Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 8 Problems/Problem 9"
m (→‎Solution 1)
 
(One intermediate revision by one other user not shown)
Line 6: Line 6:
 
==Solution 1==
 
==Solution 1==
  
−
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.  
+
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{\textbf{(B)}\ 1:2}</math> as the answer.  
  
 
~aopsav
 
~aopsav
Line 69: Line 69:
 
~Education, the Study of Everything
 
~Education, the Study of Everything
  
−
==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
+
==Video Solution by The Power of Logic(1 to 25 Full Solution)==
 
https://youtu.be/Xm4ZGND9WoY
 
https://youtu.be/Xm4ZGND9WoY
  

Latest revision as of 22:56, 9 December 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 $1/2 = \boxed{\textbf{(B)}\ 1:2}$ 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 (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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_9&oldid=236645"