GO Classes DPP | GATE CS | Digital Logic | Boolean Laws

Answer 1

Here  is the explanation

Comments

kmap method is much faster

---

Extly bro..faster and accurate as well , just master the kmap then whole digital will become cakewalk for you

Answer 2

Given $F=(A+B+C)(A+B+\overline{C})(A+\overline{B}+C)$.

Take the first two brackets.

Let $X=A+B$.

Then,

$(A+B+C)(A+B+\overline{C})=(X+C)(X+\overline{C})$

Using the law $(X+Y)(X+\overline{Y})=X$,

$(X+C)(X+\overline{C})=X$

So,

$(A+B+C)(A+B+\overline{C})=A+B$

Now the expression becomes:

$F=(A+B)(A+\overline{B}+C)$

Use the law $(X+Y)(X+Z)=X+YZ$.

Here, $X=A$, $Y=B$, and $Z=\overline{B}+C$.

So,

$F=A+B(\overline{B}+C)$

Now distribute $B$.

$F=A+B\overline{B}+BC$

Since $B\overline{B}=0$,

$F=A+0+BC$

$F=A+BC$

Answer: A

Answer 3

 

Option A