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Let's solve this problem.

The total available flight time is 5 hours, which includes both the trip with the wind and the return against the wind.

1. Let \( d \) be the one-way distance.
2. The time to travel with the wind (at 225 mph) is \( \frac{d}{225} \).
3. The time to travel against the wind (at 180 mph) is \( \frac{d}{180} \).
4. Since the total time is 5 hours, we can set up the equation:

\[
\frac{d}{225} + \frac{d}{180} = 5
\]

5. To solve for \( d \), find a common denominator and simplify:

\[
\frac{4d + 5d}{900} = 5
\]

\[
\frac{9d}{900} = 5
\]

\[
d = 500
\]

So, the one-way distance is 500 miles, making the round trip \( 2 \times 500 = 1000 \) miles.

### Answer:
The airplane can fly **1000 miles** roundtrip.

Would you like further details on any steps?

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1. What if the flight time was 6 hours instead?
2. How would the distance change if the plane flew at 200 mph with the wind?
3. What if the plane flew against the wind at 190 mph instead?
4. How does fuel efficiency relate to speed in this context?
5. What happens if wind direction changes mid-flight?

**Tip:** Always check units and time constraints to ensure calculations align with real-world constraints.

An airplane has just enough fuel for a five-hour flight. How far can it fly, roundtrip, if it flies at 225 mph with the wind during the first leg of the trip, and at 180 mph against the wind on the return?

Solve a real-world distance, rate, and time problem involving an airplane with enough fuel for a five-hour roundtrip. Calculate the maximum roundtrip distance at variable speeds of 225 mph with the wind and 180 mph against the wind. Suitable for high school students.

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