Let the distance between Alan's house and the gym be $d$ miles.

Step 1: Express the time to run and walk in terms of $d$

• Alan runs to the gym at 6 mph, so the time to run to the gym is: $t_{\text{run}} = \frac{d}{6}$
• Alan walks home at 4 mph, so the time to walk home is: $t_{\text{walk}} = \frac{d}{4}$

Step 2: Relate the times

We are told that it takes Alan 15 minutes longer to walk home than to run to the gym. Since 15 minutes is equal to $\frac{15}{60} = \frac{1}{4}$ hours, the relationship between the times is: $t_{\text{walk}} = t_{\text{run}} + \frac{1}{4}$

Step 3: Set up the equation

Substitute the expressions for $t_{\text{run}}$ and $t_{\text{walk}}$ into the equation: $\frac{d}{4} = \frac{d}{6} + \frac{1}{4}$

Step 4: Solve for $d$

To eliminate the fractions, multiply the entire equation by 12 (the least common multiple of 4 and 6): $12 \times \frac{d}{4} = 12 \times \left( \frac{d}{6} + \frac{1}{4} \right)$ $3d = 2d + 3$ Subtract $2d$ from both sides: $d = 3$

Final Answer:

The distance between Alan’s house and the gym is 3 miles.

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the solution change if Alan walked at 5 mph instead of 4 mph?
2. What if the time difference between walking and running were 20 minutes instead of 15 minutes?
3. How long does it take Alan to complete the round trip to the gym and back?
4. What if Alan increased his running speed by 1 mph? How would that affect the distance?
5. Can you express the time difference in seconds and solve the problem using that unit?

Tip: Always double-check your units (minutes vs. hours) when working with time-based problems.