Difference between revisions of "2019 AMC 8 Problems/Problem 9"
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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 | ||
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+ | ==Solution 2== | ||
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+ | 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> | ||
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+ | -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 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 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 ~~SmileKat32
Solution 2
The ratio of them is , 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 |
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