GO Classes 2023 | IIITH Mock Test 1 | Question: 35

Answer 1

The Boolean function $\text{F(A, B, C, D)}$ evaluates to $1$ for minterms $1$ and $9$, and it evaluates to $0$ for minterms $0$ and $12$.

Thus, Option D) $\mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\Sigma(1,3,4,6,7,8,9,11)$ is correct.

Comments

isn’t it evaluating to 1 on 12 ?
         A B C D
12 =  1 1 0 0

S0 S1 S2
C   B.  A
0   1.   1 = 3
So, select line will pick 3 and as its connected to 1 so, it F will evaluate to 1 ?

Can you please tell where I am wrong, or I am totally doing it wrong.

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@Arnav S2 is msb bit, so S2 S1 S0-→ A B C--->110, so select line will pick 6, there F will be 0 independent of D

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F=1 for minterms 1 and 9 

So, there are inputs on which D=1, i.e. 4 and 3.

A(S2) B(S1) C(S0) D

For 4:

1        0          0       1  = 9,      Hence, for minterm 9; F=1.

For 3:

0       1           1       1  =7,       Hence for minterm 7, F=1.

But as given answer, it should be minterm 1. Where I'm going wrong?