If n(A) = 70, n(B) = 62, n(C) = 58, n(A ∩ B) = 30, n(B ∩ C) = 24, n(C ∩ A) = 26, n(A ∩ B ∩ C) = 12, and n(U) = 130, how many elements are in none of the sets?
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Answer and explanation
Correct answer: 8
Use the inclusion–exclusion formula for three sets: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C). Substitution gives 70 + 62 + 58 − 30 − 24 − 26 + 12 = 122. The elements in none of the sets are outside the union, so their number is n(U) − n(A ∪ B ∪ C) = 130 − 122 = 8. Option A is correct.
Frequently asked questions
What is the correct answer to this question?
8
Why is this the correct answer?
Use the inclusion–exclusion formula for three sets: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(C ∩ A) + n(A ∩ B ∩ C). Substitution gives 70 + 62 + 58 − 30 − 24 − 26 + 12 = 122. The elements in none of the sets are outside the union, so their number is n(U) − n(A ∪ B ∪ C) = 130 − 122 = 8. Option A is correct.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Venn Diagrams.
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