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Question

In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 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, then number of families which buy A only is

A
3100
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B
3300
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C
2900
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D
1400
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Solution

The correct option is B 3300
n(A)=40% of 10,000=4,000
n(B)=20% of 10,000=2,000
n(C)=10% of 10,000=1,000
n(AB)=5% of 10,000=500
n(BC)=3% of 10,000=300
n(CA)=4% of 10,000=400
n(ABC)=2% of 10,000=200
We want to find n(ABcCc)
n(ABcCc)=n[A(BC)c].
=n(A)n[A(BC)]=n(A)n[(AB)(AC)]
=n(A)[n(AB)+n(AC)n(ABC)]
=4000[500+400200]=4000700=3300.



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