Two linked statements about D/E and D/W; check conclusions: Statements: C ≥ D = E ≤ F Y < D ≥ W Conclusions: I. C ≥ Y II. F ≥ Y

Introduction / Context: We combine two statements that hinge on D, then test relationships to Y via D and to C and F via D/E.

Given Data / Assumptions:

• C ≥ D = E ≤ F
• Y < D and D ≥ W
• Conclusions: (I) C ≥ Y, (II) F ≥ Y

Concept / Approach: Exploit D = E to move freely between them. Then chain inequalities that connect C and F with D, and Y with D.

Step-by-Step Solution: From Y < D and C ≥ D, we get Y < D ≤ C ⇒ Y ≤ C, i.e., C ≥ Y (true). From D = E ≤ F and Y < D, we have Y < D ≤ F ⇒ Y <= F, i.e., F ≥ Y (true).

Verification / Alternative check: Example: D = E = 10, C = 12, F = 11, Y = 7, W = 4 satisfies both conclusions. Even if C = D and F = D (tight bounds), Y < D ensures Y ≤ C and Y ≤ F remain true.

Why Other Options Are Wrong: They deny at least one necessary transitive consequence established above.

Common Pitfalls: Forgetting that equality allows substitution without changing order; mixing up strict and non-strict when chaining.

Final Answer: Both Conclusions I and II are true.