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 . How many families live in the town ?

Answer

To solve the problem, we will use the principle of inclusion-exclusion and the information given about the families in the town.

Let:
- \( x \) = total number of families in the town.

From the problem, we know:
- 25% of families own a phone: \( P = 0.25x \)
- 15% of families own a car: \( C = 0.15x \)
- 65% of families own neither a phone nor a car: \( N = 0.65x \)
- 2000 families own both a car and a phone: \( B = 2000 \)

Since 65% of families own neither a phone nor a car, the percentage of families that own either a phone or a car (or both) is:
\[ P + C - B + N = x \]
Substituting the known values:
\[ (0.25x + 0.15x - 2000 + 0.65x) = x \]

Now, we can simplify this equation:
\[ (0.25x + 0.15x + 0.65x - 2000) = x \]
\[ (1.05x - 2000) = x \]
Now, we will isolate \( x \):
\[ 1.05x - x = 2000 \]
\[ 0.05x = 2000 \]
Now, divide both sides by 0.05 to find \( x \):
\[ x = \frac{2000}{0.05} \]
\[ x = 40000 \]

Thus, the total number of families living in the town is \( 40000 \).