In a 100-metre race, A beats B by 10 metres and B beats C by 20 metres. After touching the finishing line, A turns back and starts running in the opposite direction with the same speed. He crosses B and C at points P and Q respectively. What is the approximate distance (in metres) between P and Q?

Let's analyze the given problem step by step:

### 1. Understanding the relative speeds and distances

- **A beats B by 10 meters**: This means when A finishes the 100-meter race, B is at 90 meters.
  
  The ratio of their speeds is:  
  \[
  \text{Speed of A} : \text{Speed of B} = \frac{100}{90} = \frac{10}{9}
  \]

- **B beats C by 20 meters**: This means when B reaches the 100-meter mark, C is at 80 meters.
  
  The ratio of their speeds is:  
  \[
  \text{Speed of B} : \text{Speed of C} = \frac{100}{80} = \frac{5}{4}
  \]

- From the above ratios, we can deduce that the ratio of speeds between A and C is:  
  \[
  \text{Speed of A} : \text{Speed of C} = \frac{10}{9} \times \frac{5}{4} = \frac{50}{36} = \frac{25}{18}
  \]

### 2. Time taken for A, B, and C to meet at points P and Q

Since A turns back and runs in the opposite direction at the same speed, we need to find the time taken for A to meet B and C after crossing the finish line.

- Let the speed of A be \( v \), and the speed of B is \( \frac{9}{10}v \), and C’s speed is \( \frac{18}{25}v \).
  
- The relative speed of A and B when running towards each other is:  
  \[
  v + \frac{9}{10}v = \frac{19}{10}v
  \]
  
- The relative speed of A and C is:  
  \[
  v + \frac{18}{25}v = \frac{43}{25}v
  \]

### 3. Finding the distances between P and Q

- Since A crosses B when B is 10 meters away from the finish line, A would need to cover 10 meters relative to B. The time taken for A to meet B is:  
  \[
  t_{AB} = \frac{10}{\frac{19}{10}v} = \frac{100}{19v}
  \]

- A crosses C when C is 20 meters away from the finish line. The time taken for A to meet C is:  
  \[
  t_{AC} = \frac{20}{\frac{43}{25}v} = \frac{500}{43v}
  \]

### 4. Approximate distance between P and Q

To find the distance between P and Q, we calculate how far B travels while A is moving to meet C:
  
- The time difference between the two meetings is:  
  \[
  t_{AC} - t_{AB} = \frac{500}{43v} - \frac{100}{19v}
  \]

We need a common denominator to simplify:
\[
t_{AC} - t_{AB} = \frac{500 \times 19 - 100 \times 43}{43 \times 19 \times v} = \frac{9500 - 4300}{817v} = \frac{5200}{817v}
\]

- During this time, B moves a distance of:
  \[
  \text{Distance} = \frac{9}{10}v \times \frac{5200}{817v} = \frac{46800}{8170} \approx 5.73 \text{ meters}
  \]

Thus, the approximate distance between P and Q is **5.73 meters**.

---

Do you have any questions or want further clarification?

Here are 5 related questions to consider:

1. How can we generalize this problem for different distances?
2. How does changing the speed of A affect the distance between P and Q?
3. What if A had a different speed while returning, how would the calculation change?
4. What if the distances by which A and B beat each other were different?
5. How do relative speed concepts apply to other real-world scenarios?

**Tip:** When calculating relative speeds, always ensure that the directions of motion are correctly considered to determine whether to add or subtract speeds.

Solve a 100-meter race problem where A beats B by 10 meters and B beats C by 20 meters. Explore how A, after finishing the race, crosses B and C while running back with the same speed. Learn to calculate the approximate distance between points P and Q using relative speed and time calculations. Suitable for Grades 10-12 students.

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