In the following comparison puzzle, assume all the inequality statements are true and then determine which of the given conclusions must be true. Statements: P > Q ≤ R > Z; Y > X > P ≥ U; Q ≥ S < T Conclusions: I. P > S II. Z < T

Introduction / Context:
This question is a typical inequality or comparison problem often asked in logical reasoning sections. Several variables are connected with greater than, less than, and less than or equal to symbols. Your task is to determine which conclusions must be true in every possible arrangement that satisfies the given statements. This tests your ability to combine chains of inequalities and to distinguish between what is definitely true and what is only possibly true.

Given Data / Assumptions:

• Statements:
  • P > Q
  • Q ≤ R
  • R > Z
  • Y > X
  • X > P
  • P ≥ U
  • Q ≥ S
  • S < T
• Conclusion I: P > S.
• Conclusion II: Z < T.
• All inequalities are consistent and refer to the same scale of comparison, such as height, marks, or any ordered quantity.

Concept / Approach:
The main technique is to build chains of inequalities using transitivity. If A > B and B ≥ C, then A > C. But if two elements do not share a chain, their relationship may remain undetermined. To check a conclusion, you must see whether it holds in all valid numerical assignments that satisfy the given statements. If you can construct even a single valid counterexample where a conclusion fails, that conclusion is not definitely true.

Step-by-Step Solution:
Step 1: Use P and Q. From P > Q and Q ≥ S, we can combine them: since Q is at least as large as S, P is definitely greater than Q and so also greater than S. Step 2: Formally, we have P > Q and Q ≥ S, therefore P > S. This directly supports Conclusion I as a definite truth. Step 3: Now check the parts involving Z and T. From the first chain, R > Z, so Z is less than R. From the last chain, S < T and Q ≥ S, so T is greater than S and S is at most Q. However, we are not given any link between Z and S, or Z and T, through a continuous chain of inequalities. Step 4: Try to see if Z < T must always hold. To disprove a universal conclusion, it is enough to find one possible arrangement where all the given statements are satisfied, but Z < T is not true. Step 5: Create a counterexample. Take numbers P = 9, Q = 7, R = 10, Z = 8, Y = 15, X = 11, U = 5, S = 5, T = 6. Then:

• P > Q is 9 > 7 (true)
• Q ≤ R is 7 ≤ 10 (true)
• R > Z is 10 > 8 (true)
• Y > X is 15 > 11 (true)
• X > P is 11 > 9 (true)
• P ≥ U is 9 ≥ 5 (true)
• Q ≥ S is 7 ≥ 5 (true)
• S < T is 5 < 6 (true)

Here Z = 8 and T = 6, so Z > T, which directly contradicts Conclusion II.

Verification / Alternative check:
The counterexample confirms that while all the given statements hold, Z can be greater than T. Therefore, Z < T is not guaranteed by the premises. On the other hand, the relationship P > Q ≥ S always leads to P > S, no matter what valid numbers you choose. This means Conclusion I is robust across all scenarios, but Conclusion II is not. Checking multiple sample assignments is a practical way to gain confidence in your reasoning.

Why Other Options Are Wrong:
Option a: “None of the conclusions is true” is incorrect because Conclusion I has been shown to be true in every valid arrangement. Option c: “Only conclusion II is true” is wrong because we have explicitly built a situation where Z > T, so Conclusion II fails. Option d: “Either conclusion I or conclusion II is true” is incorrect because both can be checked separately, and we find that I is always true while II is not necessarily true. Option e: “Both conclusions I and II are true” is incorrect since Conclusion II does not hold in all possible cases.

Common Pitfalls:
Many candidates mistakenly assume that if two variables appear in the same group of inequalities, there must be a fixed relationship between them. However, unless there is a continuous chain connecting two variables, their order may remain undetermined. Another frequent error is not using counterexamples; if a relationship is not obvious, trying real numbers that satisfy the conditions can quickly show whether a conclusion is always true or only sometimes true.

Final Answer:
Only conclusion I is true.