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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