T, W. We must identify the unique person who is second only to the heaviest. DS requires proving uniqueness, not just listing possibilities.

Given Data / Assumptions:

I: P > M and P > T (P heavier than M and T).
II: J > W > P and J is not the heaviest (someone heavier than J exists).

Concept / Approach:
Translate comparisons into partial orders, then test if the identity of the second heaviest is forced. If more than one person could occupy that rank under consistent completions, the data are insufficient.

Step-by-Step Solution:

1) Using I alone: We only know P is above M and T. The positions of K, J, W relative to these are unconstrained. Several people could be second heaviest; I alone is insufficient.2) Using II alone: We know there exists someone X heavier than J (since J is not heaviest) and the chain X > J > W > P. The ranks of K, M, T relative to these are unknown. The second heaviest could be J (if X is heaviest and all others are below J), or it could be someone else above J (if both K and X exceed J), etc. II alone is insufficient.3) Combining I + II: We retain X > J > W > P > {M, T}. The person K is still unconstrained—K could be heavier than J, between J and W, or below P. If K > J and X is the heaviest, then K becomes second heaviest; if K is below J, then J could be second heaviest. Multiple consistent orders remain; identity is not fixed.

Verification / Alternative check:
Construct models: (a) Let X=K>J>W>P>M>T ⇒ second heaviest = J. (b) Let X=K, and also have another Y (say unknown external) heaviest > J>K... ⇒ second heaviest could be J or K depending on placements. The variability confirms insufficiency.

Why Other Options Are Wrong:

A/B/C: Neither statement alone nor “either alone” pins a unique second heaviest.
E: Even together, multiple candidates remain.

Common Pitfalls:
Assuming “J is not heaviest” implies J is second heaviest; ignoring unconstrained elements like K that can overtake J.

Final Answer:
Both statements together are NOT sufficient.

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Among M, P, K, J, T, and W, who is lighter than only the heaviest (i.e., second heaviest)? I. P is heavier than M and T. II. W is heavier than P but lighter than J, and J is not the heaviest.

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