Q. 274.4( 11 Votes )

In a town of 10,0

Answer :

Given:


Total number of families = 10,000


Number of families buy newspaper A = 40%


Number of families buy newspaper B = 20%


Number of families buy newspaper C = 10%


Number of families buy newspaper A and B = 5%


Number of families buy newspaper B and C = 3%


Number of families buy newspaper A and C = 4%


Number of families buy all three newspapers = 2%


Let U be the total number of families, A, B and C be the number of families buy newspaper A, B and C respectively


(i) To find: number of families which buy newspaper A only


n(A) = 40%, n(B) = 20%, n(C) = 10%, n(A B) = 5%


n(B C) = 3%, n(A C) = 4%, n(A B C) = 2%


Percentage of families which buy newspaper A only


= n(A) – n(A B) – n(A C) + n(A B C)


= 40 – 5 – 4 + 2


= 33%


Number of families which buy newspaper A only



= 3300


Hence, there are 3300 families which buy newspaper A only


(b) To find: number of families which buy none of A, B and C


n(A) = 40%, n(B) = 20%, n(C) = 10%, n(A B) = 5%


n(B C) = 3%, n(A C) = 4%, n(A B C) = 2%


Percentage of families which buy either of A, B and C


= n(A B C)


= n(A) + n(B) + n(C) – n(A B) – n(B C) – n(A C) + n(A B C)


= 40 + 20 + 10 – 5 – 3 – 4 + 2


= 60%


Percentage of families which buy none of A, B and C


= Total percentage – Number of students who play either


= 100% – 60%


= 40%


Number of families which buy none of A, B and C



= 4000


Hence, there are 4000 families which buy none of A, B and C


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