1. In a | mile race, the first person is 2 miles ahead of the second person, and 4 miles ahead of the third person. If the runners maintain a constant speed throughout the race, the number of miles by which the second person will be ahead of the third person when the second person reaches the finish line is
(A) 2.
(B) $2 \frac{1}{4}$.
(C) $2 \frac{1}{2}$.
(D) $2 \frac{3}{4}$.
(E) 3.
(15th American High School Mathematics Examination, 1964)
[Solution] In the same time, the three people run different distances. According to the problem, the speed ratio of the three people is $10: 8: 6$. Since the speed remains constant, the speed ratio also remains unchanged.
When the first person runs 10 miles, the second person runs 8 miles, and the third person runs 6 miles;
When the second person runs the remaining 2 miles, let the third person run another $x$ miles. According to the previous analysis,
$$
\frac{2}{x}=\frac{8}{6} \text {, we get } \quad x=1 \frac{1}{2} \text {. }
$$

Therefore, when the second person runs 10 miles, the third person runs $7 \frac{1}{2}$ miles, which means the two are $2 \frac{1}{2}$ miles apart.
So the answer is $(C)$.