Using a 7485 four-bit magnitude comparator — specific case check: Given A = 0110 (decimal 6) and B = 1010 (decimal 10), evaluate the following claimed outputs for the 7485 comparator: A < B = 1, A = B = 0, A > B = 1. Determine whether this set is valid.

Introduction / Context:
The 7485 magnitude comparator outputs three mutually exclusive flags: A<B, A=B, and A>B. Only one of these should be asserted for any well-defined comparison. This problem tests your ability to evaluate a specific input pair and verify consistent outputs.

Given Data / Assumptions:

• A = 0110 (6 unsigned).
• B = 1010 (10 unsigned).
• Comparator is a standard 4-bit device without cascade inputs affecting the result.

Concept / Approach:
Compare A and B numerically as unsigned binary. Since 6 is less than 10, the correct flags are A<B = 1, A=B = 0, A>B = 0. The claim that both A<B and A>B are 1 simultaneously violates mutual exclusivity and is logically inconsistent.

Step-by-Step Solution:
Convert to decimal or compare bitwise from MSB: A(0110) vs B(1010).MSB comparison: A3=0, B3=1 → immediately implies A<B.Therefore set outputs: A<B = 1, A=B = 0, A>B = 0.Compare with claim: claim has A>B = 1 as well → inconsistent → false.

Verification / Alternative check:
7485 truth tables show that when the highest unequal bit indicates A<B, the lower bits do not change the outcome; only one relation output is asserted (taking into account cascade inputs which are assumed inactive here).

Why Other Options Are Wrong:
“Correct as stated” would require simultaneous greater and less, which cannot happen. Considering signed interpretation or ignoring the MSB does not apply; the device compares given bit patterns as provided.

Common Pitfalls:
Forgetting the mutual exclusivity; misreading bit order; letting lower bits override a decision already made by a higher bit.

Final Answer:
Incorrect (A < B = 1, A = B = 0, A > B = 0)