Difference between revisions of "2010 AMC 12A Problems/Problem 7"
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== Solution == | == Solution == | ||
− | The water tower holds <math>\frac{100000}{0.1} = 1000000</math> times more water than Logan's miniature. | + | The water tower holds <math>\frac{100000}{0.1} = 1000000</math> times more water than Logan's miniature. The volume of a sphere is: <math>V=\dfrac{4}{3}\pi r^3</math>. 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 <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>. |
+ | |||
+ | Note: The fact that <math>1\text{ L}=1000\text{ cm}^3</math> 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 == | ||
+ | {{AMC12 box|year=2010|num-b=6|num-a=8|ab=A}} | ||
+ | {{AMC10 box|year=2010|num-b=11|num-a=13|ab=A}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 17:08, 28 June 2021
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?
Solution
The water tower holds times more water than Logan's miniature. The volume of a sphere is: . 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 times shorter than the actual tower. This is meters high, or choice .
Note: The fact that 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.