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Difference between revisions of "2019 AMC 8 Problems/Problem 9"

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==Solution 3==
 
==Solution 3==
  
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The ratio of the numbers is <math>1/2</math>. Looking closely at the formula <math>r^2 * h * \pi</math>, we see that the <math>r * h * \pi</math> will cancel, meaning that the ratio of them will be <math>1(2)/2(2)</math> = <math>\boxed{\textbf{(B)}\ 1:2}</math>   
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The ratio of the numbers is <math>1/2</math>. Looking closely at the formula <math>r^2 * h * \pi</math>, we see that the <math>r * h * \pi</math> will cancel, meaning that the ratio of them will be <math>\frac{1(2)}{2(2)}</math> = <math>\boxed{\textbf{(B)}\ 1:2}</math>   
  
 
-Lcz
 
-Lcz

Revision as of 11:35, 14 April 2020

Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are $6$ cm in diameter and $12$ cm high. Felicia buys cat food in cylindrical cans that are $12$ cm in diameter and $6$ cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?

$\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1$

Solution 1

Using the formula for the volume of a cylinder, we get Alex, $\pi108$, and Felicia, $\pi216$. We can quickly notice that $\pi$ cancels out on both sides, and that Alex's volume is $1/2$ of Felicia's leaving $1/2 = \boxed{1:2}$ as the answer.

~aopsav

Solution 2

Using the formula for the volume of a cylinder, we get that the volume of Alex's can is $3^2\cdot12\cdot\pi$, and that the volume of Felicia's can is $6^2\cdot6\cdot\pi$. Now we divide the volume of Alex's can by the volume of Felicia's can, so we get $\frac{1}{2}$, which is $\boxed{\textbf{(B)}\ 1:2}$ lol this is something no one should be able to do.-(Algebruh123)2020

Solution 3

The ratio of the numbers is $1/2$. Looking closely at the formula $r^2 * h * \pi$, we see that the $r * h * \pi$ will cancel, meaning that the ratio of them will be $\frac{1(2)}{2(2)}$ = $\boxed{\textbf{(B)}\ 1:2}$

-Lcz

See Also

2019 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

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