Difference between revisions of "1950 AHSME Problems/Problem 34"
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\textbf{(E)}\ \dfrac{\pi}{5}\text{ in}</math> | \textbf{(E)}\ \dfrac{\pi}{5}\text{ in}</math> | ||
==Solution== | ==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 <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> | |
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==See Also== | ==See Also== | ||
Revision as of 18:28, 10 April 2013
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 
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
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| All AHSME Problems and Solutions | ||
