Difference between revisions of "2021 AMC 12A Problems/Problem 10"
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<math>\textbf{(A) }1 \qquad \textbf{(B) }\frac{47}{43} \qquad \textbf{(C) }2 \qquad \textbf{(D) }\frac{40}{13} \qquad \textbf{(E) }4</math> | <math>\textbf{(A) }1 \qquad \textbf{(B) }\frac{47}{43} \qquad \textbf{(C) }2 \qquad \textbf{(D) }\frac{40}{13} \qquad \textbf{(E) }4</math> | ||
− | ==Solution | + | ==Solution (Fraction Trick)== |
<b>Initially:</b> | <b>Initially:</b> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Quick and dirty)== | ||
+ | The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii 3 and 6 and infinitely large height. Then the base area of the wide cylinder is 4 times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise <math>\boxed{\textbf{(E) } 4}</math> times as much. | ||
+ | |||
+ | -scrabbler94 | ||
+ | |||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== |
Revision as of 14:14, 12 February 2021
Contents
Problem
Two right circular cones with vertices facing down as shown in the figure below contains the same amount of liquid. The radii of the tops of the liquid surfaces are cm and cm. Into each cone is dropped a spherical marble of radius cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
Solution (Fraction Trick)
Initially:
For the narrow cone liquid, the base radius is Let its height be By similar triangles, the ratio of base radius to height is The volume is
For the wide cone liquid, the base radius is Let its height be By similar triangles, the ratio of base radius to height is The volume is
Equating initial volumes gives from which
Finally:
For the narrow cone liquid, the base radius is where By similar triangles, it follows that its height is and its volume is
For the wide cone liquid, the base radius is where By similar triangles, it follows that its height is and its volume is
Equating final volumes simplifies to or
Lastly, the fraction we seek is
PS:
1. This problem uses the following fraction trick:
For unequal positive numbers and if then
Quick Proof
From we know that and . Therefore,
2. Most of the steps can be done through mental math. Also, drawing a table showing the initial and final measurements can effectively organize the work.
~MRENTHUSIASM
Solution 2 (Quick and dirty)
The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii 3 and 6 and infinitely large height. Then the base area of the wide cylinder is 4 times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise times as much.
-scrabbler94
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (Similar Triangles, 3D Geometry - Cones)
~ pi_is_3.14
See also
2021 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 |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.