Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
 
Line 18: Line 18:
 
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>.
 
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>
+
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 - 7.5 = 2.5</math> miles, which is answer <math>\boxed{\textbf{(C)}}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 22:23, 24 July 2019

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 $\boxed{\textbf{(C)}}$

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_26&oldid=107906"