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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% - Mathematics

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प्रश्न

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 newspaper A only.

योग

उत्तर

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 newspaper A only

= n(A) – n(A ∩ B) – n(A ∩ C) + n(A ∩ B ∩ C)

= `40/100 - 5/100 - 4/100 + 2/100`

= `33/100`

= `10000 xx 33/100`

= 3300 families

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अध्याय 1: Sets - Exercise [पृष्ठ १५]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 1 Sets
Exercise | Q 27.(i) | पृष्ठ १५

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