Difference between revisions of "1950 AHSME Problems/Problem 34"
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When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be <math>\pi</math> anymore) | When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be <math>\pi</math> anymore) | ||
We see that the circumference was increased by <math>25\%</math>. This means the radius was also increased by <math>25\%</math>. The radius of the original balloon is <math>\frac{20}{2\pi}=\frac{10}{\pi}</math>. With the <math>25\%</math> increase, it becomes <math>\frac{12.5}{\pi}</math>. The increase is <math>\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math> | We see that the circumference was increased by <math>25\%</math>. This means the radius was also increased by <math>25\%</math>. The radius of the original balloon is <math>\frac{20}{2\pi}=\frac{10}{\pi}</math>. With the <math>25\%</math> increase, it becomes <math>\frac{12.5}{\pi}</math>. The increase is <math>\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math> | ||
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+ | ==Solution 2== | ||
+ | The radii of the circle are <math>\frac{20}{2\pi}</math> and <math>\frac{25}{2\pi}</math>, respectively. The positive difference is therefore <math>\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 19:58, 20 March 2018
Contents
[hide]Problem
When the circumference of a toy balloon is increased from inches to inches, the radius is increased by:
Solution
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be anymore) We see that the circumference was increased by . This means the radius was also increased by . The radius of the original balloon is . With the increase, it becomes . The increase is
Solution 2
The radii of the circle are and , respectively. The positive difference is therefore .
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 33 |
Followed by Problem 35 | |
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