Exams › JEE Advanced › Maths
In a universal set U with n(U) = 600, two subsets A and B satisfy: n(A) = 200, n(A∩B) = 100, and n(A^c ∩ B^c) = 200. Find n(A^c ∩ B).
- 100
- 150
- 200
- 300
Correct answer: 200
Solution
By De Morgan: n(A^c ∩ B^c) = n(U) - n(A∪B) -> n(A∪B) = 600 - 200 = 400. From inclusion-exclusion: n(B) = n(A∪B) - n(A) + n(A∩B) = 400 - 200 + 100 = 300. Finally, n(A^c ∩ B) = n(B) - n(A∩B) = 300 - 100 = 200.
Related JEE Advanced Maths questions
- In a chemistry class, there are 20 students, while a physics class has 30 students. If 10 students are enrolled in both classes, and the two classes are held at separate times, determine the value of k/8, where k represents the total number of students attending either class.
- Given that A represents the divisors of 15, B contains prime numbers less than 10, and C includes even numbers less than 9, how many elements are there in the intersection of (A ∪ C) and B?
- Consider the set S = {a + b√2: a, b ∈ Z}, and the sets T₁ = {⌊(−1 + √2)ⁿ⌋: n ∈ N} and T₂ = {⌊(1 + √2)ⁿ⌋: n ∈ N}. Which of the following statements is/are correct?
- If set A contains 3 elements and set B contains 6 elements, what is the minimum number of elements in A union B?
- Let A = {1, 2, 3, 4, 5} and B = {2, 3, 6, 7}. Find the number of elements in (A x B) ∩ (B x A).
- In a school, 21 students are in the cricket team, 26 in the hockey team, and 29 in the football team. Among them, 14 play both hockey and cricket, 15 play both hockey and football, and 12 play both football and cricket. Eight students play all three games. If k is the total number of members across the three athletic teams, find the sum of digits of k.
⚔️ Practice JEE Advanced Maths free + battle 1v1 →