Difference between revisions of "2015 AMC 10A Problems/Problem 9"
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Two right circular cylinders have the same volume. The radius of the second cylinder is <math>10\%</math> more than the radius of the first. What is the relationship between the heights of the two cylinders? | Two right circular cylinders have the same volume. The radius of the second cylinder is <math>10\%</math> more than the radius of the first. What is the relationship between the heights of the two cylinders? | ||
| − | <math>\textbf{(A)}\ | + | <math> \textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ |
| + | \qquad\textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.} \\ | ||
| + | \qquad\textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ | ||
| + | \qquad\textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.} \\ | ||
| + | \qquad\textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.} </math> | ||
==Solution== | ==Solution== | ||
Let the radius of the first cylinder be <math>r_1</math> and the radius of the second cylinder be <math>r_2</math>. Also, let the height of the first cylinder be <math>h_1</math> and the height of the second cylinder be <math>h_2</math>. We are told <cmath>r_2=\frac{11r_1}{10}</cmath> <cmath>\pi r_1^2h_1=\pi r_2^2h_2</cmath> Substituting the first equation into the second and dividing both sides by <math>\pi</math>, we get <cmath>r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.</cmath> Therefore, <math>\boxed{\textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}}</math> | Let the radius of the first cylinder be <math>r_1</math> and the radius of the second cylinder be <math>r_2</math>. Also, let the height of the first cylinder be <math>h_1</math> and the height of the second cylinder be <math>h_2</math>. We are told <cmath>r_2=\frac{11r_1}{10}</cmath> <cmath>\pi r_1^2h_1=\pi r_2^2h_2</cmath> Substituting the first equation into the second and dividing both sides by <math>\pi</math>, we get <cmath>r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.</cmath> Therefore, <math>\boxed{\textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}}</math> | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/fx7bYbysB24 | ||
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| + | ~Education, the Study of Everything | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/zVCHWxfKErE | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Latest revision as of 20:54, 14 August 2024
- The following problem is from both the 2015 AMC 12A #7 and 2015 AMC 10A #9, so both problems redirect to this page.
Problem
Two right circular cylinders have the same volume. The radius of the second cylinder is 
more than the radius of the first. What is the relationship between the heights of the two cylinders?
Solution
Let the radius of the first cylinder be 
and the radius of the second cylinder be 
. Also, let the height of the first cylinder be 
and the height of the second cylinder be 
. We are told ![\[r_2=\frac{11r_1}{10}\]](http://latex.artofproblemsolving.com/4/1/7/41704dff30bf2daea06efcba03fdcd72cb507478.png)
![\[\pi r_1^2h_1=\pi r_2^2h_2\]](http://latex.artofproblemsolving.com/1/7/7/177bac559181a4d8c72c9057a70337eabab64b2f.png)
Substituting the first equation into the second and dividing both sides by 
, we get ![\[r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.\]](http://latex.artofproblemsolving.com/c/e/3/ce3ff9714fedab21b6406945de393a33855a715f.png)
Therefore, 
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2015 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
