Question

If $n\left(U\right)=48,n\left(A\right)=28,n\left(B\right)=33$ and $n(B–A)=12$, then $n(A\cap B{)}^C=$

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

Answer 1

The correct option is A $27$

$n(B–A)=n\left(B\right)–n(B\cap A)$
$\Rightarrow n(B\cap A)=33-12=21\therefore n(A\cap B{)}^C=n(U)–n(A\cap B)=48-21=27$