ISRO EC 2023 Q.33: solution and answer

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

1. Connect the full-adder inputs $A,B,C_{in}$ to the select lines of the $3:8$ active-high decoder. Output $D_i$ is then high exactly when the input combination is $i$, so $D_i$ is the minterm $m_i$.

2. Sum: $S=\sum m(1,2,4,7)$, which is the OR of four decoder outputs $D_1,D_2,D_4,D_7$.

3. Carry: $C_{out}=\sum m(3,5,6,7)$, which is the OR of four decoder outputs $D_3,D_5,D_6,D_7$.

4. A 3-input OR gate has only 3 inputs, so it cannot combine 4 signals alone. Each output therefore needs at least 2 gates.

5. Two gates are enough for each output: the first gate ORs three of the decoder outputs, and the second gate ORs that result with the fourth decoder output (its third input is unused). So each output needs exactly 2 gates.

6. The two minterm sets share only $m_7$. A gate shared by $S$ and $C_{out}$ would have to OR two or more common decoder outputs, and there is only one common output. The decoder is active high, so no complemented outputs are available to help. No OR gate can be shared, so the gate counts add.

7. Total $=2+2=4$ three-input OR gates, which is option (c).