Question

If $n(\mu) = 48 , n(A) = 28, n(B) = 33$ and $n(B - A) = 12$, then $n(A \cap B)^C$ is

• A. 27
• B. 28
• C. 29
• D. 30

Answer 1

Answer: (A)

27

Answer 2

$n\left(\mu\right) =48 , n\left(A\right)=28 , n\left(B\right)=33$ $n\left(B-A\right) =12$ $n\left(A \cap B\right) =n\left(B\right) -n\left(B -A\right) =33 -12 =21$ $n\left(A \cap B\right)^{C} = n\left(\mu\right) -n=\left(A\cap B\right)=48 -21 =27$