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

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== Problem ==
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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:  
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A running track is the ring formed by two concentric circles. It is <math>10\pi</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
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\textbf{(D)}\ 100\text{ feet} \textbf{(E)}\ \text{none of these}</math>
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\textbf{(D)}\ 100\text{ feet}\\ \textbf{(E)}\ \text{none of these}   </math>
  
 
== Solution ==
 
== Solution ==
  
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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 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 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==
 
==See Also==

Latest 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 $20\pi$ feet. 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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