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Which functions does NOT implement the Karnaugh map given below? 1. $(w + x) y$
2. $xy + yw$
3. $(w + x) (\bar{w} + y) (\bar{x} + y)$
4. None of the above

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Please derive the expression which is in option C

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See we can simplify each equation given in the option and get that all of them gives $xy +wy$. But let think in another way.

$1^{st}$ option is written in $POS$ form, as we can check we get the same if we consider the following implicants. which is $(w+x)y$

for the second one Which gives $wy+xy$ now for 3rd one, we can verify like this

which is $(w+x)(\bar w+y)(\bar x+y)$

So as we can verify each equation in a given K-Map, so the answer is option D

Thank you
I think we can even further simplify POS here by making a Octet and QUAD--> Z .(X+W) ..( Though this is not necessary for this question :P)

Solving $K$ map gives $xy +wy$

Why answer is D why not B,  xy+wy = xy + yw Why considering them different?
^^ There is NOT in question

a, b and c are same.
I am sorry, I didn't see NOT in the question.
how to get the 2nd option..??
Please derive the expression which is in option c

@Praveen Saini Sir  @Anu007  xy+wy ,   xy + yw These two are different terms..?

No, in binary logic

Answer is D.See we can simplify each equation given in the option and get that all of them gives xy +wy. But let think in other way.

1st option is written in POS form, as we can check we get the same if we consider the following impicants. which is (w+x)y

for second one Which gives wy+xy

now for 3rd one we can verify like this which is (w+x)(ˉw+y)(ˉx+y)

So as we can verify each equation in a given KMap, so answer is D

## Hi, just multiply the terms in option C, you'll get this wx+wy which we can easily derive so, ans is option D. After simplification we get WY+XY
So Ans is Option D

Answer : $Option$ $D$ by ￼￼s￼￼￼￼￼￼￼￼s￼￼￼￼￼￼￼s￼ee my answer.....

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