Difference between revisions of "2015 AMC 10A Problems/Problem 9"

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==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==
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https://youtu.be/zVCHWxfKErE
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~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 14:15, 24 July 2020

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?

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

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

https://youtu.be/zVCHWxfKErE

~savannahsolver

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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