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Given that,

In a race of three horses, the first beat the second by $11 \; \text{metres}$ and the third by $90 \; \text{metres}$.

Let the length of the racecourse be $x \; \text{metres.}$

- The first horse finished the race $ = x \; \text{metres}.$
- The second horse finished the race $ = (x-11) \; \text{metres}.$
- The third horse finished the race $ = (x-90) \; \text{metres}.$

If the second beat the third by $80 \; \text{metres} .$

- The second horse finished the race $ = x \; \text{metres}.$
- The third horse finished the race $ = (x-80) \; \text{metres}.$

Now, second horse has run $11 \; \text{metres}$ more than before and the third horse also run $10 \; \text{metres}$ more than before.

Using the formula $: \boxed{\text{Speed} = \frac{\text{Distance}}{\text{Time}}}$

If time is constant, then we can write, $ \boxed{\frac{\text{Distance}}{\text{Speed}} \propto \text{Constant} \Rightarrow \frac{\text{D}_{1}}{\text{S}_{1}} = \frac{\text{D}_{2}}{\text{S}_{2}} = \frac{\text{D}_{3}}{\text{S}_{3}}}$

Let $\text{S}_{1},\text{S}_{2},$ and $\text{S}_{3}$ be the speed of first horse, second horse, and third horse respectively.

Now, we have $ \frac{x}{\text{S}_{1}} = \frac{x-11}{\text{S}_{2}} = \frac{x-90}{\text{S}_{3}}$

The ratio of $ \frac{\text{S}_{2}} {\text{S}_{3}} = \frac{x-11}{x-90} \quad \longrightarrow (1)$

And, similarly, we have $\frac{11}{\text{S}_{2}} = \frac{10}{\text{S}_{3}}$

The ratio of $ \frac{\text{S}_{2}} {\text{S}_{3}} = \frac{11}{10} \quad \longrightarrow (2)$

Equate the equation $(1)$ and $(2)$ are equal, we get

$ \frac{x-11}{x-90} = \frac{11}{10}$

$ \Rightarrow 10x – 110 = 11x – 990$

$ \Rightarrow 990-110 = 11x – 10x$

$ \Rightarrow \boxed{x = 880}$

$\therefore$ The length of the racecourse $ = 880 \; \text{metres}.$

Correct Answer $: 880$