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

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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==
 
==Solution==
−
The volume of a cone is <math>\frac{1}{3}\pir^2h</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>\pir^2h</math> so the volume of the water in the cylinder would be <math>24\cdot24\cdot\pi\cdoth</math>.
+
The volume of a cylinder is <math>\pi\cdotr^2\cdoth</math> so the volume of the water in the cylinder would be <math>24\cdot24\cdot\pi\cdot h</math>.
  
−
We can equate this two equations like this <math>24\cdot24\cdot\pi\cdoth = 6\cdot144\pi</math>. We get <math>4h = 6</math> and <math>h=\frac{6}{4}</math>.
+
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>.
  
 
So the answer is <math>1.5 = \boxed{\textbf{(A)}}.</math>
 
So the answer is <math>1.5 = \boxed{\textbf{(A)}}.</math>

Revision as of 19:45, 11 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 $\pi\cdotr^2\cdoth$ (Error compiling LaTeX. Unknown error_msg) so the volume of the water in the cylinder would be $24\cdot24\cdot\pi\cdot h$.

We can equate this two equations like this $24\cdot24\cdot\pi\cdot h = 6\cdot144\pi$. We get $4h = 6$ and $h=\frac{6}{4}$.

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

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