Question

A restaurant serves 3 dishes to choose from - A, B and C. One day they had 275 visitors. Of these, 173 had dish A, 169 had B and 155 had C. 92 visitors had both A and B, 81 visitors had both A and C, and 107 visitors had both B and C.

How many visitors had A, B and C? And how many visitors hade no dish at all?

Hint: Venn diagram?

Answer 1

n(A) = 173

n(B) = 169

n(C) = 155

n(A^B) = 92

n(B^C) = 107

n(A^C) = 81

n(A U B U C) = 275

n(A U B U C) = n(A) + n(B) + n(C) - n(A^B) - n(B^C) - n(A^C) + n(A^B^C)

275 = 173 + 169 + 155 - 92 - 81 - 107 + n(A^B^C)

n(A^B^C) = 58

How many visitors had A, B and C?

visitors had A, B and C = 58

And how many visitors had no dish at all?

Total - visitors had A, B and C = 275 - 58 = 217