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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==
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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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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?
  
 
<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>
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==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>.
  
 
The volume of a cylinder is <math>\pi \cdot r^2 \cdot h</math> so the volume of the water in the cylinder would be <math>24\cdot24\cdot\pi\cdot h</math>.
 
The volume of a cylinder is <math>\pi \cdot r^2 \cdot h</math> so the volume of the water in the cylinder would be <math>24\cdot24\cdot\pi\cdot h</math>.
  
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We can equate this two equations 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>.
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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>.
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So the answer is <math>\boxed{\textbf{(A)} ~1.5}.</math>
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~abhinavg0627
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==Solution 2==
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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 = \boxed{\textbf{(A)} ~1.5}.</math>
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~PureSwag
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==Video Solution by Punxsutawney Phil==
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https://youtube.com/watch?v=qpvS2PVkI8A&t=509s
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== Video Solution by OmegaLearn (3D Geometry - Cones and Cylinders) ==
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https://youtu.be/4JhZLAORb8c
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 +
~ pi_is_3.14
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 +
==Video Solution by Hawk Math==
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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
  
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So the answer is <math>1.5 = \boxed{\textbf{(A)}}.</math>
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==Video Solution (Under 2 min!)==
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https://youtu.be/6O32bUIvNEo
  
 +
~Education, the Study of Everything
  
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--abhinavg0627
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==See Also==
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{{AMC12 box|year=2021|ab=B|num-b=5|num-a=7}}
 +
{{AMC10 box|year=2021|ab=B|num-b=9|num-a=11}}
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{{MAA Notice}}

Latest revision as of 22:51, 18 July 2023

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

~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 (Under 2 min!)

https://youtu.be/6O32bUIvNEo

~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. AMC logo.png

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