GATE 2020 EC Q.10: solution and answer

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

1. The EN input has an inversion bubble, so it is active-low. It is tied to 0, so the MUX is enabled and works as a normal 4-to-1 multiplexer.

2. Output of a 4-to-1 MUX (complement written with $\overline{}$): $$F=\overline{S_1}\overline{S_0}{I}_0+\overline{S_1}{S}_0{I}_1+S_1\overline{S_0}{I}_2+S_1{S}_0{I}_3$$

3. Substitute $S_1=P$, $S_0=Q$, $I_0=R$, $I_1=0$, $I_2=R$, $I_3=1$: $$F=\overline{P}\overline{Q}R+0+P\overline{Q}R+PQ$$

4. Combine the two $\overline{Q}R$ terms using $\overline{P}+P=1$: $\overline{P}\overline{Q}R+P\overline{Q}R=\overline{Q}R$.

5. So $F=PQ+\overline{Q}R$, which is option (A).

6. Checking the other options against the MUX directly: with $P=1,Q=0,R=0$ the MUX selects $I_2=R=0$, but (B) $P+Q\overline{R}=1$. With $P=1,Q=1$ the MUX selects $I_3=1$, but (C) $P\overline{Q}R+\overline{P}Q=0$. With $P=0,Q=0,R=0$ the MUX selects $I_0=R=0$, but (D) $\overline{Q}+PR=1$. So (B), (C) and (D) are all wrong, and (A) is the answer.