Question 8.

Assuming the length of race course = $x$ and the speed of three horses be $a, b, c$ respectively.

Hence,  $\frac{x}{a} = \frac{x-11}{b} \quad ...(1)$

and  $\frac{x}{a} = \frac{x-90}{c} \quad ...(2)$

Also,  $\frac{x}{b} = \frac{x-80}{c} \quad ...(3)$

From (1) and (2), we get,  

$\frac{x-11}{b} = \frac{x-90}{c} \quad ...(4)$

Dividing (3) by (4), we get,  

$\frac{x-11}{x} = \frac{x-90}{x-80}$

$\Rightarrow (x-11)(x-80) = x(x-90)$

91x-90x=880=>x=880