In a town of 10,000 families it was found

Solution:

According to the question,

Total number of families = 10,000

Number of families buying newspaper A = n(A) = 40%

Number of families buying newspaper B = n(B) = 20%

Number of families buying newspaper C = n(C) = 10%

Number of families buying newspaper A and B = n(A ∩ B) = 5%

Number of families buying newspaper B and C = n(B ∩ C) = 3%

Number of families buying newspaper A and C = n(A ∩ C) = 4%

Number of families buying all three newspapers = n(A ∩ B ∩ C) = 2%

Let the total number of families = U

Let the number of families buying newspaper A = A

Let the number of families buying newspaper B = B

Let the number of families buying newspaper C = C

(a) Number of families which buy newspaper A only

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

= ((33/100)×10000)

= 3300

Hence, there are 3300 families which buy newspaper A only

(b) Number of families which buy none of A, B and C

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

= ((40/100)×10000)

= 4000

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