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?
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?
Answers
GPT-4o mini
To find out after how many minutes Mr. Beecher and Mrs. Carter will be the same distance from school, we can set up equations based on the distances they cover over time.
Let \( t \) represent the number of minutes they travel.
**Mr. Beecher's situation:**
- He starts 8 miles away from school.
- He covers \(\frac{1}{6}\) of a mile every minute.
- Therefore, after \( t \) minutes, the distance he has left to travel to school is:
\[
8 - \frac{1}{6}t
\]
**Mrs. Carter's situation:**
- She starts 20 miles away from school.
- She covers \(\frac{1}{2}\) of a mile every minute.
- Therefore, after \( t \) minutes, the distance she has left to travel to school is:
\[
20 - \frac{1}{2}t
\]
We want to find out when they are the same distance away from school, so we set the two distances equal to each other:
\[
8 - \frac{1}{6}t = 20 - \frac{1}{2}t
\]
Now we can solve for \( t \). First, let's eliminate the fractions by finding a common denominator. The least common multiple of 6 and 2 is 6. We multiply the entire equation by 6:
\[
6(8) - 6\left(\frac{1}{6}t\right) = 6(20) - 6\left(\frac{1}{2}t\right)
\]
This simplifies to:
\[
48 - t = 120 - 3t
\]
Next, we'll rearrange the equation to isolate \( t \). Add \( 3t \) to both sides:
\[
48 - t + 3t = 120
\]
Which simplifies to:
\[
48 + 2t = 120
\]
Now subtract 48 from both sides:
\[
2t = 120 - 48
\]
This simplifies to:
\[
2t = 72
\]
Now, divide by 2:
\[
t = 36
\]
Thus, Mr. Beecher and Mrs. Carter will first be the same distance away from school after \( \boxed{36} \) minutes.
Let \( t \) represent the number of minutes they travel.
**Mr. Beecher's situation:**
- He starts 8 miles away from school.
- He covers \(\frac{1}{6}\) of a mile every minute.
- Therefore, after \( t \) minutes, the distance he has left to travel to school is:
\[
8 - \frac{1}{6}t
\]
**Mrs. Carter's situation:**
- She starts 20 miles away from school.
- She covers \(\frac{1}{2}\) of a mile every minute.
- Therefore, after \( t \) minutes, the distance she has left to travel to school is:
\[
20 - \frac{1}{2}t
\]
We want to find out when they are the same distance away from school, so we set the two distances equal to each other:
\[
8 - \frac{1}{6}t = 20 - \frac{1}{2}t
\]
Now we can solve for \( t \). First, let's eliminate the fractions by finding a common denominator. The least common multiple of 6 and 2 is 6. We multiply the entire equation by 6:
\[
6(8) - 6\left(\frac{1}{6}t\right) = 6(20) - 6\left(\frac{1}{2}t\right)
\]
This simplifies to:
\[
48 - t = 120 - 3t
\]
Next, we'll rearrange the equation to isolate \( t \). Add \( 3t \) to both sides:
\[
48 - t + 3t = 120
\]
Which simplifies to:
\[
48 + 2t = 120
\]
Now subtract 48 from both sides:
\[
2t = 120 - 48
\]
This simplifies to:
\[
2t = 72
\]
Now, divide by 2:
\[
t = 36
\]
Thus, Mr. Beecher and Mrs. Carter will first be the same distance away from school after \( \boxed{36} \) minutes.