Question

In a certain town$25\%$ of families own a phone and$15\%$ own a car,$65\%$ of families own neither a phone nor a car.$2000$ families own both a car and a phone. Consider the following statements in this regard

1.$10\%$ of families own both a car and a phone

2. $35\%$ of families own either car or a phone

3. $40,000$families live in the town?

Which of the above statements are correct?

• A. $1$ and $2$
• B. $1$ and $3$
• C. $2$ and $3$
• D. $1,2$and $3$

Answer 1

The correct option is C

$2$ and $3$

Explanation for the correct option:

Finding the correct statements:

Let $\mathrm{A}$ be the set of people who own phones and $\mathrm{B}$ be who own a car, then according to the question,
$$\begin{gathered} \mathrm{P}\left(\mathrm{A}\right)=\frac{25}{100}, \\ \mathrm{P}\left(\mathrm{B}\right)=\frac{15}{100} \end{gathered}$$

Also given,

$$\begin{gathered} \mathrm{P}(\mathrm{A}\text{'}\cap \mathrm{B}\text{'})=\frac{65}{100} \\ \mathrm{n}(\mathrm{A}\cap \mathrm{B})=2000 \end{gathered}$$

$$\begin{gathered} \mathrm{P}(\mathrm{A}\cup \mathrm{B})\text{'}=\frac{65}{100} \\ \Rightarrow \mathrm{P}(\mathrm{A}\cup \mathrm{B})=\frac{35}{100}\left[\because P\left(A\right)+P\left(A^{\text{'}}\right)=1\right] \end{gathered}$$

Now,

$\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})$

Substituting the values in the above formula we get,

$$\begin{gathered} 35=25+15–\mathrm{n}(\mathrm{P}\cap \mathrm{C}) \\ \mathrm{n}(\mathrm{A}\cap \mathrm{B})=40–35 \\ =5 \\ \mathrm{n}(\mathrm{A}\cap \mathrm{B})=5\% \end{gathered}$$

But, $\mathrm{n}(\mathrm{A}\cap \mathrm{B})=2000$

Total number of families

$$\begin{gathered} =\frac{[2000\times 100]}{5} \\ =40000 \end{gathered}$$

As, $\mathrm{n}(\mathrm{A}\cup \mathrm{B})=35\%$

Total number of families$=40000$

And $\mathrm{n}(\mathrm{A}\cap \mathrm{B})=5\%$

Hence, the correct option is C.