Difficulty: Easy
Correct Answer: 5
Explanation:
Introduction / Context:This digital logic question checks your understanding of minterms and coverage in a three-variable Boolean expression. Knowing how to count satisfying input combinations (minterms) is foundational for Karnaugh maps, simplification, and logic synthesis.
Given Data / Assumptions:
Concept / Approach:A term A in an OR expression A + X covers all cases where A = 1 (regardless of B, C). The product BC contributes additional cases when A = 0 and both B and C are 1. We must avoid double counting overlaps.
Step-by-Step Solution:
Count cases with A = 1: there are 2^2 = 4 minterms (for all combinations of B, C).Now consider A = 0. Then the expression reduces to BC. Only (B, C) = (1, 1) satisfies BC = 1, which adds exactly 1 new minterm.Total minterms = 4 (from A = 1) + 1 (from A = 0 and BC = 1) = 5.Verification / Alternative check:List explicitly: A = 1 with (B, C) ∈ {(0,0), (0,1), (1,0), (1,1)} → 4 minterms. Plus A = 0, B = 1, C = 1 → 1 minterm. No other A = 0 case works. Sum = 5.
Why Other Options Are Wrong:
2 or 3: undercount by ignoring A = 1 coverage or the extra BC case at A = 0.4: counts only A = 1 cases and misses the BC case at A = 0.6: overcounts (there is only one additional case when A = 0).Common Pitfalls:
Double counting the case A = 1, B = 1, C = 1 (already counted in A = 1 set).Forgetting that BC only matters when A = 0.Final Answer:
5
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