ISRO EC 2023 Q.34: solution and answer

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

1. The 2:1 multiplexer has select line $B$. For $B=0$ it passes the top 4:1 multiplexer output (input $i_0$). For $B=1$ it passes the bottom 4:1 multiplexer output (input $i_1$).

2. The required function is $F=A+\overline{B}$. For $B=0$: $F=A+1=1$. So the top multiplexer output must be 1 for every combination of $A$ and $C$.

3. Hence all data inputs of the top multiplexer are 1: $X_0=X_1=X_2=X_3=1$.

4. For $B=1$: $F=A+0=A$. So the bottom multiplexer output must equal $A$, whatever the value of $C$.

5. Both 4:1 multiplexers have $S_1=A$ and $S_0=C$. The inputs $X_4,X_5$ are selected when $A=0$ (with $C=0,1$), so they must be 0. The inputs $X_6,X_7$ are selected when $A=1$, so they must be 1.

6. So $(X_4,X_5,X_6,X_7)=(0,0,1,1)$.

7. The full vector is $(X_0,\dots ,X_7)=(1,1,1,1,0,0,1,1)$, which is option (b). Option (a) has 0 in $X_2,X_3$, option (c) has 0 in $X_3$ and wrong bottom inputs, and option (d) has 0 in $X_0,X_1$.