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

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{{duplicate|[[2021 AMC 10B Problems#Problem 10|2021 AMC 10B #10]] and [[2021 AMC 12B Problems#Problem 6|2021 AMC 12B #6]]}}
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==Problem==
 
==Problem==
 
An inverted cone with base radius <math>12  \mathrm{cm}</math> and height <math>18  \mathrm{cm}</math> is full of water. The water is poured into a tall cylinder whose horizontal base has radius of <math>24  \mathrm{cm}</math>. What is the height in centimeters of the water in the cylinder?
 
An inverted cone with base radius <math>12  \mathrm{cm}</math> and height <math>18  \mathrm{cm}</math> is full of water. The water is poured into a tall cylinder whose horizontal base has radius of <math>24  \mathrm{cm}</math>. What is the height in centimeters of the water in the cylinder?
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<math>\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6</math>
 
<math>\textbf{(A)} ~1.5 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~4 \qquad\textbf{(D)} ~4.5 \qquad\textbf{(E)} ~6</math>
  
==Solution==
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==Solution 1==
 
The volume of a cone is <math>\frac{1}{3} \cdot\pi \cdot r^2 \cdot h</math> where <math>r</math> is the base radius and <math>h</math> is the height. The water completely fills up the cone so the volume of the water is <math>\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi</math>.
 
The volume of a cone is <math>\frac{1}{3} \cdot\pi \cdot r^2 \cdot h</math> where <math>r</math> is the base radius and <math>h</math> is the height. The water completely fills up the cone so the volume of the water is <math>\frac{1}{3}\cdot18\cdot144\pi = 6\cdot144\pi</math>.
  
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--abhinavg0627
 
--abhinavg0627
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==Solution 2 (ratios)==
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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>).
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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>
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-PureSwag
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==
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==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
 
https://www.youtube.com/watch?v=VzwxbsuSQ80
 
https://www.youtube.com/watch?v=VzwxbsuSQ80
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==Video Solution by TheBeautyofMath==
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https://youtu.be/GYpAm8v1h-U?t=1068 (for AMC 10B)
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https://youtu.be/kuZXQYHycdk (for AMC 12B)
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~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/DvpN56Ob6Zw?t=897
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~Interstigation
  
 
==See Also==
 
==See Also==

Latest revision as of 08:38, 2 March 2021

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


--abhinavg0627

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 $\frac{1}{3}$ 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 $18 \div 3 \div 4 = 1.5 = \boxed{\textbf{(A)}}.$

-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

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

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