In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B, 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. Find

(i) The number of families which buy newspaper A only.

(ii) The number of families which buy none of A, B and C

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8 months ago

$\begin{array}{l} T o t a l n u m b e r o f f a m i l i e s = 10000 \Rightarrow n (U) = 10000 \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r A = 40 % \Rightarrow n (A) = 40 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r B = 20 % \Rightarrow n (B) = 20 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r C = 10 % \Rightarrow n (C) = 10 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r A a n d B = 5 % \Rightarrow n (A \cap B) = 5 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r B a n d C = 3 % \Rightarrow n (B \cap C) = 3 % \\ N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r A a n d C = 4 % \Rightarrow n (A \cap C) = 4 % \\ N u m b e r o f f a m i l i e s w h o b u y a l l t h e t h r e e N e w s p a p e r s = 2 % \Rightarrow n (A \cap B \cap C) = 2 % \\ (i) N u m b e r o f f a m i l i e s w h o b u y N e w s p a p e r A o n l y \\ = n (A) - n (A \cap B) - n (A \cap C) + n (A \cap B \cap C) \\ \Rightarrow = \frac{40}{100} - \frac{5}{100} - \frac{4}{100} + \frac{2}{100} = \frac{33}{100} \\ \Rightarrow = 10000 * \frac{33}{100} = 3300 f a m i l i e s \\ (i i) N u m b e r o f f a m i l i e s w h o b u y n o n e o f A, B a n d C \\ N e w s p a p e r i n p e r c e n t (%) = n (U) - n (A \cup B \cup C) \\ \Rightarrow n (U) - [n (A) + n (B) + n (C) - n (A \cap B) - n (B \cap C) - n (A \cap C) + n (A \cap B \cap C)] \\ \Rightarrow [100 - (40 + 20 + 10 - 5 - 3 - 4 + 2) %] \\ \Rightarrow (100 - 60) % = 40 % \\ ∴ N u m b e r o f f a m i l i e s w h o b u y n o n e o f A, B a n d C n e w s p a p e r o u t o f 10000 f a m i l i e s a r e \\ = 10000 * \frac{40}{100} = 4000 f a m i l i e s . \end{array}$

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