Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1953 AHSME Problems/Problem 11: Difference between revisions

← Older edit
Fortytwok (talk | contribs)
56 edits
No edit summary
Fortytwok (talk | contribs)
56 edits
 
(One intermediate revision by the same user not shown)
Line 7: Line 7:
== Solution ==
== Solution ==


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.
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>.


==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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_11&oldid=121478"