A cyclist left Town A at 08:00 and traveled to Town B at a speed of 20 km/h throughout the entire journey. A scooter left Town B at 08:30 and traveled toward Town A. The scooter maintained the same speed until it passed the cyclist at 11:00. The cyclist had covered 4/9   of the journey when they passed each other. The speed of the scooter decreased by 10 km/h after passing the cyclist, and it continued at this new speed for the remainder of the journey. What was the speed of the scooter before it passed the cyclist?

Let’s break down the problem step-by-step.

1. **Define Variables and Information Given:**
   - The cyclist travels from Town A to Town B starting at 08:00.
   - The scooter leaves Town B for Town A at 08:30 and they meet at 11:00.
   - When they meet, the cyclist has covered \( \frac{4}{9} \) of the total journey.
   - The speed of the cyclist is \( 20 \ \text{km/h} \).
   - After meeting the cyclist, the scooter's speed decreases by \( 10 \ \text{km/h} \).

2. **Determine the Time the Cyclist Traveled Before Meeting:**
   - The cyclist started at 08:00 and met the scooter at 11:00, so the cyclist traveled for \( 3 \) hours.
   - At a speed of \( 20 \ \text{km/h} \), the cyclist covered:
     \[
     20 \ \text{km/h} \times 3 \ \text{hours} = 60 \ \text{km}
     \]
   - Since this \( 60 \ \text{km} \) represents \( \frac{4}{9} \) of the total journey, we can find the full distance \( D \) as follows:
     \[
     \frac{4}{9} D = 60 \implies D = 60 \times \frac{9}{4} = 135 \ \text{km}
     \]

3. **Determine the Time the Scooter Traveled Before Meeting:**
   - The scooter started at 08:30 and met the cyclist at 11:00, so it traveled for \( 2.5 \) hours.

4. **Calculate the Speed of the Scooter Before Passing the Cyclist:**
   - Since they met at a distance of \( 60 \ \text{km} \) from Town A (where the cyclist started), the scooter must have covered the remaining distance from Town B to that point:
     \[
     135 \ \text{km} - 60 \ \text{km} = 75 \ \text{km}
     \]
   - The scooter traveled \( 75 \ \text{km} \) in \( 2.5 \) hours, so its speed before passing the cyclist was:
     \[
     \text{Speed of scooter} = \frac{75 \ \text{km}}{2.5 \ \text{hours}} = 30 \ \text{km/h}
     \]

5. **Conclusion:**
   The speed of the scooter before passing the cyclist was **30 km/h**.

Would you like further details on any part of the solution, or have additional questions? 

### Related Questions:
1. What distance did the scooter travel after passing the cyclist?
2. How much time did the cyclist take to complete the entire journey?
3. If the scooter’s speed had not decreased after passing the cyclist, how much earlier would it have reached Town A?
4. What was the remaining distance each vehicle had to travel after they met?
5. How would the solution change if the speed reduction was different?

#### Tip:
When two objects move toward each other, you can use their combined speed to calculate when and where they’ll meet if given both speeds and starting times.

A cyclist left Town A at 08:00 and traveled to Town B at a speed of 20 km/h throughout the entire journey. A scooter left Town B at 08:30 and traveled toward Town A. The scooter maintained the same speed until it passed the cyclist at 11:00. The cyclist had covered 4/9 of the journey when they passed each other. The speed of the scooter decreased by 10 km/h after passing the cyclist, and it continued at this new speed for the remainder of the journey. What was the speed of the scooter before it passed the cyclist?

Explore a detailed solution to calculate the speed of a scooter before it passed a cyclist, given specific time, speed, and distance ratio data. Learn concepts like distance-speed-time relations and relative motion. Suitable for students in Grades 8-10.

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