1953 AHSME Problems/Problem 11

Problem

A running track is the ring formed by two concentric circles. It is $10\pi$ feet wide. The circumference of the two circles differ by about:

$\textbf{(A)}\ 10\text{ feet} \qquad \textbf{(B)}\ 30\text{ feet} \qquad \textbf{(C)}\ 60\text{ feet} \qquad \textbf{(D)}\ 100\text{ feet} \textbf{(E)}\ \text{none of these}$

Solution

We notice that since the running track is simply the area of the outer circle that is outside of the inner circle, the radius of the larger circle must be precisely $10$ feet larger than the radius of the smaller circle.

Since the circumference of a circle is calculated as $2\pi{r}$ where $r$ is the radius, we know that the circumference of the smaller circle is $2\pi{r}$ and the circumference of the larger circle is $2\pi(r+10)=2\pi{r}+20\pi$ .

The difference between the circumferences is $2\pi{r}+20\pi-2\pi{r}=20\pi\approx20\cdot3=\boxed{\textbf{(C) } 60\text{ feet}}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 11
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1953 AHSME Problems/Problem 11

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