Difference between revisions of "2010 AMC 12A Problems/Problem 7"

m (apologies for the testing)
(Solution)
Line 6: Line 6:
 
== Solution ==
 
== Solution ==
 
The water tower holds <math>\frac{100000}{0.1} = 1000000</math> times more water than Logan's miniature. Therefore, Logan should make his tower <math>\sqrt[3]{1000000} = 100</math> times shorter than the actual tower. This is <math>\frac{40}{100} = \boxed{0.4}</math> meters high, or choice <math>\textbf{(C)}</math>.
 
The water tower holds <math>\frac{100000}{0.1} = 1000000</math> times more water than Logan's miniature. Therefore, Logan should make his tower <math>\sqrt[3]{1000000} = 100</math> times shorter than the actual tower. This is <math>\frac{40}{100} = \boxed{0.4}</math> meters high, or choice <math>\textbf{(C)}</math>.
 +
 +
Also, the fact that <math>1 L=1dm^3</math> doesn't matter since only the ratios are important.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=6|num-a=8|ab=A}}
 
{{AMC12 box|year=2010|num-b=6|num-a=8|ab=A}}

Revision as of 09:02, 5 August 2012

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. Therefore, 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 $\textbf{(C)}$.

Also, the fact that $1 L=1dm^3$ doesn't matter since only the ratios are important.

See also

2010 AMC 12A (ProblemsAnswer KeyResources)
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