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Difference between revisions of "2015 AMC 10A Problems/Problem 9"
(Copied the problem from the problems page for this year because it was almost impossible to read)
 
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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?
  
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<math>\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}</math>
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<math> \textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\
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\qquad\textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.} \\
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\qquad\textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\
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\qquad\textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.} \\
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\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)==
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https://youtu.be/fx7bYbysB24
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/zVCHWxfKErE
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~savannahsolver
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2015|ab=A|num-b=6|num-a=8}}
 
{{AMC12 box|year=2015|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Geometry Problems]]

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 $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?

$\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \qquad\textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.} \\ \qquad\textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \qquad\textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.} \\ \qquad\textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}$

Solution

Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$. Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$. We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and dividing both sides by $\pi$, we get \[r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.\] Therefore, $\boxed{\textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}}$

Video Solution (CREATIVE THINKING)

https://youtu.be/fx7bYbysB24

~Education, the Study of Everything



Video Solution

https://youtu.be/zVCHWxfKErE

~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

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