Let U = {x ∈ ℕ : x ≤ 120}, A = {x ∈ U : 8 divides x}, B = {x ∈ U : 15 divides x}, and C = {x ∈ U : 20 divides x}. What is n((A ∪ B ∪ C)′)?
Answer and explanation
Correct answer: 96
Count the union by inclusion–exclusion. There are 120/8 = 15 multiples of 8, 120/15 = 8 multiples of 15, and 120/20 = 6 multiples of 20. The pairwise intersections contain 2 multiples of lcm(8,15)=120, 1 multiple of lcm(8,20)=40, and 2 multiples of lcm(15,20)=60. The triple intersection contains only 120. Hence n(A ∪ B ∪ C) = 15 + 8 + 6 - 1 - 3 - 2 + 1 = 24. Therefore, its complement has 120 - 24 = 96 elements.
Frequently asked questions
What is the correct answer to this question?
96
Why is this the correct answer?
Count the union by inclusion–exclusion. There are 120/8 = 15 multiples of 8, 120/15 = 8 multiples of 15, and 120/20 = 6 multiples of 20. The pairwise intersections contain 2 multiples of lcm(8,15)=120, 1 multiple of lcm(8,20)=40, and 2 multiples of lcm(15,20)=60. The triple intersection contains only 120. Hence n(A ∪ B ∪ C) = 15 + 8 + 6 - 1 - 3 - 2 + 1 = 24. Therefore, its complement has 120 - 24 = 96 elements.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Sets. Topic: Complement of a Set and Its Properties.
