In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C.5% families buy A and B, 3% buy B and C, 4% buy A and C. If 2% buy all the three newspapers, the number of families which buy none of A, B, C is

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5000

2000

4000

3000

answer is C.

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Detailed Solution

Given that

$\begin{array}{l} n (A) = 40, n (B) = 20, n (C) = 10, n (A \cap B) = 5 \\ n (B \cap C) = 3, n (A \cap C) = 4, n (A \cap B), n (A \cap B \cap C) = 2 \\ \Rightarrow n (A \cup B \cup C) = 60, n (\bar{A \cup B \cup C}) = 40 \end{array}$

$∴$ The number of families which buy none of A, B, C is $\frac{40}{100} \times 10000 = 4000$

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