View source for 1964 AHSME Problems/Problem 26

==Problem==

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

<math>\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 </math>

==Solution==

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

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

If <math>c</math> runs <math>0.75</math> times as fast as <math>b</math>, then when <math>b</math> has finished the race at <math>10</math> miles, <math>c</math> will run <math>10 \cdot 0.75 = 7.5</math> miles.  This is a difference of <math>10 - 75. = 2.5</math> miles, which is answer <math>\boxed{\textbf{(C)}}</math>

==See Also==
{{AHSME 40p box|year=1964|num-b=25|num-a=27}}

[[Category:Introductory Algebra Problems]]

{{MAA Notice}}