Difference between revisions of "2021 AMC 12A Problems/Problem 10"
Icematrix2 (talk | contribs) (Created page with "==Problem== These problems will not be posted until the 2021 AMC12A is released on Thursday, February 4, 2021. ==Solution== The solutions will be posted once the problems are...") |
Sugar rush (talk | contribs) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | + | Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are <math>3</math> and <math>6</math>, respectively. Into each cone is dropped a spherical marble of radius <math>1</math>, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone? | |
| + | |||
| + | <math>\textbf{(A) } 1\qquad\textbf{(B) } \frac{47}{43}\qquad\textbf{(C) } 2\qquad\textbf{(D) } \frac{40}{13}\qquad\textbf{(E) } 4\qquad</math> | ||
| + | |||
==Solution== | ==Solution== | ||
| − | The | + | The answer is <math>\boxed{\textbf{(E) } 4</math> |
| − | + | ||
| − | |||
==See also== | ==See also== | ||
{{AMC12 box|year=2021|ab=A|num-b=9|num-a=11}} | {{AMC12 box|year=2021|ab=A|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:07, 11 February 2021
Problem
Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are 
and 
, respectively. Into each cone is dropped a spherical marble of radius 
, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
Solution
The answer is $\boxed{\textbf{(E) } 4$ (Error compiling LaTeX. Unknown error_msg)
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
