ISRO2007-01

Question

The Boolean expression $\text{Y = (A}+\overline{\text{B}}+\overline{\text{A}}\text{B})\overline{\text{C}}$ is given by

• A. $\text{A}\overline{\text{C}}$
• B. $\text{B}\overline{\text{C}}$
• C. $\overline{\text{C}}$
• D. $\text{AB}$

Answer 1

$ 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) $\bar{C}$

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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. 

Answer 2

Y=(A+B'+A'B) C'

  =(A+(B'+A').(B'+B) )C'

 =(A+B'+A')C'

= (1+B')C'

=C'

OPTION C

Comments

u r correct shivani and very fast than me..

Answer 3

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 4

answer is(C) that is C bar