Difference between revisions of "1953 AHSME Problems/Problem 11"
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A running track is the ring formed by two concentric circles. It is <math>10</math> feet wide. The circumference of the two circles differ by about: | A running track is the ring formed by two concentric circles. It is <math>10</math> feet wide. The circumference of the two circles differ by about: | ||
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<math>\textbf{(A)}\ 10\text{ feet} \qquad | <math>\textbf{(A)}\ 10\text{ feet} \qquad | ||
\textbf{(B)}\ 30\text{ feet} \qquad | \textbf{(B)}\ 30\text{ feet} \qquad | ||
\textbf{(C)}\ 60\text{ feet} \qquad | \textbf{(C)}\ 60\text{ feet} \qquad | ||
− | \textbf{(D)}\ 100\text{ feet} \textbf{(E)}\ \text{none of these}</math> | + | \textbf{(D)}\ 100\text{ feet}\ \textbf{(E)}\ \text{none of these} </math> |
== Solution == | == 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 <math>20\pi </math> feet. Using <math>\pi \approx 3</math>, the answer is <math>\fbox{C}</math>. | |
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− | Since the circumference of a circle is | ||
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==See Also== | ==See Also== | ||
− | {{AHSME 50p box|year=1953|num-b=10|num-a= | + | {{AHSME 50p box|year=1953|num-b=10|num-a=12}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:23, 22 April 2020
A running track is the ring formed by two concentric circles. It is feet wide. The circumference of the two circles differ by about:
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 feet. Using , the answer is .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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