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MHT CET 2026 (14 May, Shift 2) · Logic Gates

Logic Gates question from MHT CET 2026 (14 May, Shift 2)

Q11MHT CET 2026 · 14 MAY · SHIFT 2Medium

Logic Gates

The output Y, when all the three inputs are first high and then low, will respectively be

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Answer: (d)

AND of first two inputs, then NAND with the third: $Y = \overline{(A \cdot B) \cdot C}$ .
All HIGH: $Y = \overline{1} = 0$ .
All LOW: $Y = \overline{0} = 1$ .
So the answer is (d) $0, 1$ .

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