Transitivity of strict order: Given L > M, M > N, and N > P, determine which conclusions definitely follow. Conclusions: I) L > P; II) M > P.

Introduction / Context:This question checks transitivity in a chain of strict inequalities. When each comparison is strict (">"), we can pass the relation forward to deduce farther-apart elements. The task is to test two conclusions derived from L > M > N > P.

Given Data / Assumptions:

• L > M
• M > N
• N > P
• Conclusions to test: I) L > P; II) M > P.

Concept / Approach:Use transitivity of strict order. If a > b and b > c, then a > c. Chaining multiple strict inequalities preserves strictness end-to-end. No equalities or non-strict signs (≥, ≤) are present to weaken the conclusion.

Step-by-Step Solution:From L > M and M > N, we obtain L > N (by transitivity).From L > N and N > P, we obtain L > P (I is true).Separately, from M > N and N > P, we obtain M > P (II is true).

Verification / Alternative check:Create a numeric model that satisfies the premises: let L=7, M=5, N=3, P=1. Then L > P (7 > 1) and M > P (5 > 1) both hold, confirming the deductions.

Why Other Options Are Wrong:Options a/b assert only one conclusion; d says “either”, and e says “neither”. Since both I and II are compelled by transitivity, these alternatives understate the certainty or contradict the logical result.

Common Pitfalls:Mistaking strict chains for non-strict and concluding L ≥ P or M ≥ P instead of L > P and M > P. Another error is skipping a link (e.g., comparing L to P without explicitly chaining via M and N), which can obscure the role of transitivity.

Final Answer:If both conclusions I and II follow