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In a town of 20,000 families it was found that 40% familienewspaper A, 20% families buy newspaper B and 10% familiesnewspaper C, 5% families buy A and B, 3% buy B and C and 4%band C. If 2% families buy all the three newspapers, then the numberfamilies which buy A only is: |
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Answer» Answer: Step-by-step explanation: Number of families=20000 N(A)= 40×20000/100 =8000 n(B)= 20×20000/100 =4000 n(C)= 10×20000/100 =2000 n(A∩B)= 5×20000/100 =1000 n(A∩C)= 4×20000/100 =800 n(B∩C)= 3×20000/100 =600 n(A∩B∩C)= 2×20000/100 =400 no. of people buy only A, only B & only C=n(A)-[n(A^B)+n(A^C)-n(A^B^C)] =8000-(1000+800-400) =8000-1400 =6600 Hope this will be helpful to you and please mark my answer as the brainleast answer.plzplz plz plz......... |
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