Question

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 C and A.If 2% families buy all the three newspapers,find the number of families which buy atleast one of the newspapers.​

Answer 1

Answer:

Hint: In this question, we first need to draw the venn diagram using the given conditions so that we get a clear picture of what we have to find. Then subtracting the number of families reading A & C , A & B and then adding the families reading all three gives the value of A only. In the same way subtract families reading B & C, B & A and then add those reading three to get the value of B only. Now, subtracting the whole things inside the venn diagram from 10000 gives the families reading none.

Complete step-by-step answer:

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

Now, given that 40% of them buy newspaper A

⇒40100×10000

⇒4000

Thus, 4000 families read A

Now, given that 20% of them buy newspaper B

⇒20100×10000

⇒2000

Thus, 2000 families buy B

Now, given that 10% of them buy newspaper C

⇒10100×10000

⇒1000

Thus, 1000 families buy C

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

⇒5100×10000⇒500

Thus, 500 families buy both A & B.

Now, given that 3% of them buy newspaper B & C

⇒3100×10000⇒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

Thus, 200 families buy all A & B & C

Now, let us draw the venn diagram

(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

Now, on further simplification we get,

⇒3300

Thus, 3300 families buy only A

(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

⇒10000−(A+B+C−(A∩B)−(B∩C)−(C∩A)+(A∩B∩C))

Now, on substituting the respective values we get,

⇒10000−(4000+2000+1000−500−300−400+200)

Now, on further simplification we get,

⇒10000−6000

Now, this can be further written as

⇒4000

Thus, 4000 families buy none of the newspaper

Hence, the correct option is (b).