TRB 2006 - Question 5

Boolean function of $A+BC$ is reduced form of

a) $AB + BC$
b) $(A+B) (A+C)$
c) $A'B+AB'C$
d) $(A+C) B$

Correct Answer: Option B

Option A: $AB + BC$ can't be reduced further
Option B: $$=(A+B) (A+C)$$ $$=AA+AC+AB+BC$$ $$=A+AC+AB+BC$$ $$=A(1+C)+AB+BC$$ since $1+C=1$ $$=A+AB+BC$$ $$=A(1+B)+BC$$ $$=A+BC$$ Option C: $A'B+AB'C$ can't be reduced further
Option D: $(A+C) B$ can't be reduced further