# Digital-Grandtest1-Q1

Given $A\bar B + \bar A B = C, A\bar C + \bar A C = ?$

$(A). B$

$(B).\bar B$

$(C).AC$

$(D).\bar A\bar C$

asked Jun 19 in Digital
reshown Jun 20

Answer : option A) -> B

Solution :

Put given C= AB' +A'B

Hence, AC'+A'C=  A(AB'+A'B)' + A'(AB'+A'B)

->Now by rule : (X+Y)'= X'Y'

Hence = A( (AB')'.(A'B)' )  + A'(AB'+A'B)

-> Again similar rule :  (XY)' = X'+Y'     (DeMorgan's)

Hence =  A((A'+B).(A+B')) + A'(AB'+A'B)

-> Rule: X (Y+Z)= XY + XZ

Hence = A(A'B'+AB)+A'(AB'+A'B')

= AA'B+AAB+A'AB'+A'A'B

-> Since AA' =0

Hence = AB+A'B    ,Taking B out= B(A+A')   = B.1  =B
answered Jun 21 by (73,870 points)
selected Jun 21 by gbmentor
+1 vote
 $A$ $B$ $\bar A$ $\bar B$ $A \bar B$ $\bar A B$ $C$ $\bar C$ $A\bar C$ $\bar A C$ $A\bar C+\bar AC$ 0 0 1 1 0 0 0 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 1 0 1 0 0 0 0 1 1 0 0 0 0 0 1 1 0 1

So the option is A

answered Jun 20 by (6,630 points)
+1 vote
Here c is nothing but XOR gate

And complement of XOR is XNOR

-->  A.XNOR+A'.XOR

--> A(A'B'+AB)+A'(AB'+A'B')

--> AA'B+AAB+A'AB'+A'A'B

--> 0+AB+0+A'B=AB+A'B

--> B(A+A')=B.1=B

SO OPTION A) IS CORRECT.
answered Jun 20 by (24,690 points)
edited Jun 20