If n(A ∪ B ∪ C) = 155, n(A ∪ B) = 118, n(B ∪ C) = 124, n(C ∪ A) = 121, and n(B) = 64, then what is n((A ∩ C) − B)?
Answer and explanation
Correct answer: 23
Let the seven disjoint Venn regions be the only-A, only-B, only-C, AB-only, BC-only, AC-only, and ABC regions. Since the total union is 155 and A ∪ B is 118, only-C = 155 − 118 = 37. Similarly, only-A = 155 − 124 = 31 and only-B = 155 − 121 = 34. The set B contains only-B plus the three regions involving B, so the sum of B-only, AB-only, BC-only, and ABC is 64. Combining these region equations with the total gives AC-only = n((A ∩ C) − B) = 23.
Frequently asked questions
What is the correct answer to this question?
23
Why is this the correct answer?
Let the seven disjoint Venn regions be the only-A, only-B, only-C, AB-only, BC-only, AC-only, and ABC regions. Since the total union is 155 and A ∪ B is 118, only-C = 155 − 118 = 37. Similarly, only-A = 155 − 124 = 31 and only-B = 155 − 121 = 34. The set B contains only-B plus the three regions involving B, so the sum of B-only, AB-only, BC-only, and ABC is 64. Combining these region equations with the total gives AC-only = n((A ∩ C) − B) = 23.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Venn Diagrams.
