Difference between revisions of "1952 AHSME Problems/Problem 25"
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| + | == Problem== | ||
| + | A powderman set a fuse for a blast to take place in <math>30</math> seconds. He ran away at a rate of <math>8</math> yards per second. Sound travels at the rate of <math>1080</math> feet per second. When the powderman heard the blast, he had run approximately: | ||
| + | <math> \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.} </math> | ||
| + | ==Solution== | ||
| + | Let <math>p(t)=24t</math> be the number of feet the powderman is from the blast at <math>t</math> seconds after the fuse is lit, and let <math>q(t)=1080t-32400</math> be the number of feet the sound has traveled. We want to solve for <math>p(t)=q(t)</math>. | ||
| + | <cmath>24t=1080t-32400</cmath> | ||
| + | <cmath>1056t=32400</cmath> | ||
| + | <cmath>t=\frac{32400}{1056}</cmath> | ||
| + | <cmath>t=\frac{675}{22}=30.6\overline{81}</cmath> | ||
| + | The number of <b>yards</b> the powderman is from the blast at time <math>t</math> is <math>\frac{24t}3=8t</math>, so the answer is <math>8(30.6\overline{81})</math>, which is about <math>245</math> yards. <math>\boxed{\textbf{(D)}}</math> | ||
| + | |||
| + | ==See also== | ||
| + | {{AHSME 50p box|year=1952|num-b=24|num-a=26}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | [[Category:Rate Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:11, 18 July 2015
Problem
A powderman set a fuse for a blast to take place in 
seconds. He ran away at a rate of 
yards per second. Sound travels at the rate of 
feet per second. When the powderman heard the blast, he had run approximately:
Solution
Let 
be the number of feet the powderman is from the blast at 
seconds after the fuse is lit, and let 
be the number of feet the sound has traveled. We want to solve for 
.
![\[24t=1080t-32400\]](http://latex.artofproblemsolving.com/a/d/3/ad3c9d14d2b96ee5aaac9dfa8b0989be5311134a.png)
![\[1056t=32400\]](http://latex.artofproblemsolving.com/b/e/0/be000171fd9bb68b1817375e2f9e3f101b108fee.png)
![\[t=\frac{32400}{1056}\]](http://latex.artofproblemsolving.com/e/5/6/e5618305edf9017bd3a6fd6f0806d9d9e47f9c76.png)
![\[t=\frac{675}{22}=30.6\overline{81}\]](http://latex.artofproblemsolving.com/b/8/f/b8fcfb451ef0754eeb229ff8ba4c4a4c776f23f9.png)
The number of yards the powderman is from the blast at time is 
, so the answer is 
, which is about 
yards. 
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. 
