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Question
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 the number of families which buy none of A, B and C
Solution
Total number of families = 10000
⇒ n(U) = 10000
Number of families who buy newspaper A = 40%
⇒ n(A) = 40%
Number of families who buy newspaper B = 20%
⇒ n(B) = 20%
Number of families who buy newspaper C = 10%
⇒ n(C) = 10%
Number of families who buy newspapers A and B = 5%
⇒ n(A ∩ B) = 5%
Number of families who buy newspapers B and C = 3%
⇒ n(B ∩ C) = 3%
Number of families who buy newspapers A and C = 4%
⇒ n(A ∩ C) = 4%
Number of families who buy all the three newspapers = 2%
⇒ n(A ∩ B ∩ C) = 2%
Number of families who buy none of A, B and C
Newspaper in percent (%) = 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)]
⇒ [100 – (40 + 20 + 10 – 5 – 3 – 4 + 2)]%
⇒ (100 – 60)% = 40%
∴ Number of families, who buy none of A, B and C newspaper out of 10000 families are
= `10000 xx 40/100`
= 4000 families.
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