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MHT CET 2024 (11 May, Shift 1) · Logic Gates

Logic Gates question from MHT CET 2024 (11 May, Shift 1)

Q76MHT CET 2024 · 11 MAY · SHIFT 1Medium

Logic Gates

Following combination of gates is equivalent to

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Answer: (d)

First NAND: $C = \overline{AB}$ ; middle NANDs give $\overline{AC}$ and $\overline{BC}$ .
Output: $Y = AC + BC = (A+B)\overline{AB}$ .
$Y = A\bar B + \bar A B$ , which is 1 only when the inputs differ.
So the answer is (d) X-OR gate.

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