2021 AMC 12B Problems/Problem 6
- The following problem is from both the 2021 AMC 10B #10 and 2021 AMC 12B #6, so both problems redirect to this page.
Contents
Problem
An inverted cone with base radius and height is full of water. The water is poured into a tall cylinder whose horizontal base has radius of . What is the height in centimeters of the water in the cylinder?
Solution 1
The volume of a cone is where is the base radius and is the height. The water completely fills up the cone so the volume of the water is .
The volume of a cylinder is so the volume of the water in the cylinder would be .
We can equate these two expressions because the water volume stays the same like this . We get and .
So the answer is
~abhinavg0627
Solution 2
The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has of the volume of the cylinder, and so the height is divided by . Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since ).
Therefore, the height is divided by and divided by , which is
~PureSwag
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=qpvS2PVkI8A&t=509s
Video Solution by OmegaLearn (3D Geometry - Cones and Cylinders)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
Video Solution by TheBeautyofMath
https://youtu.be/GYpAm8v1h-U?t=1068 (for AMC 10B)
https://youtu.be/kuZXQYHycdk (for AMC 12B)
~IceMatrix
Video Solution by Interstigation
https://youtu.be/DvpN56Ob6Zw?t=897
~Interstigation
Video Solution
~Education, the Study of Everything
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 5 |
Followed by Problem 7 |
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 |
2021 AMC 10B (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.