Question

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

• A. 3100
• B. 3300
• C. 2900
• D. 1400

Answer 1

Answer: (B)

3300

Answer 2

n(1) = 40% of 10,000 = 4,000

n(2) = 20% of 10,000 = 2,000

n(3) = 10% of 10,000 = 1,000

n (A ∩ B) = 5% of 10,000 = 500, n (B ∩ C) = 3% of 10,000

= 300

n(C ∩ A) = 4% of 10,000 = 400, n(A ∩ B ∩ C) = 2% of 10,000 = 200

We want to find n(A ∩ B^c ∩ C^c) = n[A ∩ (B ∪ C)^c]

= n(1) – n[A ∩ (B ∪ C)] = n(1) – n[(A ∩ B) ∪ (A ∩ C)]

= n(1) – [n(A ∩ B) + n(A ∩ C) – n(A ∩ B ∩ C)]

= 4000 – [500 + 400 – 200] = 4000 – 700 = 3300.