1952 AHSME Problems/Problem 25

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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