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in Digital Logic by (157 points) | 91 views

3 Answers

+3 votes

Answer : D 

by Active (4.4k points)
+1 vote
A\BC 00 01 11 10
0   1   1
1 1   1  

We can't make any grouping so we need to minimize by taking individual ones

A'B'C+A'BC'+AB'C'+ABC=$A\bigoplus B\bigoplus C$

$A\bigoplus B\bigoplus C \\ \\ =(A\bigoplus B)\bigoplus C\\ =(A'B+AB')\bigoplus C\\ =(A'B+AB')'C+(A'B+AB')C'\\ =(A'B'+AB)C+(A'B+AB')C'\\ =A'B'C+ABC+A'BC'+AB'C'$

Therefore ans should be D

by Junior (969 points)
0 votes

It's option (d)

It can be solved using two methods

1)  all variables are independent cannot be minimised further so the resul has to of 4 minterm expressions when you expand the options only option d would match

2) A'B'C+A'BC'+ABC+AB'C'

     A'(B'C+BC')+A(BC+B'C')

     A'(B XOR C)+A(B XNOR C)

     A XOR (B XOR C)

      A XOR B XOR C

by (27 points)
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