In the following symbolic inequality problem, assume all statements are true and decide which of the given conclusions must follow. Statements: Y ≤ K < D = S; D < B < O; A ≥ D < Z Conclusions: I. A > B II. Y < Z

Introduction / Context:
This question belongs to the family of symbolic inequality reasoning problems. You are given several chained comparisons using symbols such as less than, greater than, and equal to. From these, you must determine which conclusions are definitely true in every possible arrangement of the variables. This tests your ability to track relative positions and to distinguish between what must be true and what may or may not be true.

Given Data / Assumptions:

• Y ≤ K < D = S
• D < B < O
• A ≥ D and D < Z
• Conclusion I: A > B
• Conclusion II: Y < Z
• All variables represent comparable quantities on the same scale, and the inequality relations are consistent.

Concept / Approach:
To solve such problems, we form chains by linking inequalities that share common variables. This allows us to infer new relationships. However, it is important to check whether a conclusion is forced by the chain or if there is freedom to assign values such that the conclusion may fail. If we can construct even one valid example where the conclusion does not hold, then that conclusion is not definitely true. When dealing with symbols like ≤ and ≥, remember that equality is also possible, not just strict inequality.

Step-by-Step Solution:
Step 1: Start from Y ≤ K < D. We know D = S, but for the conclusions we mainly need D and Z. From the third statement we have D < Z, so Z is greater than D and everything that is less than or equal to D. Step 2: Link Y to Z. Since Y ≤ K and K < D, we can write Y ≤ K < D < Z. This means that Y is less than D, which is itself less than Z. Therefore Y must be less than Z. That directly supports Conclusion II: Y < Z is definitely true. Step 3: Now analyze Conclusion I, which claims A > B. From the data we know A ≥ D and D < B. This implies that B is greater than D, while A is at least as large as D. There are two possible cases:

• Case 1: A = D. Then we have D < B and A = D, which gives A < B.
• Case 2: A > D. In this case A could be greater than B or still less than B, depending on the actual values, because we only know that B is above D, not how far above it.

Since we can easily choose numbers where A is not greater than B, Conclusion I is not guaranteed. Step 4: Construct a concrete counterexample for Conclusion I. Take D = 5, so K = 4, Y = 3, S = 5, B = 7, O = 9, Z = 8, and A = 5. Then:

• Y ≤ K < D = S gives 3 ≤ 4 < 5 = 5, which is true.
• D < B < O gives 5 < 7 < 9, which is true.
• A ≥ D < Z gives 5 ≥ 5 < 8, which is true.

In this arrangement, A = 5 and B = 7, so A > B is false. However, Y = 3 and Z = 8, hence Y < Z is true.

Verification / Alternative check:
The counterexample satisfies all the given statements but makes Conclusion I false. Therefore, Conclusion I cannot be considered definitely true. In contrast, the chain Y ≤ K < D < Z always leads to Y < Z no matter which valid values you choose. There is no way to make Y greater than or equal to Z while keeping all the inequalities valid. This comparison confirms that only Conclusion II is logically guaranteed by the premises.

Why Other Options Are Wrong:
Option a: “None of the conclusions is true” is incorrect because Conclusion II always holds. Option b: “Only conclusion I is true” is clearly wrong because we have constructed an example where A is not greater than B. Option d: “Either conclusion I or conclusion II is true” is incorrect because there is no such either or situation; Conclusion II is always true, while Conclusion I is not. Option e: “Both conclusions I and II are true” fails because Conclusion I can be false in some valid arrangements.

Common Pitfalls:
A typical mistake is to assume that A ≥ D and D < B automatically means A > B. This would only be justified if A were strictly greater than D and D were very close to B in a way that forced A to exceed B, which is not given. Another pitfall is ignoring the possibility of equality in symbols like ≥ and ≤. Always consider the equality case when these symbols appear; it often provides the counterexample needed to disprove a conclusion.

Final Answer:
Only conclusion II is true.