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Q. In a town of 10000 families, it was found that 40% families buy newspaper A,20% families buy newspaper B and 10% families buy newspaper C,5% buy A and B,3% buy B and C and 4% buy A and C. If 2% families buy all of three newspapers, then the number of families which buy A only, is

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Solution:

n(A)=40% of 10000=4000,n(B)=2000
n(C)=1000,n(AB)=500,n(BC)=300
n(CA)=400,n(ABC)=200
n(AˉBˉC)=n{A(BC)}
=n(A)n{A(BC)}
=n(A)n(AB)n(AC)+n(ABC)
=4000500400+200=3300