Difference between revisions of "2019 AMC 8 Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
− | 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> lol this is something no one should be able to do.-(Algebruh123)2020 | + | 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> |
+ | <math>\small{lol this is something no one should be able to do.-(Algebruh123)2020}</math> | ||
==Solution 3== | ==Solution 3== |
Revision as of 18:56, 5 July 2020
Problem 9
Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are cm in diameter and cm high. Felicia buys cat food in cylindrical cans that are cm in diameter and cm high. What is the ratio of the volume one of Alex's cans to the volume one of Felicia's cans?
Solution 1
Using the formula for the volume of a cylinder, we get Alex, , and Felicia, . We can quickly notice that cancels out on both sides, and that Alex's volume is of Felicia's leaving 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 , and that the volume of Felicia's can is . Now we divide the volume of Alex's can by the volume of Felicia's can, so we get , which is
Solution 3
The ratio of the numbers is . Looking closely at the formula , we see that the will cancel, meaning that the ratio of them will be =
-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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.