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.