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,
$n(P)=\frac{25x}{100}=\frac{x}{4}\text{, }$
$n(C)=\frac{15x}{100}=\frac{3x}{20}$
and $n(\bar{P}\cap \bar{C})=\frac{65x}{100}=\frac{13x}{20}$
Also, $n(P\cap C)=2000$
Now, $n(\bar{P}\cap \bar{C})=\frac{13x}{20}$
$\Rightarrow n(\overline{P\cup C})=\frac{13x}{20}$
$\Rightarrow n(U)-n(P\cup C)=\frac{13x}{20}$
$\Rightarrow x-\{n(P)+n(C)-n(P\cap C)\}=\frac{13x}{20}$
$\Rightarrow x-\left(\frac{x}{4}+\frac{3x}{20}-2000\right)=\frac{13x}{20}$
$\Rightarrow \frac{x}{20}=2000$
$\Rightarrow x=40000$
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, $n(P\cup C)=n(P)+n(C)-n(P\cap C)$
$\Rightarrow n(P\cup C)=\frac{x}{4}+\frac{3x}{20}-2000$
$\Rightarrow n(P\cup C)=10000+6000-2000=14000$
$\Rightarrow n(P\cup C)=35\%\text{ of }40000$
Thus, Statement II is correct.