ISRO CS 2023 Q.38: solution and answer

Official answer: DSource: ISRO's revised and final answer key of 16 February 2024

1. The first gate has both inputs tied to $A$, so its output is $\overline{A\cdot A}=\overline{A}$. It works as an inverter.

2. The second gate has both inputs tied to $B$, so its output is $\overline{B\cdot B}=\overline{B}$.

3. The third gate is a NAND of these two outputs: $Y=\overline{\overline{A}\cdot \overline{B}}$.

4. By De Morgan's theorem, $\overline{X\cdot Z}=\overline{X}+\overline{Z}$. With $X=\overline{A}$ and $Z=\overline{B}$, $Y=\overline{\overline{A}}+\overline{\overline{B}}=A+B$.

5. So $Y=A+B$, which is the OR operation. Check with the truth table: $A=B=0$ gives $Y=\overline{1\cdot 1}=0$. $A=1,B=0$ gives $Y=\overline{0\cdot 1}=1$. $A=0,B=1$ gives $Y=1$. $A=B=1$ gives $Y=\overline{0\cdot 0}=1$.

6. The other options fail. XOR gives 0 for $A=B=1$, but here $Y=1$. NAND gives 1 for $A=B=0$, but here $Y=0$. AND gives 0 for $A=1,B=0$, but here $Y=1$.