In the following question, the relationships between A, B, C, D and E are given as: A ≤ B > C and D ≥ B < E. Based only on these statements, determine which of the conclusions D ≥ A and E > C must be true.

Introduction / Context:
This question checks understanding of chained inequalities involving several variables. The goal is to see whether certain comparisons between pairs of variables are guaranteed to hold, given the initial inequality statements. Such questions appear in many banking and management entrance tests as part of logical and analytical reasoning.

Given Data / Assumptions:

• A ≤ B.
• B > C.
• D ≥ B.
• B < E.
• Conclusion I: D ≥ A.
• Conclusion II: E > C.

Concept / Approach:
We combine inequalities using transitivity. When one variable is greater than or equal to another, and that second variable is greater than a third, we can chain the relations. We must be cautious not to assume any relation that is not supported by a continuous chain of inequalities. To check whether a conclusion must be true, we can try to construct a counter example. If no counter example can be found without violating the original statements, then the conclusion is logically valid.

Step-by-Step Solution:
Step 1: Work on conclusion I, D ≥ A. We know A ≤ B, so B is at least as large as A. We also know D ≥ B, so D is at least as large as B. Combining these, we get D ≥ B ≥ A. Hence D ≥ A must always hold. So conclusion I is definitely true. Step 2: Work on conclusion II, E > C. We know B > C, meaning B is strictly greater than C. We also know B < E, so E is strictly greater than B. Now we have C < B < E. By transitivity, C < E. Therefore E is always greater than C, and conclusion II must also be true.

Verification / Alternative check:
Take a simple numeric example that satisfies the inequality chain. Choose C = 2, B = 5, A = 4, D = 7, and E = 9. Then A ≤ B (4 ≤ 5) and B > C (5 > 2). Also D ≥ B (7 ≥ 5) and B < E (5 < 9). Now compare D and A: D = 7 and A = 4, so D ≥ A holds. Compare E and C: E = 9 and C = 2, so E > C holds. Any other numeric assignment that respects the original inequalities will keep D at or to the right of B, and B at or to the right of A, and will keep E to the right of B and B to the right of C. Hence the final relations D ≥ A and E > C remain true in all valid cases.

Why Other Options Are Wrong:

• Options a and b, which claim that only one conclusion is true, are incorrect because both D ≥ A and E > C always hold.
• Option c (either I or II) and option d (neither) are also inconsistent with the clear chained ordering.

Common Pitfalls:
Learners sometimes misread A ≤ B as A < B and overlook the equality case. Even then, the conclusion D ≥ A holds whether A equals B or not. Another error is to think that if there is a mix of strict and non strict inequalities, no definite conclusion can be drawn. In reality, as long as the chain from one variable to another is complete and consistent, solid conclusions like D ≥ A and E > C can be made.

Final Answer:
Both conclusions I and II are true, so option e is correct.