Strict and equality mix: Given A > B, B = H, and H > G, determine which conclusions definitely follow. Conclusions: I) A > G; II) A > H.

Introduction / Context:Here we combine equality with strict inequalities. Equalities allow substitution (B = H), which lets us realign the chain and apply transitivity correctly to test A against both G and H.

Given Data / Assumptions:

• A > B
• B = H
• H > G
• Conclusions to test: I) A > G; II) A > H.

Concept / Approach:Replace B by H wherever B appears (since B = H). Then apply transitivity: if A > H and H > G, we can deduce A > G. We must verify both conclusions independently.

Step-by-Step Solution:From A > B and B = H ⇒ A > H (II true).From A > H and H > G ⇒ A > G (I true).

Verification / Alternative check:Let H = B = 10, G = 8, and A = 12. Then A > H (12 > 10) and A > G (12 > 8). Both conclusions hold in this concrete model.

Why Other Options Are Wrong:Options a/b assert only one conclusion; option d (“either”) understates the certainty; option e (“neither”) contradicts the established chain.

Common Pitfalls:Failing to substitute B = H before chaining; or assuming A ≥ H rather than the required A > H. Here, because A > B and B = H, the comparison remains strict.

Final Answer:If both conclusions I and II follow