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The problem describes a race between Ravi, Sumit, and Vikas, and it gives information about the distances by which Ravi and Sumit beat Vikas. Specifically:

- Ravi beats Sumit by 40 meters.
- Sumit beats Vikas by 50 meters.
- Ravi beats Vikas by 80 meters.

We are asked to find the length of the race.

### Step-by-step approach:

Let the total length of the race be \( L \) meters.

1. **Sumit vs. Vikas:**
   When Sumit finishes the race, Vikas is 50 meters behind. This means that when Sumit runs the entire \( L \) meters, Vikas runs \( L - 50 \) meters. Therefore, the ratio of their speeds (Sumit to Vikas) is:
   \[
   \frac{\text{Speed of Sumit}}{\text{Speed of Vikas}} = \frac{L}{L - 50}
   \]

2. **Ravi vs. Vikas:**
   When Ravi finishes the race, Vikas is 80 meters behind. This means that when Ravi runs \( L \) meters, Vikas runs \( L - 80 \) meters. Therefore, the ratio of their speeds (Ravi to Vikas) is:
   \[
   \frac{\text{Speed of Ravi}}{\text{Speed of Vikas}} = \frac{L}{L - 80}
   \]

3. **Ravi vs. Sumit:**
   When Ravi finishes the race, Sumit is 40 meters behind. This means that when Ravi runs \( L \) meters, Sumit runs \( L - 40 \) meters. Therefore, the ratio of their speeds (Ravi to Sumit) is:
   \[
   \frac{\text{Speed of Ravi}}{\text{Speed of Sumit}} = \frac{L}{L - 40}
   \]

4. **Equating the two ratios:**
   Using the ratios from Ravi vs. Sumit and Sumit vs. Vikas:
   \[
   \frac{\text{Speed of Ravi}}{\text{Speed of Vikas}} = \frac{\text{Speed of Ravi}}{\text{Speed of Sumit}} \times \frac{\text{Speed of Sumit}}{\text{Speed of Vikas}}
   \]
   Substituting the expressions:
   \[
   \frac{L}{L - 80} = \frac{L}{L - 40} \times \frac{L}{L - 50}
   \]
   Simplifying this equation will allow us to solve for \( L \).

### Simplifying:
\[
\frac{L}{L - 80} = \frac{L^2}{(L - 40)(L - 50)}
\]
Multiplying both sides by \((L - 80)(L - 40)(L - 50)\) to eliminate the denominators:
\[
L(L - 40)(L - 50) = L^2 (L - 80)
\]
Cancelling \(L\) on both sides and expanding:
\[
(L - 40)(L - 50) = L(L - 80)
\]
Expanding both sides:
\[
L^2 - 90L + 2000 = L^2 - 80L
\]
Cancelling \(L^2\) and simplifying the equation:
\[
-90L + 2000 = -80L
\]
\[
-10L = -2000
\]
\[
L = 200
\]

### Conclusion:
The length of the race is \( \boxed{200} \) meters.

Would you like further details or have any other questions?

Here are five related questions:
1. How does speed ratio help solve relative distance problems?
2. What would happen if Ravi beat Sumit by 50 meters instead?
3. Can you explain how different distances affect speed ratios?
4. How would the problem change if Sumit beat Ravi?
5. What if the race were twice as long?

**Tip:** Always express the relationship between distance and speed when tackling problems like this.

The sports events were conducted at a Kendriya Vidyalaya, where three finalists—Ravi, Sumit, and Vikas—competed in a race. In the race, Ravi beat Sumit by 40 meters, Sumit beat Vikas by 50 meters, and Ravi beat Vikas by 80 meters. What was the length of the race?

Learn how to calculate the length of a race by using the speed and distance ratios of Ravi, Sumit, and Vikas. This problem uses concepts of algebra and proportionality. Suitable for students in Grades 9-10.

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