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4 Answers

Best answer
13 votes
13 votes

$ Y = ( A + \bar{B} + \bar{A} B) \bar{C} $

$ Y = (A + (\bar{B} + \bar{A}) . (\bar{B} + B) ) \bar{C} $

$ Y = (A + \bar{B} + \bar{A} ) \bar{C} $

$ Y = ( (A + \bar{A}) + \bar{B}) \bar{C} $

$ Y = (1 + \bar{B}) \bar{C} $

$ Y = \bar{C} $

Answer would be Option C) $\bar{C}$


We have $A + \bar B$. When both are false, we get $\bar A . B$ which is given as the third term of logical OR. So, this part always return TRUE. 

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Another Solution

At exam time, if you become clueless to reduce A+B'+A'B just make a truth table

A B A+B'+A'B
0 0 1
0 1 1
1 0 1
1 1 1

Clearly we can say A+B'+A'B=1 hence (A+B'+A'B)C'= 1.C' =C'

Answer:

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