Suppose n(A)=70, n(B)=65, n(C)=60, n(A ∪ B ∪ C)=128, n(A ∩ B)=27, n(B ∩ C)=24, and n(C ∩ A)=22. Which conclusion is mathematically correct?
Answer and explanation
Correct answer: Such sets cannot exist
Let x=n(A ∩ B ∩ C). Inclusion–exclusion gives 128=70+65+60−27−24−22+x, so x=52. However, the triple intersection must be contained in every pairwise intersection, so it cannot exceed n(C ∩ A)=22, n(B ∩ C)=24, or n(A ∩ B)=27. Since 52 is impossible, the supplied data are inconsistent and no such sets exist.
Frequently asked questions
What is the correct answer to this question?
Such sets cannot exist
Why is this the correct answer?
Let x=n(A ∩ B ∩ C). Inclusion–exclusion gives 128=70+65+60−27−24−22+x, so x=52. However, the triple intersection must be contained in every pairwise intersection, so it cannot exceed n(C ∩ A)=22, n(B ∩ C)=24, or n(A ∩ B)=27. Since 52 is impossible, the supplied data are inconsistent and no such sets exist.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Venn Diagrams.
