GATE ECE 2014 Set 1 | Question: 15

Question

The Boolean expression $(X+Y)(X+\overline{Y}) + \overline{(X\;\overline{Y}) + \overline{X}}$ simplifies to

• A. $X$
• B. $Y$
• C. $XY$
• D. $X+Y$

Comments

Detailed Video Solution: https://youtu.be/3hahE-VIteQ?feature=shared 

NOTE: $A + A’B = A+B$

So, $\overline{X} + X \overline{Y} = \overline{X} + \overline{Y}.$

Now, $\overline{X’+Y’} = XY.$

$(X+Y)(X+\overline{Y}) = X + YY’ $ (By Distributive Property)

& $X + YY’ = X$

Finally, we have $X + XY = X$ (By Absorption Law)

Detailed Video Solution: Video Solution (Click Here)

Answer 1

(x+y)(x+!y)+!((x!y)+!x)

=x+xy+x!y+xy

=x

Answer 2

Given boolean expression: $(X+Y)(X+\bar Y)+\overline{(X\bar Y)+\bar X}$

$\implies$ $(X.X+X\bar Y+X.Y+Y.\bar Y)+(\overline{(X\bar Y)}*\overline{\bar X})$

$\left [ \because \textrm{Demorgon’s law= $\overline{A+B}=\bar A+\bar B$, Idempotent law= A*A=A}\right]$

$\implies$ $(X+X\bar Y+XY)+(\bar X+\overline{\bar Y})*X)$

taking $X$ as common:

$\implies$ $X(1+\bar Y+Y)+(\bar X+Y)*X$

 $\left [ \because A+\bar A=1\textrm{( Complement law)}\right ]$

$\implies$ $X+X*\bar X+X*Y$

$\implies$ $X+XY$

$\implies$ $X(1+Y)$

$\implies$ $X$

Option A is correct.

Ref: Properties of Boolean Algebra