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

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

~ pi_is_3.14

## Video Solution by TheBeautyofMath

https://youtu.be/GYpAm8v1h-U?t=1068 (for AMC 10B)

https://youtu.be/kuZXQYHycdk (for AMC 12B)

~IceMatrix

~Interstigation