Problem 1200

1. Let's denote the speeds of First, Second, and Third as $v_1$, $v_2$, and $v_3$ respectively.
2. Given that First beats Second by 2 miles in a 10-mile race, we can write the relationship:
$$\frac{10}{v_1} = \frac{8}{v_2}$$
This implies:
$$v_1 = \frac{10}{8} v_2 = \frac{5}{4} v_2$$
3. Similarly, given that First beats Third by 4 miles in a 10-mile race, we can write:
$$\frac{10}{v_1} = \frac{6}{v_3}$$
This implies:
$$v_1 = \frac{10}{6} v_3 = \frac{5}{3} v_3$$
4. We need to find the distance by which Second beats Third. First, we express $v_2$ and $v_3$ in terms of $v_1$:
$$v_2 = \frac{4}{5} v_1$$
$$v_3 = \frac{3}{5} v_1$$
5. The time taken by Second to finish the race is:
$$t_2 = \frac{10}{v_2} = \frac{10}{\frac{4}{5} v_1} = \frac{10 \cdot 5}{4 v_1} = \frac{50}{4 v_1} = \frac{25}{2 v_1}$$
6. The distance covered by Third in this time is:
$$d_3 = v_3 \cdot t_2 = \left( \frac{3}{5} v_1 \right) \left( \frac{25}{2 v_1} \right) = \frac{3}{5} \cdot \frac{25}{2} = \frac{75}{10} = 7.5 \text{ miles}$$
7. Therefore, the distance by which Second beats Third is:
$$10 - 7.5 = 2.5 \text{ miles}$$

The final answer is $\boxed{2.5}$.