In the following question, the relationships between several variables are given: A > B ≤ C < D, C > E, and F ≥ D. Using only this information, determine which of the conclusions A > E and F > E must be true and select the correct option.

Introduction / Context:
This problem is a standard inequalities and conclusions question from logical reasoning. We are given chained inequalities relating A, B, C, D, E, and F. The task is to decide which conclusions definitely follow from the statements and which do not. The challenge is to read the inequality chain correctly and use transitivity rules without adding any extra assumptions.

Given Data / Assumptions:

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

Concept / Approach:
For chained inequalities, we rely on transitivity. If X > Y and Y > Z, then X > Z. If X > Y and Y ≥ Z, we can still say X > Z. Similarly, if X ≥ Y and Y > Z, then X > Z. We must see what we can deduce about positions of A, D, and F with respect to E by combining the given relations. We must also check if any relation about A and E can be broken by a counter example while still respecting all given inequalities.

Step-by-Step Solution:
Step 1: Start with C and E. We are told C > E. Step 2: We also have C < D. Hence D is greater than C, and therefore D is also greater than E. So D > E. Step 3: From F ≥ D and D > E, we can conclude F > E. This is because F is at least D, and D is already strictly greater than E. Step 4: Therefore conclusion II (F > E) must always be true. Step 5: Now examine A and E. We know A > B and B ≤ C, and C > E. Step 6: Since B ≤ C, B can be less than C or equal to C. Also, E is less than C, but there is no direct relation between B and E. Step 7: Construct an example where A > B, B ≤ C, and C > E but A is not greater than E. Let E = 5, C = 10, B = 9, and A = 9.5. All inequalities hold, but here A = 9.5 is greater than E = 5, so this example still supports A > E. Step 8: Now choose E = 5, C = 10, B = 9, and A = 6. Then A > B would fail, so we need B smaller. Take E = 5, C = 10, B = 1, A = 2. All relations hold, and A > E is false since 2 is not greater than 5. This shows that A > E is not guaranteed.

Verification / Alternative check:
When examining conclusion I, we only know that A is to the right of B on the number line, and C is to the right of E. There is no direct link between A and E, so both A > E and A ≤ E are possible. For conclusion II, D is definitely to the right of E, and F is at or to the right of D. Hence E is always to the left of F, so F > E is forced in every allowed configuration.

Why Other Options Are Wrong:

• Option a (only conclusion I is true) is wrong because conclusion I is not always true.
• Option c (either I or II) is wrong since II is always true, not an either or situation.
• Option d (neither) is wrong because conclusion II definitely follows.
• Option e (both) is wrong since conclusion I can fail for some values.

Common Pitfalls:
A frequent error is to treat B ≤ C as if it forced B < C, and then assume a clear ordering among all variables without checking each link. Another mistake is to assume that if A is linked through B and C to E, then A must be greater than E, even though the inequalities do not give a strict chain from A to E. Always check whether you can build at least one valid counter example before declaring a conclusion definitely true.

Final Answer:
Only conclusion II is true, so option b is correct.