Difference between revisions of "1953 AHSME Problems/Problem 11"

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== Solution ==
 
== 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 <math>10</math> feet larger than the radius of the smaller circle. Since the circumference of a circle is calculated as <math>2\pi{r}</math> where <math>r</math> is the radius, we know that the circumference of the smaller circle is <math>2\pi{r}</math> and the circumference of the larger circle is <math>2\pi(r+10)=2\pi{r}+20\pi</math>. The difference between the circumferences is <math>2\pi{r}+20\pi-2\pi{r}=20\pi\approx20\cdot3=\boxed{\textbf{(C) } 60\text{ feet}}</math>.
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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 <math>10</math> feet larger than the radius of the smaller circle.  
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Since the circumference of a circle is calculated as <math>2\pi{r}</math> where <math>r</math> is the radius, we know that the circumference of the smaller circle is <math>2\pi{r}</math> and the circumference of the larger circle is <math>2\pi(r+10)=2\pi{r}+20\pi</math>.  
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 +
The difference between the circumferences is <math>2\pi{r}+20\pi-2\pi{r}=20\pi\approx20\cdot3=\boxed{\textbf{(C) } 60\text{ feet}}</math>.
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==See Also==
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{{AHSME 50p box|year=1953|num-b=10|num-a=11}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 20:48, 1 April 2017

Problem

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

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 $2\pi{r}+20\pi-2\pi{r}=20\pi\approx20\cdot3=\boxed{\textbf{(C) } 60\text{ feet}}$.


See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 11
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All AHSME Problems and Solutions

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