# 1953 AHSME Problems/Problem 31

## Problem

The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click. The speed of the train in miles per hour is approximately the number of clicks heard in: $\textbf{(A)}\ 20\text{ seconds} \qquad \textbf{(B)}\ 2\text{ minutes} \qquad \textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad \textbf{(D)}\ 5\text{ minutes}\\ \textbf{(E)}\ \text{none of these}$

## Solution

We assume that the clicks are heard at the head of the train. Then if the train's speed in miles per hour is $x$, we can convert it to clicks per minute: $$\frac{x\text{ mile}}{\text{hr}}\cdot\left(\frac{1\text{ hr}}{60\text{ min}}\right)\cdot\left(\frac{5280\text{ ft}}{1\text{ mile}}\right)\cdot\left(\frac{1\text{ click}}{30\text{ ft}}\right)=\frac{5280x}{1800}\cdot\frac{\text{click}}{\text{min}}.$$ Therefore every minute, on average, $5280x/1800$ clicks are heard. The number of clicks heard in $y$ minutes is $y\cdot 5280x/1800$, so the number of clicks heard in $1800/5280$ minutes is equal to $x$. In other words, the speed of the train in miles per hour is equal to the number of clicks heard in $1800/5280$ minutes, which is approximately one-third of a minute, or $\boxed{\textbf{(A)}\ 20\text{ seconds}}$.

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