Difficulty: Easy
Correct Answer: true
Explanation:
Introduction / Context:
This is a classic transitivity problem in inequality reasoning. When comparing quantities of three categories, we can sometimes chain the comparisons to reach a necessary conclusion about two of them. The challenge is to recognize that “more than” is a transitive relation for strict comparisons.
Given Data / Assumptions:
Concept / Approach:
For strict inequalities on real numbers (and therefore on nonnegative integers), transitivity holds: if x > y and y > z, then x > z. We simply apply this chain to A, L, and O to determine A vs. O. No additional assumptions are required.
Step-by-Step Solution:
Verification / Alternative check:
Pick sample numbers consistent with the premises: let A = 10, L = 7, O = 3. Both premises hold, and indeed 10 > 3, confirming the derived statement. Any valid triple that satisfies A > L and L > O will automatically satisfy A > O.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing strict “more than” with “at least as many as”, which is non-strict and can break simple transitivity patterns in edge cases. Here, “more than” is strict, so the chain is safe.
Final Answer:
true
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