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
Answer: Option B
:
B
n(A) = 40% of 10,000 = 4,000
n(B) = 20% of 10,000 = 2,000
n(C) = 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 the number of families which buy only A= n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B∩ C)]
=4000 - [500 + 400 - 200] = 4000 - 700 = 3300
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:
B
n(A) = 40% of 10,000 = 4,000
n(B) = 20% of 10,000 = 2,000
n(C) = 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 the number of families which buy only A= n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B∩ C)]
=4000 - [500 + 400 - 200] = 4000 - 700 = 3300
Was this answer helpful ?
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