Integers A, B, C, D, E, F satisfy: E < F, B > A, and A < D < B. C is the greatest of all six. Is A the smallest? I. E + B < A + D. II. D < F.

Introduction / Context:
The question asks whether A is the smallest, given several inequalities. In data sufficiency, we must see whether each statement (alone or together) allows a definitive Yes/No answer to that question.

Given Data / Assumptions:

• Base: E < F; B > A; and A < D < B. Also, C is the greatest of all six (so C > {A,B,D,E,F}).
• Statement I: E + B < A + D.
• Statement II: D < F.

Concept / Approach:
Manipulate Statement I to compare E and A, using A < D and B > A to extract ordering implications. If we can deduce some element is strictly below A, then A cannot be the smallest.

Step-by-Step Solution:
From I: E + B < A + D ⇒ E − A < D − B.But from the base, D − B < 0 (since D < B). Hence E − A < (a negative number) ⇒ E − A < 0 ⇒ E < A.We already know A < D and A < B, and C is the greatest. With E < A, there exists at least one integer smaller than A. Therefore the answer to “Is A the smallest?” is a definitive No.So Statement I alone is sufficient. Statement II (D < F) merely says F is above D and gives no comparison placing anyone below A; alone it is not sufficient.

Verification / Alternative check:
Try sample values consistent with the base and I; E comes out below A regardless of the exact magnitudes.

Why Other Options Are Wrong:
II alone does not answer the question; both-together is unnecessary because I alone already resolves it.

Common Pitfalls:
Missing the sign of D − B, which is negative; that is the key to deducing E < A.

Final Answer:
Statement I alone is sufficient (A is not the smallest).