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In a hotel, 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 ?
15
18
21
24
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.
People who had neither type of lunch = (96 - 72)= 24.
Hence, option 4 is the correct answer.
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