Difference between revisions of "2021 AMC 12B Problems/Problem 6"

Line 14: Line 14:
  
 
--abhinavg0627
 
--abhinavg0627
 +
 +
==Video Solution by Punxsutawney Phil==
 +
https://youtube.com/watch?v=qpvS2PVkI8A&t=509s
  
 
== Video Solution by OmegaLearn (3D Geometry - Cones and Cylinders) ==
 
== Video Solution by OmegaLearn (3D Geometry - Cones and Cylinders) ==

Revision as of 18:37, 12 February 2021

Problem

An inverted cone with base radius $12\mathrm{cm}$ and height $18\mathrm{cm}$ is full of water. The water is poured into a tall cylinder whose horizontal base has radius of $24\mathrm{cm}$. What is the height in centimeters of the water in the cylinder?

$\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6$

Solution

The volume of a cone is $\frac{1}{3} \cdot\pi \cdot r^2 \cdot h$ where $r$ is the base radius and $h$ is the height. The water completely fills up the cone so the volume of the water is $\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi$.

The volume of a cylinder is $\pi \cdot r^2 \cdot h$ so the volume of the water in the cylinder would be $24\cdot24\cdot\pi\cdot h$.

We can equate these two expressions because the water volume stays the same like this $24\cdot24\cdot\pi\cdot h = 6\cdot144\pi$. We get $4h = 6$ and $h=\frac{6}{4}$.

So the answer is $1.5 = \boxed{\textbf{(A)}}.$


--abhinavg0627

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=qpvS2PVkI8A&t=509s

Video Solution by OmegaLearn (3D Geometry - Cones and Cylinders)

https://youtu.be/4JhZLAORb8c

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=VzwxbsuSQ80

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png