Two linked chains: Given P > M > Q and Q > Z > N, determine which conclusions definitely follow. Conclusions: I) M ≥ Z; II) N < P.

Introduction / Context:
Here two strict chains share Q. We must compare M to Z and N to P. Because all relations are strict and connect through Q, both conclusions become straightforward via transitivity.

Given Data / Assumptions:

• P > M > Q
• Q > Z > N
• Conclusions: I) M ≥ Z; II) N < P.

Concept / Approach:
Chaining strict inequalities preserves strictness. From M > Q and Q > Z, deduce M > Z (which implies M ≥ Z). From P > M and M > Q > Z > N, deduce P > N.

Step-by-Step Solution:
M > Q and Q > Z ⇒ M > Z ⇒ M ≥ Z (I holds).P > M and M > Q > Z > N ⇒ P > N ⇒ N < P (II holds).

Verification / Alternative check:
Let P=10, M=8, Q=6, Z=4, N=2. Then M ≥ Z (8 ≥ 4) and N < P (2 < 10) both true.

Why Other Options Are Wrong:
a/b recognize only one of the two truths; d/e are inconsistent with the strict chains.

Common Pitfalls:
Interpreting ≥ as necessary; note we actually get a stronger result M > Z, which still satisfies ≥. Also beware reversing links through Q incorrectly.

Final Answer:
If both conclusions I and II follow