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

Question

Answer 1

10% do not either the type hence 9.6

Answer 2

10% do not either the type hence 9.6

Answer 3

The answer is a result of a basic set theory knowledge. Let A and B be the Vegetarian and Non vegetarian population respectively.
Now, we have the following data with us:
n(A) = 60
n(B) = 30
U = 96
n(A Union B) = n(A) + n(B) - n(A inter section B) = 75
Now,
U - n(A U B) = 96-75 = 21

Therefore 21 people didn't ate.

Answer 4

The answer is a result of a basic set theory knowledge. Let A and B be the Vegetarian and Non vegetarian population respectively.
Now, we have the following data with us:
n(A) = 60
n(B) = 30
U = 96
n(A Union B) = n(A) + n(B) - n(A inter section B) = 75
Now,
U - n(A U B) = 96-75 = 21

Therefore 21 people didn't ate.

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