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

m (Solution 2)
 
(15 intermediate revisions by 7 users not shown)
Line 1: Line 1:
==Problem 9==
+
==Problem==
 
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?
  
Line 6: Line 6:
 
==Solution 1==
 
==Solution 1==
  
Using the formula for the volume of a cylinder, we get Alex, <math>\pi108</math>, and Felicia, <math>\pi216</math>. We can quickly notice that <math>\pi</math> cancels out on both sides, and that Alex's volume is <math>1/2</math> of Felicia's leaving <math>1/2 = \boxed{1:2}</math> as the answer.  
+
Using the formula for the volume of a cylinder, we get Alex, <math>108\pi</math>, and Felicia, <math>216\pi</math>. We can quickly notice that <math>\pi</math> cancels out on both sides and that Alex's volume is <math>1/2</math> of Felicia's leaving <math>1/2 = \boxed{1:2}</math> as the answer.  
  
 
~aopsav
 
~aopsav
Line 21: Line 21:
  
 
-Lcz
 
-Lcz
 +
 +
 +
== Solution 4 ==
 +
 +
The second can is <math>\cdot 2</math> size in each of 2 dimensions, and <math>\cdot 1/2</math> size in 1 dimension. <math> 2^2/2
 +
= \boxed{\textbf{(B)}\ 1:2}</math>.
 +
 +
~oinava
 +
 +
== Solution 5 ==
 +
 +
Without calculating much, you can do
 +
(<math>\pi ra^2) \cdot ha</math> <-- which is Alex's volume, with ra being Alex's radius<math> (1/2 \cdot</math> diameter), and <math>ha</math> being her cylinders height<math>
 +
(\pi rf^2) \cdot hf <-- </math>which is Felicia's volume, with <math>rf</math> being Felicia's radius, and <math>hf</math> being her cylinders height.
 +
Since we need the ratio between Alexa's and Felicias, we can do <math>(\pi ra^2)\cdot ha/(\pi rf^2)\cdot hf</math> The <math>\pi</math> cancel out, then substitute back in the numbers, which gives you:
 +
 +
<math>(3^2 \cdot 12)/(6^2 \cdot 6) = (9 \cdot 12)/(36 \cdot 6) = 18/36 = 1/2 = 1:2</math>
 +
 +
-wahahaqueenie
  
 
== Video Solution ==
 
== Video Solution ==
 +
 +
==Video Solution by Math-X (Extremely simple approach!!!)==
 +
https://youtu.be/IgpayYB48C4?si=wsD8LhZK8hsWd9wu&t=2773
 +
 +
~Math-X
 +
 +
 
The Learning Royal : https://youtu.be/8njQzoztDGc
 
The Learning Royal : https://youtu.be/8njQzoztDGc
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==
Line 37: Line 63:
  
 
~savannahsolver
 
~savannahsolver
 +
 +
== Video Solution ==
 +
https://youtu.be/ChwC1Hnk_pw
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 +
https://youtu.be/Xm4ZGND9WoY
 +
 +
~Hayabusa1
  
 
==See also==
 
==See also==

Latest revision as of 09:31, 9 November 2024

Problem

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, $108\pi$, and Felicia, $216\pi$. 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 $\boxed{\textbf{(B)}\ 1:2}$.

-(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


Solution 4

The second can is $\cdot 2$ size in each of 2 dimensions, and $\cdot 1/2$ size in 1 dimension. $2^2/2 = \boxed{\textbf{(B)}\ 1:2}$.

~oinava

Solution 5

Without calculating much, you can do ($\pi ra^2) \cdot ha$ <-- which is Alex's volume, with ra being Alex's radius$(1/2 \cdot$ diameter), and $ha$ being her cylinders height$(\pi rf^2) \cdot hf <--$which is Felicia's volume, with $rf$ being Felicia's radius, and $hf$ being her cylinders height. Since we need the ratio between Alexa's and Felicias, we can do $(\pi ra^2)\cdot ha/(\pi rf^2)\cdot hf$ The $\pi$ cancel out, then substitute back in the numbers, which gives you:

$(3^2 \cdot 12)/(6^2 \cdot 6) = (9 \cdot 12)/(36 \cdot 6) = 18/36 = 1/2 = 1:2$

-wahahaqueenie

Video Solution

Video Solution by Math-X (Extremely simple approach!!!)

https://youtu.be/IgpayYB48C4?si=wsD8LhZK8hsWd9wu&t=2773

~Math-X


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

Video Solution by OmegaLearn

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

Video Solution

https://youtu.be/ChwC1Hnk_pw

~Education, the Study of Everything

Video Solution by The Power of Logic(1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

See also

2019 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png