For three sets, n(A) = 38, n(A ∩ B) = 16, n(A ∩ C) = 13, and n(A ∩ B ∩ C) = 5. How many elements belong only to A?
Answer and explanation
Correct answer: 14
The total n(A) includes three kinds of elements: those only in A, those in A ∩ B but not C, and those in A ∩ C but not B; the triple intersection is included in both pairwise intersections. Thus the correct inclusion–exclusion expression is only A = n(A) − n(A ∩ B) − n(A ∩ C) + n(A ∩ B ∩ C). Hence only A = 38 − 16 − 13 + 5 = 14.
Frequently asked questions
What is the correct answer to this question?
14
Why is this the correct answer?
The total n(A) includes three kinds of elements: those only in A, those in A ∩ B but not C, and those in A ∩ C but not B; the triple intersection is included in both pairwise intersections. Thus the correct inclusion–exclusion expression is only A = n(A) − n(A ∩ B) − n(A ∩ C) + n(A ∩ B ∩ C). Hence only A = 38 − 16 − 13 + 5 = 14.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Venn Diagrams.
