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

Given

Total number of families = 10000

Percentage of families that buy newspaper A = 40

Percentage of families that buy newspaper B = 20

Percentage of families that buy newspaper C = 10

Percentage of families that buy newspaper A and B = 5

Percentage of families that buy newspaper B and C = 3

Percentage of families that buy newspaper A and C = 4

Percentage of families that buy all three newspapers = 2

Find out

We have to determine the number of families which buy

(i) A only,
(ii) B only,
(iii) none of A, B, and C

### Solution

Number of families that buy newspaper A = n(A) = 40% of 10000 = 4000

Number of families that buy newspaper B = n(B) = 20% of 10000 = 2000

Number of families that buy newspaper C = n(C) = 10% of 10000 = 1000

Number of families that buy newspaper A and B = n(A ∩ B) = 5% of 10000 = 500

Number of families that buy newspaper B and C = n(B ∩ C) = 3% of 10000 = 300

Number of families that buy newspaper A and C = n(A ∩ C) = 4% of 10000 = 400

Number of families that buys all three newspapers = n(A ∩ B ∩ C)=v = 2% of 10000 = 200

We have, n(A ∩ B) = v + t 500 = 200 + t

t = 500 – 200 = 300

n(B ∩ C) = v + s

300 = 200 + s

s = 300 – 200 = 100

n(A ∩ C) = v + u

400 = 200 + u

u = 400 – 200 = 200

p = Number of families that buy newspaper A only

We have, A = p + t + v + u

4000 = p + 300 + 200 + 200 p = 4000 – 700

p = 3300

Therefore, Number of families that buy newspaper A only = 3300

(ii) Number of families that buy newspaper B only q = Number of families that buy newspaper B only B = q + s + v + t 2000 = q + 100 + 200 + 300 q = 2000 – 600 =1400 Therefore, Number of families that buy newspaper B only = 1400

(iii) Number of families that buys none of the newspaper

Number of families that buy none of the newspaper = 10000 – {n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)}

= 10000 – (4000 + 2000 + 1000 – 500 – 300 – 400 + 200)

= 10000 – 6000

= 4000

Therefore, Number of families that buy none of the newspaper = 4000

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