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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

0 5 Views | Posted 4 months ago
Asked by Shiksha User

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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    T o t a l     n u m b e r     o f     f a m i l i e s = 1 0 0 0 0     ⇒ n ( U ) = 1 0 0 0 0 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 = 4 0 %     ⇒ n ( A ) = 4 0 % 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 = 2 0 %     ⇒ n ( B ) = 2 0 % 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 = 1 0 %     ⇒ n ( C ) = 1 0 % 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 %     ⇒ n ( A ∩ 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 %     ⇒ n ( B ∩ 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 %     ⇒ n ( A ∩ 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 %     ⇒ n ( A ∩ B ∩ 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 ∩ B ) − n ( A ∩ C ) + n ( A ∩ B ∩ C ) ⇒             = 4 0 1 0 0 − 5 1 0 0 − 4 1 0 0 + 2 1 0 0 = 3 3 1 0 0 ⇒             = 1 0 0 0 0 × 3 3 1 0 0 = 3 3 0 0     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 ∪ B ∪ C ) ⇒     n ( U ) − [ n ( A ) + n ( B ) + n ( C ) − n ( A ∩ B ) − n ( B ∩ C ) − n ( A ∩ C ) + n ( A ∩ B ∩ C ) ] ⇒     [ 1 0 0 − ( 4 0 + 2 0 + 1 0 − 5 − 3 − 4 + 2 ) % ] ⇒     ( 1 0 0 − 6 0 ) % = 4 0 % ∴     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     1 0 0 0 0     f a m i l i e s     a r e             = 1 0 0 0 0 × 4 0 1 0 0 = 4 0 0 0     f a m i l i e s .

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