GATE CSE 2000 | Question: 2.11

Question

Which functions does NOT implement the Karnaugh map given below?

                                        

• A. $(w + x) y$
• B. $xy + yw$
• C. $(w + x) (\bar{w} + y) (\bar{x} + y)$
• D. None of the above

Comments

Please derive the expression which is in option C

Answer 1

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 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

Comments

this is Best answer

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Thank you

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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)

Answer 2

Answer - D.

Solving $K$ map gives $xy +wy$

Comments

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

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Hi, just multiply the terms in option C, you'll get this wx+wy which we can easily derive so, ans is option D.

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can you do the simplification please. I can't get to reduce x'yw + xyw' to 0

Answer 3

Answer : $Option$ $D$

Comments

Thanks for the simplification, it really helped me see a silly mistake of mine.

Answer 4

After simplification we get WY+XY
So Ans is Option D