Question

In a town of 10,000 families it was found that 40% families buy

newspaper A, 20% families buy newspaper B and 10% families

by 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 no of families which buy(1) A only (2) B only (3) none of

A, B and C (4) exactly two newspapers (5) exactly one

newspaper (6) A and C but not B (7) at least one of A,B, C.

What is the importance of reading newspaper?
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Answer 1

Total Number of Families (N) = 10,000

 a) Number of families which buy Newspaper A = 40%*N +5%N + 4%N +2%N

                       = 40/100*10000 + 5/100*10000 + 4*100*10000 + 2/100*10000

                      = 10000/100(40+5+4+2) = 100(51) = 5100

b) Number of Families which doesn't buy Any NewsPaper = (100 -    (40+20+10+5+3+4+2))%N

= (100 - 84)%N

  = 16%N

  = 16/100*10000 = 16*100= 1600

Hope This helps You!!

Answer 2

Answer:

Step-by-step explanation:

Total Number of Families (N) = 10,000

 a) Number of families which buy Newspaper A = 40%*N +5%N + 4%N +2%N

                       = 40/100*10000 + 5/100*10000 + 4*100*10000 + 2/100*10000

                      = 10000/100(40+5+4+2) = 100(51) = 5100

b) Number of Families which doesn't buy Any NewsPaper = (100 -    (40+20+10+5+3+4+2))%N

= (100 - 84)%N

  = 16%N

  = 16/100*10000 = 16*100= 1600

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