Question

If $n\left(U\right)=48,n\left(A\right)=28,n\left(B\right)=33$ and $n(B-A)=12$ then $n{(A\cap B)}^C$ is (where $U$ denotes the universal set)

• A. $27$
• B. $28$
• C. $29$
• D. $30$

Answer 1

The correct option is A $27$

$n(A\cap B{)}^c=n(A)-n(A\cap B)$

$n(B-A)=n\left(B\right)-n(A\cap B)$

$12=33-n(A\cap B)$

$\Rightarrow n(A\cap B)=33-12$

$=21$

$\therefore n(A\cap B{)}^c=48-21$

$=27$