Logic Gates question from MHT CET 2024 (3 May, Shift 2)

Answer: (b)

1. (I) Each input NAND has shorted inputs, so it is a NOT: $\overline{0} = 1$ and $\overline{0} = 1$.
2. Final NAND: $\overline{1 \cdot 1} = 0$.
3. (II) NAND followed by a NAND with shorted inputs (NOT) gives AND: $Y = A \cdot B$.
4. With both inputs HIGH, $Y = 1 \cdot 1 = 1$.
5. So the answer is (b) 0, 1.