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Difference between revisions of "1953 AHSME Problems/Problem 11"
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== Solution ==
 
== Solution ==
  
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Call the radius of the outer circle <math>r_1</math> and that of the inner circle <math>r_2</math>. The width of the track is <math>r_1-r_2</math>. The circumference of a circle is <math>2\pi</math> times the radius, so the difference in circumferences is <math>2\pi r_1-2\pi r_2=10\pi</math> feet. If we divide each side by <math>2\pi</math>, we get <math>r_1-r_2=\boxed{5}</math> feet.
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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 <math>20\pi</math>. Using <math>\pi \approx 3</math>, the answer is <math>\fbox{C}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 11:23, 22 April 2020

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 $\fbox{C}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AHSME Problems and Solutions

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