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.

Let's analyze the information and solve for the unknowns.

1. **Determine the Distance Between Towns A and B**

   Let \( D \) represent the total distance between Town A and Town B.

   The cyclist travels at a constant speed of \( 20 \) km/h, and by 11:00, they had traveled for:
   \[
   11:00 - 08:00 = 3 \text{ hours}
   \]

   During this time, the cyclist covered \( \frac{4}{9} \) of the journey, so:
   \[
   \text{Distance covered by cyclist} = \frac{4}{9} D
   \]

   Since speed \( = \frac{\text{distance}}{\text{time}} \), we have:
   \[
   20 \times 3 = \frac{4}{9} D
   \]
   Simplifying:
   \[
   60 = \frac{4}{9} D
   \]
   \[
   D = 60 \times \frac{9}{4} = 135 \text{ km}
   \]

2. **Calculate the Speed of the Scooter**

   The scooter left Town B at 08:30 and passed the cyclist at 11:00, meaning it traveled for:
   \[
   11:00 - 08:30 = 2.5 \text{ hours}
   \]

   Let \( v \) be the speed of the scooter before it decreased. The distance covered by the scooter in this time is \( \frac{5}{9} D \) because the cyclist covered \( \frac{4}{9} D \).

   Therefore:
   \[
   v \times 2.5 = \frac{5}{9} \times 135
   \]
   Simplifying:
   \[
   v \times 2.5 = 75
   \]
   \[
   v = \frac{75}{2.5} = 30 \text{ km/h}
   \]

3. **Determine the New Speed of the Scooter After Passing the Cyclist**

   The speed of the scooter decreased by \( 10 \) km/h after passing the cyclist, so its new speed is:
   \[
   30 - 10 = 20 \text{ km/h}
   \]

4. **Time for the Scooter to Complete the Journey**

   After passing the cyclist, the scooter still had \( \frac{4}{9} \) of the journey remaining to reach Town A, a distance of:
   \[
   \frac{4}{9} \times 135 = 60 \text{ km}
   \]

   With the new speed of \( 20 \) km/h, the time taken to cover this distance is:
   \[
   \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{60}{20} = 3 \text{ hours}
   \]

   Therefore, the scooter will reach Town A at:
   \[
   11:00 + 3 \text{ hours} = 14:00
   \]

### Summary of Answers:
1. **Distance between Town A and Town B:** \( 135 \) km
2. **Initial speed of the scooter:** \( 30 \) km/h
3. **New speed of the scooter after passing the cyclist:** \( 20 \) km/h
4. **Time the scooter reaches Town A:** \( 14:00 \)

Would you like any part of this broken down further, or do you have any other questions?

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### Related Questions:
1. How would the solution change if the cyclist’s speed increased by 5 km/h?
2. What would happen if the scooter reduced its speed by 15 km/h instead of 10 km/h?
3. If the scooter encountered a stop midway, how would this affect the arrival time?
4. How would the solution change if the cyclist left 30 minutes earlier?
5. How can we calculate the point of meeting if speeds and start times are different?

**Tip:** Drawing a timeline or diagram can often simplify problems involving moving objects and multiple events.

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.

Explore a detailed solution for finding the distance and speed in a scenario where a cyclist and a scooter travel between two towns, with the scooter's speed changing mid-journey. This problem covers core concepts like distance, speed, and proportions, suitable for Grades 9-11.

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