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 "2010 AMC 12A Problems/Problem 7"

m
Line 9: Line 9:
 
Note: The fact that <math>1\text{ L}=1000\text{ cm}^3</math> doesn't matter since only the ratios are important.
 
Note: The fact that <math>1\text{ L}=1000\text{ cm}^3</math> doesn't matter since only the ratios are important.
  
−
==Video Solution==
+
==Video Solution by the Beauty of Math==
 
https://youtu.be/kU70k1-ONgM?t=154
 
https://youtu.be/kU70k1-ONgM?t=154
−
 
−
~IceMatrix
 
  
 
== See also ==
 
== See also ==

Revision as of 20:19, 6 December 2020

Problem

Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?

$\textbf{(A)}\ 0.04 \qquad \textbf{(B)}\ \frac{0.4}{\pi} \qquad \textbf{(C)}\ 0.4 \qquad \textbf{(D)}\ \frac{4}{\pi} \qquad \textbf{(E)}\ 4$

Solution

The water tower holds $\frac{100000}{0.1} = 1000000$ times more water than Logan's miniature. The volume of a sphere is: $V=\dfrac{4}{3}\pi r^3$. Since we are comparing the heights (m), we should compare the radii (m) to find the ratio. Since, the radius is cubed, Logan should make his tower $\sqrt[3]{1000000} = 100$ times shorter than the actual tower. This is $\frac{40}{100} = \boxed{0.4}$ meters high, or choice $\textbf{(C)}$.

Note: The fact that $1\text{ L}=1000\text{ cm}^3$ doesn't matter since only the ratios are important.

Video Solution by the Beauty of Math

https://youtu.be/kU70k1-ONgM?t=154

See also

2010 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_12A_Problems/Problem_7&oldid=139172"