In a town of 10,000 families it was found that 40 % families buy newspaper A, 20 % buy newspaper B and 10 % buy newspaper C .also 5 % families buy newspaper A and B 3 % buy newspaper B and C and 4 % buy newspaper A can C ,If 2 % families buy all the three newspaper , then number of families which buy newspaper A only is

Options

A. 3100
B. 3300
C. 2900
D. 1400

Answer

B

Explanation

n(A) = 40% of 10000 = 4000
n(B) = 20% of 10000 = 2000
n(C ) = 10% of 10000 = 1000
`n(AnnB)` = 5% of 10000 = 500
`n(BnnC)` = 3% of 10000 = 300
`n(CnnA)` = 4% of 10000 = 400
`n(AnnBnnC)` = 2% of 10000 = 200
We want to find `n(AnnB^(c)nnC^(c))=n[Ann(BuuC)^(c)]`
`=n(A)-n[Ann(BnnC)]=n(A)-n[(AnnB)uu(AnnC)]`
`=n(A)-[n(AnnB)+n(AnnC)-n(AnnBnnC)]`
`=4000-[500+400-200]=4000-700=3300`