Question

8. In a town of 10,000 families 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 and 4% buy A and
C. If 2% families buy all the three
newspapers. Find:

(a) The number of families
which buy newspaper A only.

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

Answer 1

Step-by-step explanation:

Given data:

the total number of families in the town are =10,000

A. 20% families buy newspaper

B. 10% families buy newspaper

Answer:

⇒40/100×10000

=⇒4000

Thus, 4000 families read A

Now, given that 20% of them buy newspaper B

⇒20/100×10000

⇒2000

Thus, 2000 families buy B

Now, given that 10% of them buy newspaper C

⇒10/100×10000

⇒1000

Thus, 1000 families buy C

Now, given that 5% of them buy newspaper A & B

⇒3/100×10000

⇒300

Thus, 300 families buy both B & C.

Now, given that 4% of them buy newspaper A & C

⇒4100×10000

⇒400

Thus, 400 families buy both A & C.

Now, given that 2% of them buy newspaper A & B & C

⇒2100×10000

⇒200

(1) Now, let us find the number of families that buy only A

(1) Now, let us find the number of families that buy only A⇒A−(A∩B)−(A∩C)+(A∩B∩C)

(1) Now, let us find the number of families that buy only A⇒A−(A∩B)−(A∩C)+(A∩B∩C)Now, on substituting the respective values we get,

(1) Now, let us find the number of families that buy only A⇒A−(A∩B)−(A∩C)+(A∩B∩C)Now, on substituting the respective values we get,⇒4000−500−400+200

Thus, 3300 families buy only A

(2) Now, let us find the number of families that buy only B

(2) Now, let us find the number of families that buy only B⇒B−(B∩C)−(B∩A)+(A∩B∩C)

)Now, on substituting the respective values we get,

⇒2000−300−500+200 Now, on further simplification we get,⇒1400

Thus, 1400 families buy only B

(3) Let us now find the families that buy none of A, B, C

(3) Let us now find the families that buy none of A, B, C⇒10000−(A+B+C−(A∩B)−(B∩C)−(C∩A)+(A∩B∩C))

Now, on substituting the respective values we get,

Now, on substituting the respective values we get,⇒10000−(4000+2000+1000−500−300−400+200)

Now, on substituting the respective values we get,⇒10000−(4000+2000+1000−500−300−400+200)Now, on further simplification we get,

⇒4000

⇒4000

⇒4000 Thus, 4000 families buy none of the newspaper

Hence, the correct option is (b)

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