Question

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

Answer 1

Let x be the people did not eat either type of lunch

60% of 96 + 30% of 96 - 15% of 96 = 96-x

75% of 96 = 96 - x

72 = 96 - x

x = 24

alternatively, you can calculate the total value separately, n(a) = 288/5, n(b) = 144/5 and n(anb) = 72/5

n(aUb) = n(a) + n(b) – n(anb) = 288/5 + 144/5 – 72/5 = 72

Hence, 96 - 72 = 24 that have neither of them.