If n(A) = 94, n(A ∩ B) = 40, n(A ∩ C) = 37, and n(A ∩ B ∩ C) = 16, how many elements belong only to A?
Answer and explanation
Correct answer: 33
The elements in A that also lie in B or C are counted by n(A ∩ (B ∪ C)). Using the two-set union rule inside A, n(A ∩ (B ∪ C)) = n(A ∩ B) + n(A ∩ C) − n(A ∩ B ∩ C) = 40 + 37 − 16 = 61. The triple intersection is subtracted once because it appears in both pairwise intersections. Hence, elements only in A = n(A) − 61 = 94 − 61 = 33. Therefore, option C is correct.
Frequently asked questions
What is the correct answer to this question?
33
Why is this the correct answer?
The elements in A that also lie in B or C are counted by n(A ∩ (B ∪ C)). Using the two-set union rule inside A, n(A ∩ (B ∪ C)) = n(A ∩ B) + n(A ∩ C) − n(A ∩ B ∩ C) = 40 + 37 − 16 = 61. The triple intersection is subtracted once because it appears in both pairwise intersections. Hence, elements only in A = n(A) − 61 = 94 − 61 = 33. Therefore, option C is correct.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Operations on Sets (Union, Intersection, Difference).
