# Difference between revisions of "1964 AHSME Problems/Problem 26"

## Problem

In a ten-mile race $\textit{First}$ beats $\textit{Second}$ by $2$ miles and $\textit{First}$ beats $\textit{Third}$ by $4$ miles. If the runners maintain constant speeds throughout the race, by how many miles does $\textit{Second}$ beat $\textit{Third}$?

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2\frac{1}{4}\qquad \textbf{(C)}\ 2\frac{1}{2}\qquad \textbf{(D)}\ 2\frac{3}{4}\qquad \textbf{(E)}\ 3$

## Solution

Let the speeds of the runners in miles per hour be $a, b, c$, with $a>b>c$. If person $a$ reaches the finish line in $h$ hours, then we have $10 = ah, 8 = bh, 6 = bh$.

Thus, the speeds of the second and third place people are $\frac{8}{h}$ and $\frac{6}{h}$. The ratio of those speeds is $\frac{\frac{8}{h}}{\frac{6}{h}} = \frac{4}{3}$, meaning person $b$ runs $\frac{4}{3}$ times as fast as person $c$, or conversely that person $c$ is $\frac{3}{4}$ times slower than person $b$.

If $c$ runs $0.75$ times as fast as $b$, then when $b$ has finished the race at $10$ miles, $c$ will run $10 \cdot 0.75 = 7.5$ miles. This is a difference of $10 - 7.5 = 2.5$ miles, which is answer $\boxed{\textbf{(C)}}$