Q. 145.0( 2 Votes )

# 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.

Answer :

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

To find:

(i) Number of families that buy newspaper A only

Consider the Venn Diagram below:

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

Rate this question :

Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities:

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × (B – C) = (A × B) – (A × C)

(iii) (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)

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If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find.

(i) n(A × B)

(ii) n(B × A)

(iii) n(A × B) ∩ (B × A)

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If A × B ⊆ C × D and A × B ≠ ϕ, prove that A ⊆ C and B ⊆ D.

RS Aggarwal - Mathematics(i) If A ⊆ B, prove that A × C ⊆ B × C for any set C.

(ii) If A ⊆ B and C ⊆ D then prove that A × C ⊆ B × D.

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Using properties of sets prove the statements given

For all sets A and B, (A ∪ B) – B = A – B

Mathematics - ExemplarIf A and B are two sets such that n(A) = 23, n(b) = 37 and n(A – B) = 8 then find n(A ∪ B).

Hint n(A) = n(A – B) + n(A ∩ B) n(A ∩ B) = (23 – 8) = 15.

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