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

(Video Solution by Punxsutawney Phil)
m
Line 13: Line 13:
 
We can equate these two expressions because the water volume stays the same like this <math>24\cdot24\cdot\pi\cdot h = 6\cdot144\pi</math>. We get <math>4h = 6</math> and <math>h=\frac{6}{4}</math>.
 
We can equate these two expressions because the water volume stays the same like this <math>24\cdot24\cdot\pi\cdot h = 6\cdot144\pi</math>. We get <math>4h = 6</math> and <math>h=\frac{6}{4}</math>.
  
So the answer is <math>1.5 = \boxed{\textbf{(A)}}.</math>
+
So the answer is <math>\boxed{\textbf{(A)} ~1.5}.</math>
  
 +
-abhinavg0627
  
--abhinavg0627
+
==Solution 2==
 
 
==Solution 2 (ratios)==
 
 
The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has <math>\frac{1}{3}</math> of the volume of the cylinder, and so the height is divided by <math>3</math>. Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since <math>2^2 = 4</math>).
 
The water completely fills up the cone. For now, assume the radius of both cone and cylinder are the same. Then the cone has <math>\frac{1}{3}</math> of the volume of the cylinder, and so the height is divided by <math>3</math>. Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since <math>2^2 = 4</math>).
  
Therefore, the height is divided by <math>3</math> and divided by <math>4</math>, which is <math>18 \div 3 \div 4 = 1.5 = \boxed{\textbf{(A)}}.</math>
+
Therefore, the height is divided by <math>3</math> and divided by <math>4</math>, which is <math>18 \div 3 \div 4 = 1.5 = \boxed{\textbf{(A)} ~1.5}.</math>
 
 
-PureSwag
 
 
 
==Video Solution 1==
 
https://youtu.be/6O32bUIvNEo
 
 
 
~Education, the Study of Everything
 
 
 
 
 
  
 +
~PureSwag
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==
Line 50: Line 41:
  
 
~IceMatrix
 
~IceMatrix
 +
 
==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/DvpN56Ob6Zw?t=897
 
https://youtu.be/DvpN56Ob6Zw?t=897
  
 
~Interstigation
 
~Interstigation
 +
 +
==Video Solution==
 +
https://youtu.be/6O32bUIvNEo
 +
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==

Revision as of 02:18, 19 August 2022

The following problem is from both the 2021 AMC 10B #10 and 2021 AMC 12B #6, so both problems redirect to this page.

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 1

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 $\boxed{\textbf{(A)} ~1.5}.$

-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 $\frac{1}{3}$ of the volume of the cylinder, and so the height is divided by $3$. Then, from the problem statement, the radius is doubled, meaning the area of the base is quadrupled (since $2^2 = 4$).

Therefore, the height is divided by $3$ and divided by $4$, which is $18 \div 3 \div 4 = 1.5 = \boxed{\textbf{(A)} ~1.5}.$

~PureSwag

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

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

https://youtu.be/6O32bUIvNEo

~Education, the Study of Everything

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