First slide

Operations on Sets

Question

Let $U$ be a Universal set and $n (U) = 12 .$ If $A, B \subseteq U$ are such that $n (B) = 6$ and $n (A \cap B) = 2$ then $n (A \cup B')$ is equal to

Moderate

A

6
B

10
C

7
D

8

Solution

$n (A \cup B') = n (A) + n (B') - n (A \cap B')$
$= n (A) + n (U) - n (B) - (n (A) - n (A \cap B)) = n (U) - n (B) + n (A \cap B) = 12 - 6 + 2 = 8 .$

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