Question

Mr. Beecher and Mrs. Carter are teachers at the same school. They leave their houses at the same time in the morning to get to school.
Mr. Beecher lives 8 miles away from school and rides his bicycle to work. Every minute, he gets 1/6


of a mile closer to school.
Mrs. Carter lives 20 miles away from school and drives her car to work. Every minute, she gets 1/2


of a mile closer to school.
After how many minutes will Mr. Beecher and Mrs. Carter first be the same distance away from school?

Answer 1

To find out after how many minutes Mr. Beecher and Mrs. Carter will first be the same distance away from school, we can set up equations based on their distances from the school over time.

Let \( t \) be the number of minutes after they leave their houses.

1. **Distance of Mr. Beecher from the school:**
He starts 8 miles away and gets \( \frac{1}{6} \) miles closer each minute.
\[
\text{Distance from school} = 8 - \frac{1}{6}t
\]

2. **Distance of Mrs. Carter from the school:**
She starts 20 miles away and gets \( \frac{1}{2} \) miles closer each minute.
\[
\text{Distance from school} = 20 - \frac{1}{2}t
\]

We need to find when these two distances are equal:
\[
8 - \frac{1}{6}t = 20 - \frac{1}{2}t
\]

To solve this equation, we'll first eliminate the fractions by multiplying through by 6 (the least common multiple of the denominators 6 and 2):
\[
6(8) - 6\left(\frac{1}{6}t\right) = 6(20) - 6\left(\frac{1}{2}t\right)
\]

This simplifies to:
\[
48 - t = 120 - 3t
\]

Now, rearranging gives us:
\[
48 - t + 3t = 120
\]
\[
48 + 2t = 120
\]
\[
2t = 120 - 48
\]
\[
2t = 72
\]
\[
t = 36
\]

Therefore, Mr. Beecher and Mrs. Carter will first be the same distance away from school after **36 minutes**.