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

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Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are <math>6</math> cm in diameter and <math>12</math> cm high. Felicia buys cat food in cylindrical cans that are <math>12</math> cm in diameter and <math>6</math> cm high. What is the ratio of the volume of one of Alex's cans to the volume of one of Felicia's cans?
 
Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are <math>6</math> cm in diameter and <math>12</math> cm high. Felicia buys cat food in cylindrical cans that are <math>12</math> cm in diameter and <math>6</math> cm high. What is the ratio of the volume of one of Alex's cans to the volume of one of Felicia's cans?
  
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<math>\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }(B)2:1\qquad\textbf{(E) }4:1</math>
+
<math>\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1</math>
  
 
==Solution 1==
 
==Solution 1==

Revision as of 16:59, 31 December 2022

Problem 9

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 of one of Alex's cans to the volume of 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

Video Solution

The Learning Royal : https://youtu.be/8njQzoztDGc

Video Solution 2

https://youtu.be/FDgcLW4frg8?t=2440

~ pi_is_3.14

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=G-gEdWP0S9M&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=10

Video Solution

https://youtu.be/FLT3iOKBC8c

~savannahsolver

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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