In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse?
(Assume that the speed of a given horse does not change during the course of the race)
Enter the answer as an integer. Round to the nearest integer, if needed

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

1. **Define the variables**:
   - Let \( L \) be the length of the racecourse in meters.
   - Let the speeds of the first, second, and third horses be \( S_1 \), \( S_2 \), and \( S_3 \) respectively.

2. **Translate the information**:
   - Since the first horse beats the second by 11 meters, it means when the first horse covers the entire distance \( L \), the second horse has covered \( L - 11 \) meters.
   - Similarly, since the first horse beats the third by 90 meters, it means when the first horse covers \( L \), the third horse has covered \( L - 90 \) meters.
   - The second horse beats the third by 80 meters, so when the second horse covers \( L \), the third horse has covered \( L - 80 \) meters.

3. **Set up equations using the relative distances**:
   - Since the speeds of the horses are constant, the ratios of their distances covered will equal the ratios of their speeds.

   \[
   \frac{S_2}{S_1} = \frac{L - 11}{L}
   \]

   \[
   \frac{S_3}{S_1} = \frac{L - 90}{L}
   \]

   \[
   \frac{S_3}{S_2} = \frac{L - 80}{L}
   \]

4. **Solve for \( L \)**:
   To solve for \( L \), equate the expressions for \( S_3 \) from the first and second equations:

   \[
   \frac{L - 90}{L} = \frac{L - 80}{L - 11}
   \]

5. **Cross-multiply to simplify**:

   \[
   (L - 90)(L - 11) = L(L - 80)
   \]

6. **Expand and simplify**:

   \[
   L^2 - 101L + 990 = L^2 - 80L
   \]

   \[
   -21L + 990 = 0
   \]

   \[
   L = \frac{990}{21} \approx 47.14
   \]

Rounding to the nearest integer, we find:

\[
\boxed{47}
\]

The length of the racecourse is approximately **47 meters**.

Would you like a breakdown of any specific step?

Here are five related questions to expand understanding:
1. How would the answer change if the second horse beat the third by a different distance?
2. What assumptions are we making about the speeds of the horses in this problem?
3. How can we confirm our answer by checking all horses' distances at the finish line?
4. What if the first horse only beat the second by half the current difference?
5. How would this approach work with more than three horses?

**Tip**: Setting up equations based on ratios is a useful method when dealing with relative speeds and distances in race problems.

In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse?

Learn how to determine the length of a racecourse when given the distances by which horses beat each other in a three-horse race. This problem covers algebraic techniques in ratios, proportions, and relative speed, suitable for Grades 9-10.

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