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

Line 1: Line 1:
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:
 +
<math>\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}    </math>
  
 
== Solution ==
 
== Solution ==

Revision as of 11:19, 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

Call the radius of the outer circle $r_1$ and that of the inner circle $r_2$. The width of the track is $r_1-r_2$. The circumference of a circle is $2\pi$ times the radius, so the difference in circumferences is $2\pi r_1-2\pi r_2=10\pi$ feet. If we divide each side by $2\pi$, we get $r_1-r_2=\boxed{5}$ feet.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png