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 "1952 AHSME Problems/Problem 25"
m (Added categories: Introductory Algebra Problems, Rate Problems)
 
Line 13: Line 13:
 
==See also==
 
==See also==
 
{{AHSME 50p box|year=1952|num-b=24|num-a=26}}
 
{{AHSME 50p box|year=1952|num-b=24|num-a=26}}
 +
 +
[[Category:Introductory Algebra Problems]]
 +
[[Category:Rate Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:11, 18 July 2015

Problem

A powderman set a fuse for a blast to take place in $30$ seconds. He ran away at a rate of $8$ yards per second. Sound travels at the rate of $1080$ feet per second. When the powderman heard the blast, he had run approximately: $\textbf{(A)}\ \text{200 yd.}\qquad\textbf{(B)}\ \text{352 yd.}\qquad\textbf{(C)}\ \text{300 yd.}\qquad\textbf{(D)}\ \text{245 yd.}\qquad\textbf{(E)}\ \text{512 yd.}$

Solution

Let $p(t)=24t$ be the number of feet the powderman is from the blast at $t$ seconds after the fuse is lit, and let $q(t)=1080t-32400$ be the number of feet the sound has traveled. We want to solve for $p(t)=q(t)$. \[24t=1080t-32400\] \[1056t=32400\] \[t=\frac{32400}{1056}\] \[t=\frac{675}{22}=30.6\overline{81}\] The number of yards the powderman is from the blast at time $t$ is $\frac{24t}3=8t$, so the answer is $8(30.6\overline{81})$, which is about $245$ yards. $\boxed{\textbf{(D)}}$

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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=1952_AHSME_Problems/Problem_25&oldid=71232"