Terrence walks at a pace of 2 miles per hour to the theater and watches a movie for 2 hours and 15 minutes. He rides back home, taking the same route, on a bus that travels at a rate of 40 miles per hour. The entire trip takes 3.5 hours. How far along this route is Terrence’s house from the theater?

To find the distance from Terrence's house to the theater, we can break down the problem into two main parts: the walking time and the bus ride time.

### Let's define the variables:
- Let \( d \) be the distance between Terrence's house and the theater (in miles).
- The walking speed is 2 miles per hour.
- The bus speed is 40 miles per hour.
- The total time spent watching the movie is 2 hours and 15 minutes, which is 2.25 hours.
- The entire trip takes 3.5 hours.

### Step 1: Determine the time spent traveling
The total time for the entire trip is 3.5 hours. Out of this, 2.25 hours were spent watching the movie. So, the time spent traveling is:
\[
\text{Travel time} = 3.5 - 2.25 = 1.25 \text{ hours}
\]

### Step 2: Set up the equation
Terrence walks to the theater, which takes him \( \frac{d}{2} \) hours. He then rides back on the bus, which takes him \( \frac{d}{40} \) hours. The sum of these travel times is equal to the total travel time of 1.25 hours:
\[
\frac{d}{2} + \frac{d}{40} = 1.25
\]

### Step 3: Solve for \( d \)
To solve for \( d \), we first find a common denominator for the fractions. The common denominator of 2 and 40 is 40, so:
\[
\frac{20d}{40} + \frac{d}{40} = 1.25
\]
\[
\frac{21d}{40} = 1.25
\]
Next, multiply both sides by 40 to eliminate the denominator:
\[
21d = 1.25 \times 40
\]
\[
21d = 50
\]
Finally, divide by 21 to find \( d \):
\[
d = \frac{50}{21} \approx 2.38 \text{ miles}
\]

### Conclusion
The distance from Terrence’s house to the theater is approximately 2.38 miles.

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. If Terrence had walked both ways, how long would the entire trip have taken?
2. How much time would Terrence save if the bus traveled at 60 miles per hour instead of 40 miles per hour?
3. What would be the travel time if Terrence had taken a taxi back home traveling at 50 miles per hour?
4. If the walking speed was reduced to 1.5 miles per hour, how would that affect the total time for the trip?
5. How long would Terrence have spent traveling if the movie had lasted 3 hours?

**Tip:** When solving problems involving multiple parts of a journey, it can be helpful to break down the problem by first determining the time or distance associated with each segment of the trip.

Solve how far Terrence’s house is from the theater based on his walking and bus travel times. This problem applies algebraic concepts and time-speed-distance relationships. Suitable for Grades 6-8, it involves breaking down the total trip time and calculating the distance using linear equations.

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