Question

In a certain town, $25 \%$ families own a phone and $15 \%$ own a car, $65 \%$ families own neither a phone nor a car, 2000 families own both a car and a phone.
Consider the following statements in this regard
I. $10 \%$ families own both a car and a phone.
II. $35 \%$ families own either a car or a phone.
III. 40000 families live in the town.
Which of the following are correct?

• A. I and II are correct
• B. I and III are correct
• C. II and III are correct
• D. All are correct

Answer 1

Answer: (C)

Let there be $x$ families in the town. Let $P$ and $C$ denote the set of families using phone and car, respectively. Then,
$ n(P)=\frac{25 x}{100}=\frac{x}{4} \text {, }$
$ n(C)=\frac{15 x}{100}=\frac{3 x}{20}$
and $ n(\bar{P} \cap \bar{C})=\frac{65 x}{100}=\frac{13 x}{20} $
Also, $ n(P \cap C)=2000$
Now, $ n(\bar{P} \cap \bar{C})=\frac{13 x}{20}$
$ \Rightarrow n(\overline{P \cup C})=\frac{13 x}{20} $
$ \Rightarrow n(U)-n(P \cup C)=\frac{13 x}{20} $
$ \Rightarrow x-\{n(P)+n(C)-n(P \cap C)\}=\frac{13 x}{20}$
$ \Rightarrow x-\left(\frac{x}{4}+\frac{3 x}{20}-2000\right)=\frac{13 x}{20} $
$\Rightarrow \frac{x}{20}=2000 $
$\Rightarrow x=40000$
Thus, Statement III is correct
$10 \%$ of the total families in the town is 4000 and it is given that 2000 families own both a car and a phone. So, Statement I is not correct.
Now, $ n(P \cup C)=n(P)+n(C)-n(P \cap C) $
$\Rightarrow n(P \cup C)=\frac{x}{4}+\frac{3 x}{20}-2000$
$\Rightarrow n(P \cup C)=10000+6000-2000=14000$
$\Rightarrow n(P \cup C)=35 \% \text { of } 40000$
Thus, Statement II is correct.