In a hoel, 60% had vegetarian lunch while 30% had non-vegetaian lunch and 15% had both types of lunch. If 96 people were present, how many did not eat either type of lunch ?

A 24

B 30

C 36

D 42

Correct Answer: Option A

n (A) = (60/100 * 96) = 288/5, n (B) = (30/100 * 96) = 144/5, n (A ∩ B) = (15/100 * 96) = 72/5.

n(A ∪ B) = n (A) + n (B) - n (A ∩ B) = 72.

So, people who had either or both types of lunch = 72.

Hence, people who had neither type of lunch = (96 - 72)= 24.