Solve 4th problem.

Option (c) is the correct option

below is the solution

Using Boolean Algebra won’t give you the answer directly, I will show the result obtained using the Boolean Algebra

Z = ABCD + AB’CD + A’CB’D

Z = ACD(B+B’) + A’CB’D {taking ACD common from the first two terms}

Z = ACD + A’CB’D {X+X’ = 1}

Z = CD(A + A’B’) {taking CD common}

Z = CD(A+B’) {X + X’Y = X+Y, known as distributive law of Boolean Algebra}

Z = ACD + B’CD {Distributive law}

And, if you try to get the minimized result using the K’Map you will get the same result i.e. Z = ACD + B’CD

Thanks

But how can i recognize such type of problem where Boolean alzebra not gives correct answer?

@chaituvarma there is mistake in your calculations, A + A’CB’D {=}\llap{/\,} CB’D

you said A + A’ is 1, which is true but you have A + A’CB’D

use distributive law

**A + A’X = (A + A’)(A + X)** *{ where X = CB’D}*

**A + A’X = 1.(A + X)** *{since A + A’ = 1}*

**A + A’CB’D = A + CB’D** *{substituting X = CB’D}*