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In a town, it was found that 40% families buy newspaper A, 20% families buy newspaper B, 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C, 4% buy A and C and 40% buy none of A, B and C. If 200 families buy all the three newspapers, Find.
a) The number of families which buy newspaper A only.
b) The number of families which buy at least one of the three newspapers​

Sagot :

Semsee45idn

Answer:

a) 3,300

b) 6,000

Step-by-step explanation:

We are given the following information:

  • 40% of families buy newspaper A.
  • 20% of families buy newspaper B.
  • 10% of families buy newspaper C.
  • 5% of families buy both A and B.
  • 3% of families buy both B and C.
  • 4% of families buy both A and C.
  • 40% of families buy none of A, B, and C.
  • 200 families buy all three newspapers.

Let's denote the following sets:

  • A: families that buy newspaper A.
  • B: families that buy newspaper B.
  • C: families that buy newspaper C.

Let n represent the total number of families in the town. Therefore:

  • A = 0.4n
  • B = 0.2n
  • C = 0.1n
  • (A ∩ B) = 0.05n
  • (B ∩ C) = 0.03n
  • (A ∩ C) = 0.04n
  • (A' ∩ B' ∩ C') = 0.4n
  • (A ∩ B ∩ C) = 200

Since 40% of the families buy none of the newspapers, 60% of the families buy at least one newspaper. Therefore, 0.6n families buy at least one newspaper:

(A ∪ B ∪ C) = 0.6n

We can also express the number of families that buy at least one newspaper using the principle of inclusion-exclusion for three sets:

(A ∪ B ∪ C) = A + B + C - (A ∩ B) - (B ∩ C) - (A ∩ C) + (A ∩ B ∩ C)

Substitute the expressions:

0.6n = 0.4n + 0.2n + 0.1n - 0.05n - 0.03n - 0.04n + (A ∩ B ∩ C)

Solve for (A ∩ B ∩ C):

0.6n = 0.58n + (A ∩ B ∩ C)

(A ∩ B ∩ C) = 0.6n - 0.58n

(A ∩ B ∩ C) = 0.02n

Since 200 families buy all three newspapers, then (A ∩ B ∩ C) = 200. Therefore:

0.02n = 200

n = 200 / 0.02

n = 10000

So, the total number of families (n) is 10,000.

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Part (a)

To find the number of families that buy newspaper A only, we can use the formula:

A only = A - (A ∩ B) - (A ∩ C) + (A ∩ B ∩ C)

Therefore:

A only = 0.4n - 0.05n - 0.04n + 200

A only = 0.31n + 200

Substitute n = 10000:

A only = 0.31(10000) + 200

A only = 3100 + 200

A only = 3300

So, the number of families which buy newspaper A only is 3,300.

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Part (b)

We have already calculated that 0.6n families buy at least one newspaper:

(A ∪ B ∪ C) = 0.6n

Substitute n = 10000:

(A ∪ B ∪ C) = 0.6(10000)

(A ∪ B ∪ C) = 6000

Therefore, the number of families which buy at least one of the three newspapers​ is 6,000.

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