In the following set of symbolic inequalities, assume all statements are true and then decide which of the given conclusions must follow. Statements: C ≤ R ≤ N = M ≥ F; Q ≥ M < O; D ≥ L; C ≥ D ≤ S ≥ Z Conclusions: I. R ≤ F II. C < Q

Introduction / Context:
This is a more involved symbolic inequality question that includes several variables and mixed relations using ≤, ≥, and = symbols. You are asked to identify which of the proposed conclusions must be true in all possible arrangements that satisfy the given statements. The challenge comes from properly chaining only the relevant inequalities and recognizing when the available information is not strong enough to fix the relative order of two variables.

Given Data / Assumptions:

• C ≤ R ≤ N = M ≥ F
• Q ≥ M and M < O
• D ≥ L
• C ≥ D and D ≤ S and S ≥ Z
• Conclusion I: R ≤ F
• Conclusion II: C < Q
• All inequalities refer to the same underlying ordered quantity.

Concept / Approach:
When working with many variables, it is helpful to focus on the ones that appear in the conclusions. For Conclusion I, we care about R and F; for Conclusion II, we care about C and Q. We use transitivity to form chains involving these variables. A conclusion is definite only if every allowed assignment of numerical values keeps it true. If even one assignment satisfies all statements but breaks a conclusion, that conclusion is not definitely true.

Step-by-Step Solution:
Step 1: Analyze the chain involving R and F. From C ≤ R ≤ N = M ≥ F, we know R is less than or equal to N, and N equals M, which is greater than or equal to F. Step 2: Write this explicitly: R ≤ N and N = M and M ≥ F. This simplifies to R ≤ M and M ≥ F. There is no direct relation that compares R and F. They are both related to M but in different ways: R is at most M, F is at most M, yet either one could be smaller than the other. Step 3: Construct a counterexample for Conclusion I (R ≤ F). Take M = N = 10, R = 9, C = 8, F = 5. Then C ≤ R ≤ N = M ≥ F becomes 8 ≤ 9 ≤ 10 = 10 ≥ 5, which is true. In this example R = 9 and F = 5, so R is greater than F, and thus R ≤ F is false. This shows that Conclusion I is not always true. Step 4: Now look at Conclusion II involving C and Q. From C ≤ R ≤ N = M we know C ≤ M. From C ≥ D and D ≤ S and S ≥ Z, we know C is at least as large as D, but this does not directly affect Q. From Q ≥ M we know Q is at least as large as M, and M is at least as large as C. Step 5: Combine what we have: C ≤ M and Q ≥ M. This gives C ≤ M ≤ Q, so C ≤ Q. However, the conclusion claims C < Q, which is a strict inequality. Because equality is allowed, it is possible for C and Q to be equal. Step 6: Construct a counterexample for Conclusion II. Let M = N = Q = C = 10, R = 10, F = 8, O = 12, D = 9, L = 7, S = 11, Z = 6. Check:

• C ≤ R ≤ N = M ≥ F becomes 10 ≤ 10 ≤ 10 = 10 ≥ 8, true.
• Q ≥ M < O becomes 10 ≥ 10 < 12, true.
• D ≥ L is 9 ≥ 7, true.
• C ≥ D ≤ S ≥ Z becomes 10 ≥ 9 ≤ 11 ≥ 6, true.

In this arrangement, C = Q = 10, so C < Q is false. This confirms that Conclusion II is not definite.

Verification / Alternative check:
The counterexamples prove that neither conclusion is forced by the premises. For Conclusion I, you can always let F be much smaller than both M and R while still keeping M ≥ F and R ≤ M valid. For Conclusion II, setting C and Q equal to M automatically satisfies all the inequalities but violates the strict inequality C < Q. Therefore, the safest and correct judgment is that none of the proposed conclusions must always hold.

Why Other Options Are Wrong:
Option b: “Only conclusion I is true” is wrong because we found a configuration where R is greater than F. Option c: “Only conclusion II is true” is incorrect due to the equal value example where C equals Q. Option d: “Either conclusion I or conclusion II is true” is not valid since we demonstrated cases where both conclusions fail simultaneously. Option e: “Both conclusions I and II are true” is clearly false; neither holds in all valid assignments.

Common Pitfalls:
A common mistake is to assume that if C ≤ M and Q ≥ M, then C must be strictly less than Q, ignoring the possibility of equality. Another pitfall is overextending chains and inferring relationships that are not supported, such as assuming R is less than or equal to F just because both are tied to M. Always test your intuition with sample values to verify whether a conclusion is unavoidable or if alternative arrangements exist that break it.

Final Answer:
None of the conclusions is true.