# 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}\ | + | 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>\ | + | 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\ | + | 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 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 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 $\pi\cdotr^2\cdoth$ (Error compiling LaTeX. ! Undefined control sequence.) so the volume of the water in the cylinder would be .

We can equate this two equations like this . We get and .

So the answer is