Difficulty: Easy
Correct Answer: E > C
Explanation:
Introduction / Context:This problem checks deterministic reasoning with chained inequalities. From a given set of relations between five variables A, B, C, D, and E, we must identify the statement that is guaranteed to be true for every numeric assignment that satisfies the chain. The core skill is reading chained comparisons and propagating them without making unsupported leaps.
Given Data / Assumptions:
Concept / Approach:Break the long chain into safe implications. From A > B and B ≥ C, it follows that A > C. From C < D and D < E, we know D > C and E > C as well as E > D. However, the chain gives no direct comparison between A and D or E, nor between D and B (beyond both being on opposite sides of C). Therefore, only relations supported purely by transitivity on shared elements (especially anchored at C) are certain.
Step-by-Step Solution:
From A > B ≥ C, infer A > C.From C < D < E, infer D > C and E > C.Thus, E > C is definitely true.No definite order between D and B is known because B could be either less than, equal to, or greater than D while still respecting B ≥ C and D > C.No definite order between A and D is known because A could be just above B (close to C) or far larger; both are consistent with the given statements.Verification / Alternative check:Construct counterexamples for the dubious options. Let C = 0, B = 0, A = 1, D = 0.1, E = 0.2. Then D ≥ B is true (0.1 ≥ 0), but choose C = 0, B = 5, A = 6, D = 4, E = 7 and D ≥ B is false (4 ≥ 5 fails). So D ≥ B is not definite. Similarly, A > D can be true or false depending on chosen values consistent with the chain. Meanwhile, E > C always holds because E > D > C is baked into the chain.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:E > C
Discussion & Comments