In a town of 10,000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy A and B, 3% families buy B and C and 4% buy A and C. If 2% families buy all three newspapers, find the number of families which buy (i) A only (ii) B only (iii) none A, B or C.

Asked by mesamit | 30th Jul, 2010, 10:26: AM

Expert Answer:

n(A)= no of people reading newspaper A=40% of 10,000=4000n(B)= no of people reading newspaper B=20% of 10,000=2000n(C)= no of people reading newspaper C=10% of 10,000=1000n(AB)= no of people reading newspaper A and B=5% of 10,000=500n(BC)= no of people reading newspaper B and C=3% of 10,000=300n(AC)= no of people reading newspaper A and C=4% of 10,000=400n(ABC)= no of people reading newspaper A,B and C=2% of 10,000=200Total number of families reading newspapers=n(ABC)=n(A)+n(B)+n(C)-n(AB)-n(CB)-n(AC)+n(ABC)=4000+2000+1000-500-300-400+200=6000(i) Families reading only paper An(BC)=n(B)+n(C)--n(CB)=2000+1000-300=2700Number of families reading A= total no of families reading paper -no of families reading B and C=6000-2700=3300(ii) Families reading only paper Bn(AC)=n(A)+n(C)--n(CA)=4000+1000-400=4600Number of families reading only B= total no of families reading paper -no of families reading A and C=6000-4600=1400(iii) Families reading none of A,B and C=10,000-6000=4000">

Answered by  | 30th Jul, 2010, 09:59: PM

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