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2019 AMC 8 Problems/Problem 9: Difference between revisions

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Using the formula for the volume of a cylinder, we get that the volume of Alex's can is <math>3^2\cdot12\cdot\pi</math>, and that the volume of Felicia's can is <math>6^2\cdot6\cdot\pi</math>. Now we divide the volume of Alex's can by the volume of Felicia's can, so we get <math>\frac{1}{2}</math>, which is <math>\boxed{\textbf{(B)}\ 1:2}</math>                  ~~SmileKat32
Using the formula for the volume of a cylinder, we get that the volume of Alex's can is <math>3^2\cdot12\cdot\pi</math>, and that the volume of Felicia's can is <math>6^2\cdot6\cdot\pi</math>. Now we divide the volume of Alex's can by the volume of Felicia's can, so we get <math>\frac{1}{2}</math>, which is <math>\boxed{\textbf{(B)}\ 1:2}</math>                  ~~SmileKat32
==Solution 2==
The ratio of them is <math>1/2</math>, meaning that the ratio of them will be <math>1(2)/2(2)<cmath>=</cmath>\boxed{\textbf{(B)}\ 1:2}</math> 
-Lcz


==See Also==
==See Also==

Revision as of 15:40, 21 November 2019

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 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 $\boxed{\textbf{(B)}\ 1:2}$ ~~SmileKat32

Solution 2

The ratio of them is $1/2$, meaning that the ratio of them will be $1(2)/2(2)<cmath>=</cmath>\boxed{\textbf{(B)}\ 1:2}$

-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

These problems are copyrighted © by the Mathematical Association of America.

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