Order-relations with equality and strict inequality: Statements: L = M ≥ N; M > P; L < K Conclusions: I. K > P II. N < K Choose the correct option.

Introduction / Context: We are given a small system of inequalities/equality and asked which conclusions must always hold. Such questions assess transitive reasoning, paying attention to whether inequalities are strict or non-strict.

Given Data / Assumptions:

• L = M ≥ N
• M > P
• L < K
• Conclusions: (I) K > P, (II) N < K

Concept / Approach: Use substitution via L = M, then chain comparisons carefully. Convert statements into clean implications: from M > P and L = M, we get L > P; from L < K we get K > L. Combine to test each conclusion.

Step-by-Step Solution: Because L = M and M > P, we obtain L > P. Also, L < K ⇒ K > L. For (I): K > L and L > P ⇒ K > P (transitivity). Hence (I) is always true. For (II): From L = M ≥ N we have L ≥ N (hence N ≤ L). With L < K, we chain N ≤ L < K ⇒ N < K. Thus (II) is also always true.

Verification / Alternative check: Pick any numbers consistent with the statements, e.g., N = 3, M = L = 5, P = 2, K = 7. Then K > P (7 > 2) and N < K (3 < 7). Try edge case N = L = M: N = 5, P = 4, K = 6 also satisfies both.

Why Other Options Are Wrong: Options claiming only one is true or neither are inconsistent with the guaranteed chains above. “Either I or II” implies exclusivity that does not exist here.

Common Pitfalls: Forgetting that equality L = M lets us move freely between L and M; also, mixing up directions when combining a non-strict with a strict inequality.

Final Answer: Both Conclusions I and II are true.