Question

In a town of 10,000 families, it was found that 40% families buy newspaper A, 20% families buy newspaper B and 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 newspapers, then

• A. 3,300 families buy A only
• B. 1,400 families buy B only.
• C. 4000 families buy none of A, B and C
• D. All are correct

Answer 1

The Correct Option is D

Let P = set of families buying A, Q = set of families buying B and R = set of families buying C. $\therefore n\left(P\right) 40\%$ of $10,000 = 4,000$, similarly $n\left(Q\right) = 2,000 , n\left(R\right) = 1,000$ $n\left(P \cap Q\right) = 500, n \left(Q \cap R\right) = 300$ $n \left(P \cap R\right) = 400$ and n $\left(P \cap Q\cap R\right) = 200$ $\left(i\right)$ Number of families buying only $A = n\left(P \cap Q' \cap R'\right)$ $= n \left(P \cap \left(Q \cup R\right)'\right) = n\left(P\right) -n\left(P \cap \left(Q \cup R\right)\right)$ $= n\left(P\right) -\left[n\left(P\cap Q\right) + n\left(P\cap R\right) - n\left(\left(P\cap Q\right)I\left(P\cap R\right)\right)\right]$ $= n\left(P\right) - n\left(P\cap Q\right) - n\left(P\cap R\right) + n\left(P\cap Q\cap R\right)$. $= 4,000 - 500 - 400 + 200 = 3,300$. $\left(ii\right)$ Number of families buying only B $= n\left(Q\right) - n\left(P\cap Q\right) - n\left(Q\cap R\right) + n\left(P\cap Q\cap R\right)$ $\left[see \left(i\right)\right]$ $= 2,000 - 500 - 300 + 200 = 1,400$. $\left(iii\right)$ Number of families buying none of A, B and $C = n\left(P'\cap Q'\cap R'\right) = n\left(P'\cap \left(Q\cup R\right) '\right)$ $= n\left(P\cup \left(Q\cup R\right)\right) ' =10000 - n\left(P\cup Q\cup R\right)$ $= 10,000- \left[n\left(P\right) +n\left(Q\right) +n\left(R\right)- n\left(P \cap Q\right)-n\left(Q \cap R\right) - n\left(P \cap R\right) + \left(P\cap Q \cap R\right)\right]$ $= 10,000 - \left[4,000 + 2,000 + 1,000 - 500 - 300 - 400 + 200\right]$ $= 10, 000 - 6,000 = 4,000.$ Note : For sets A, B, we have $\left(A\cap B\right)\cup \left(A\cap B'\right) = A\cap \left(B\cup B'\right) = A\cap U = A$ and $\left(A\cap B\right)\cap \left(A\cap B'\right) = A\cap \left(B\cap B'\right) = A\cap f = f$ $\therefore n\left(A\right) = n\left(A\cap B\right) + n\left(A\cap B'\right) or n\left(A\cap B'\right)$ $= n\left(A\right) - n\left(A\cap B\right)$ Replacing A by P and B by $Q \cup R$, we have $n \left(P\cap\left(Q\cup R\right) '\right) = n\left(P\right) - n\left(P\cap \left(Q\cup R\right)\right)$ etc. Hence all options are correct.