1953 AHSME Problems/Problem 11

A running track is the ring formed by two concentric circles. It is $10$ 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

Since the track is 10 feet wide, the diameter of the outer circle will be 20 feet more than the inner circle. Since the circumference of a circle is directly proportional to its diameter, the difference in the circles' diameters is simply $20\pi$ . Using $\pi \approx 3$ , the answer is $\fbox{C}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1953 AHSME Problems/Problem 11

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