Roger can run one mile in 21 minutes. Jeff can run one mile in 18 minutes. If Jeff gives Roger a 1 minute head​ start, how long will it take before Jeff catches up to​ Roger? How far will each have​ run?

Let's break down the problem step by step.

### Step 1: Calculate their speeds
- Roger's time to run 1 mile: 21 minutes.
- Jeff's time to run 1 mile: 18 minutes.

The speed of each runner is the distance divided by the time it takes them to run that distance.

- **Roger's speed**: \( \text{Speed of Roger} = \frac{1 \text{ mile}}{21 \text{ minutes}} = \frac{1}{21} \text{ miles per minute} \)
- **Jeff's speed**: \( \text{Speed of Jeff} = \frac{1 \text{ mile}}{18 \text{ minutes}} = \frac{1}{18} \text{ miles per minute} \)

### Step 2: Set up the equation
Jeff gives Roger a 1-minute head start. In that time, Roger will have run:
- \( \text{Distance Roger runs in 1 minute} = \frac{1}{21} \text{ miles} \)

Let \( t \) be the time in minutes after Jeff starts running when he catches up to Roger. By this time:
- Roger will have run \( \text{Distance Roger runs} = \frac{1}{21} \times (t + 1) \text{ miles} \)
- Jeff will have run \( \text{Distance Jeff runs} = \frac{1}{18} \times t \text{ miles} \)

When Jeff catches up to Roger, the distances they've run will be equal:
\[
\frac{1}{21} \times (t + 1) = \frac{1}{18} \times t
\]

### Step 3: Solve the equation
Let's solve for \( t \):

1. Multiply both sides by 21 to eliminate the fraction on the left:
   \[
   t + 1 = \frac{21}{18} \times t
   \]

2. Simplify \( \frac{21}{18} \) to \( \frac{7}{6} \):
   \[
   t + 1 = \frac{7}{6} \times t
   \]

3. Multiply both sides by 6 to eliminate the fraction:
   \[
   6t + 6 = 7t
   \]

4. Subtract \( 6t \) from both sides:
   \[
   6 = t
   \]

So, it will take Jeff **6 minutes** to catch up to Roger.

### Step 4: Calculate the distance each has run
Now, find the distance each has run when Jeff catches up to Roger:

- **Roger's distance**: 
  \[
  \text{Distance Roger runs} = \frac{1}{21} \times (6 + 1) = \frac{7}{21} = \frac{1}{3} \text{ miles}
  \]

- **Jeff's distance**: 
  \[
  \text{Distance Jeff runs} = \frac{1}{18} \times 6 = \frac{6}{18} = \frac{1}{3} \text{ miles}
  \]

### Final Answer:
- It will take Jeff **6 minutes** to catch up to Roger.
- Both Jeff and Roger will have run **\(\frac{1}{3}\) miles** when Jeff catches up.

Would you like more details or have any questions about this problem?

Here are some related questions you might find interesting:
1. If Roger increased his speed, how much faster would he need to run to keep Jeff from catching him?
2. How long would it take Jeff to complete 5 miles at his current speed?
3. If Jeff gave Roger a 2-minute head start, how long would it take to catch up then?
4. What if Roger and Jeff were running on a track—how many laps would each complete before Jeff caught up?
5. If Jeff slowed down by 10%, how much longer would it take him to catch up?

**Tip**: When working with rates, setting up equations to represent distances or speeds over time can help simplify the problem.

Explore a math problem involving Roger and Jeff's running speeds. Calculate how long it takes for Jeff to catch Roger after giving him a 1-minute head start, and determine the distance each has run when they meet. Suitable for Grades 6-8 students learning about speed, distance, and equations.

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