Difference between revisions of "2019 AMC 8 Problems/Problem 9"
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+ | ==Problem 9== | ||
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 one of Alex's cans to the volume 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 one of Alex's cans to the volume one of Felicia's cans? | ||
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==Solution 1== | ==Solution 1== | ||
− | Using the formula for the volume of a cylinder, we get | + | 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. |
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+ | ~aopsav | ||
==Solution 2== | ==Solution 2== | ||
− | The ratio of the numbers is <math>1/2</math>. Looking closely at the formula <math>r^2 * h * \pi</math>, we see that the <math>r * h * \pi</math> will cancel, meaning that the ratio of them will be <math>1(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> |
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+ | lol this is something no one should be able to do.-(Algebruh123)2020 | ||
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+ | ==Solution 3== | ||
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+ | The ratio of the numbers is <math>1/2</math>. Looking closely at the formula <math>r^2 * h * \pi</math>, we see that the <math>r * h * \pi</math> will cancel, meaning that the ratio of them will be <math>\frac{1(2)}{2(2)}</math> = <math>\boxed{\textbf{(B)}\ 1:2}</math> | ||
-Lcz | -Lcz | ||
− | ==See | + | == Video Solution == |
+ | 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 | ||
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+ | ==See also== | ||
{{AMC8 box|year=2019|num-b=8|num-a=10}} | {{AMC8 box|year=2019|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:49, 23 April 2021
Contents
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
lol this is something no one should be able to do.-(Algebruh123)2020
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
Video Solution
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
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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