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

Answer: (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 $A=n\left(P\cap Q^{\prime }\cap R^{\prime }\right)$
$=n\left(P\cap {\left(Q\cup R\right)}^{\prime }\right)=n\left(P\right)-n\left(P\cap \left(Q\cup R\right)\right)$
$=n\left(P\right)-\left[n\left(P\cap Q\right)+n\left(P\cap R\right)-n\left(\left(P\cap Q\right)I\left(P\cap R\right)\right)\right]$
$=n\left(P\right)-n\left(P\cap Q\right)-n\left(P\cap R\right)+n\left(P\cap Q\cap R\right)$.
$=4,000-500-400+200=3,300$.
$\left(ii\right)$ 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
$C=n\left(P^{\prime }\cap Q^{\prime }\cap R^{\prime }\right)=n\left(P^{\prime }\cap {\left(Q\cup R\right)}^{\prime }\right)$
$=n{\left(P\cup \left(Q\cup R\right)\right)}^{\prime }=10000-n\left(P\cup Q\cup R\right)$
$=10,000-\left[n\left(P\right)+n\left(Q\right)+n\left(R\right)-n\left(P\cap Q\right)-n\left(Q\cap R\right)-n\left(P\cap R\right)+\left(P\cap Q\cap R\right)\right]$
$=10,000-\left[4,000+2,000+1,000-500-300-400+200\right]$
$=10,000-6,000=4,000.$
Note : For sets A, B, we have
$\left(A\cap B\right)\cup \left(A\cap B^{\prime }\right)=A\cap \left(B\cup B^{\prime }\right)=A\cap U=A$
and $\left(A\cap B\right)\cap \left(A\cap B^{\prime }\right)=A\cap \left(B\cap B^{\prime }\right)=A\cap f=f$
$\therefore n\left(A\right)=n\left(A\cap B\right)+n\left(A\cap B^{\prime }\right)orn\left(A\cap B^{\prime }\right)$
$=n\left(A\right)-n\left(A\cap B\right)$
Replacing A by P and B by $Q\cup R$, we have
$n\left(P\cap {\left(Q\cup R\right)}^{\prime }\right)=n\left(P\right)-n\left(P\cap \left(Q\cup R\right)\right)$ etc.
Hence all options are correct.