GATE 2026 EC Q.48: solution and answer

Official answer: CSource: Official answer key, GATE 2026 EC (IIT Guwahati)

1. Read the data inputs off the figure. A vertical wire leaves the 1 line at a T-junction and runs down to $I_2$, so it carries 1. The 0 line hops over that wire (the arc means no connection) and goes to $I_1$. So $I_0=1$, $I_1=0$, $I_2=1$, $I_3=y$, and the select lines are $S_1=x$, $S_0=z$. Complement is written with $\overline{}$ throughout.

2. Output equation of a 4:1 MUX: $$f=\overline{S_1}\overline{S_0}{I}_0+\overline{S_1}{S}_0{I}_1+S_1\overline{S_0}{I}_2+S_1{S}_0{I}_3$$

3. Substitute $S_1=x$, $S_0=z$ and the inputs: $$f=\overline{x}\overline{z}\cdot 1+\overline{x}z\cdot 0+x\overline{z}\cdot 1+xz\cdot y=\overline{x}\overline{z}+x\overline{z}+xyz$$

4. Combine the first two terms: $\overline{x}\overline{z}+x\overline{z}=\overline{z}(\overline{x}+x)=\overline{z}$. This leaves $f=\overline{z}+xyz$.

5. Apply the absorption rule $A+\overline{A}B=A+B$ with $A=\overline{z}$ (so $\overline{A}=z$) and $B=xy$: $$f=\overline{z}+z(xy)=\overline{z}+xy$$

6. Check with the truth table. When $z=0$ the MUX selects $I_0=1$ or $I_2=1$, so $f=1$ (minterms 0, 2, 4, 6). When $z=1$ it selects $I_1=0$ (for $x=0$) or $I_3=y$ (for $x=1$), so the only 1 is at $xyz=111$ (minterm 7). So $f=\sum m(0,2,4,6,7)=xy+\overline{z}$, which is option (C). The other options fail: (A) and (B) give $f=0$ at $000$, and (D) gives $f=1$ at $011$ and $f=0$ at $111$.