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Difference between revisions of "2010 AMC 12A Problems/Problem 7"
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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>.
  
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Also, the fact that <math>1\text{ L}=1\text{ dm}^3</math> doesn't matter since only the ratios are important.
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Also, the fact that <math>1\text{ L}=1\text{ cm}^3</math> doesn't matter since only the ratios are important.
  
 
== See also ==
 
== See also ==

Revision as of 19:12, 31 July 2016

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 $1\text{ L}=1\text{ cm}^3$ doesn't matter since only the ratios are important.

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

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