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(a) The number of families which buy newspaper A only.(b) The number of families which buy exactly one of A, B and C

Answer 1

Step-by-step explanation:

there are 10,000 families in a town

(a)

40%x10,000

40/100×10,000

=4000

therefore 4000 families bye news paper A

(b)

no of families which bye newspaper A and B are 5% of 10,000

=5/100×10000

=500

no of families which bye newspaper B and C are 3% of 10000

=3/100×10000

=300

no of families which bye newspaper A and C are 4% of 10000

=4/100×10000

=400

no of families which bye newspapers A,B and C are 2% of 10000

=2/100×10000

=200

no of families another than which bye exactly one of newspaper A,B and C are,

200+400+300=900

no of families which bye exactly one of newspaper A,B and C are,

10,000+900

=9,100

therefore no of families which buy exactly one of A,B and C newspapers are 9,100