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+10 votes

Simplified Boolean expression for A'BC+AB'C'+A'B'C'+AB'C+ABC

A . AB

B . B'C

C . AB+(A'+AB')C

D . AB'+BC+B'C'

+8 votes

Best answer

+2 votes

D option is correct ans

A'BC +AB'C'+A'B'C'+ AB'C+ ABC

A'BC+B'C'(1)+AB'C+ ABC

A'BC+B'(C'+AC)+ABC

BC (A'+A)+B'(C'+A)

BC(1)+B'(C'+A)

BC+ B'C'+ B'A

A'BC +AB'C'+A'B'C'+ AB'C+ ABC

A'BC+B'C'(1)+AB'C+ ABC

A'BC+B'(C'+AC)+ABC

BC (A'+A)+B'(C'+A)

BC(1)+B'(C'+A)

BC+ B'C'+ B'A

0

Can you derive option D from this ? please write it how ...because i tried but haven't got anything.

0

you have derived it from wrong question there is a term in it which is A'B'C' and you did not use it .

+1

see if we have a+a'b just simply cut the a' literal from a'b term so we will get a+b hence a+a'b is equivalent to a+b :) rt ?

+2 votes

0

As a is most significant bit in the case of what you have taken here most significant should be taken in the place of c the above is not correct thing u have done according to me correct me of iam going wrong

0

@Vishal , Its not always true that Most significan bit should be in place of variable C . Point is final output must be valid and correct . So you can do in both way, you can use this way(attached fig) and you'll get the same output . But its good thing Vishal, you can remind and use only one method . But in exam you can't say they will follow your method , right?? So, .focus on basics.

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