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== | ||
− | <math>\boxed{\textbf{(A)} | + | 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 cylinder is <math>\pir^2h</math> so the volume of the water in the cylinder would be <math>24\cdot24\cdot\pi\cdoth</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>. | ||
+ | |||
+ | So the answer is <math>1.5 = \boxed{\textbf{(A)}}.</math> |
Revision as of 18:44, 11 February 2021
Problem
An inverted cone with base radius and height is full of water. The water is poured into a tall cylinder whose horizontal base has radius of . What is the height in centimeters of the water in the cylinder?
Solution
The volume of a cone is $\frac{1}{3}\pir^2h$ (Error compiling LaTeX. Unknown error_msg) where is the base radius and is the height. The water completely fills up the cone so the volume of the water is .
The volume of a cylinder is $\pir^2h$ (Error compiling LaTeX. Unknown error_msg) so the volume of the water in the cylinder would be $24\cdot24\cdot\pi\cdoth$ (Error compiling LaTeX. Unknown error_msg).
We can equate this two equations like this $24\cdot24\cdot\pi\cdoth = 6\cdot144\pi$ (Error compiling LaTeX. Unknown error_msg). We get and .
So the answer is