Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 8: Line 8:
 
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>.
  
−
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>.
+
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>.
  
 
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 18:46, 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 $\pi \cdot r^2 \cdot h$ 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_6&oldid=145425"