Among five boys — Arun, Vinay, Suraj, Raju, and Pratap — whose body weight is the highest? I. The average weight of Arun, Suraj, and Vinay is 68 kg; the average of Raju and Pratap is 72 kg. Also, Suraj = 78 kg, Raju = 68 kg, Vinay = 46 kg. II. The average weight of Arun, Suraj, Vinay, and Raju is 68 kg; also, Suraj = 78 kg, Raju = 68 kg, Vinay = 46 kg. All five have distinct weights.

Introduction / Context:
This data-sufficiency problem asks us to decide whose weight is the highest using partial aggregate information (averages) and a few exact weights. We must judge sufficiency — i.e., whether each statement (alone or together) is enough to identify the unique heaviest boy — without necessarily computing every person’s weight under each case.

Given Data / Assumptions:

• Persons: Arun (A), Vinay (V), Suraj (S), Raju (Rj), Pratap (P).
• I: (A + S + V) / 3 = 68; (Rj + P) / 2 = 72; and S = 78, Rj = 68, V = 46.
• II: (A + S + V + Rj) / 4 = 68; S = 78, Rj = 68, V = 46; all five have distinct weights.

Concept / Approach:
Translate averages to totals and use the known individuals to solve for unknowns when possible. If from a statement we can uniquely determine each relevant candidate (or at least the maximum), that statement is sufficient.

Step-by-Step Solution:
From I: A + S + V = 3 * 68 = 204 ⇒ A = 204 − (78 + 46) = 80.From I: Rj + P = 2 * 72 = 144 ⇒ P = 144 − 68 = 76.Thus weights are A = 80, S = 78, V = 46, Rj = 68, P = 76. The heaviest is Arun (80 kg). Therefore Statement I alone is sufficient.From II: A + S + V + Rj = 4 * 68 = 272 ⇒ A = 272 − (78 + 46 + 68) = 80.With II we know A = 80, S = 78, V = 46, Rj = 68, but P is unknown other than being distinct. P could be heavier than 80 (e.g., 82) or lighter; hence the identity of the heaviest is not guaranteed. Statement II alone is insufficient.

Verification / Alternative check:
Construct two scenarios under II with different P (e.g., 82 vs. 70) to see the heaviest flip (Arun vs. Pratap). This confirms insufficiency of II alone.

Why Other Options Are Wrong:
II alone does not fix P; “either alone” is false; combining both is unnecessary because I already settles it.

Common Pitfalls:
Assuming the “distinct weights” clause in II forces P below A; it does not.

Final Answer:
Statement I alone is sufficient.