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Difference between revisions of "2003 AMC 10B Problems/Problem 17"

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==Solution==
 
==Solution==
  
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Let <math>r</math> be the radius of both the cone and the sphere of ice cream. The cone can hold a maximum of <math>\pi r^2h/3</math> melted ice cream. The volume of the frozen ice cream is <math>4\pi r^3/3.</math> The volume of the melted ice cream is <math>75 \%</math> of that, which is <math>\pi r^3.</math> Since the melted ice cream fills the cone exactly,
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Let <math>r</math> be the radius of both the cone and the sphere of ice cream, and let <math>h</math> be the height of the cone. The cone can hold a maximum of <math>\pi r^2h/3</math> melted ice cream. The volume of the frozen ice cream is <math>4\pi r^3/3.</math> The volume of the melted ice cream is <math>75 \%</math> of that, which is <math>\pi r^3.</math> Since the melted ice cream fills the cone exactly,
  
 
<cmath>\begin{align*}\pi r^3&=\pi r^2h/3\\
 
<cmath>\begin{align*}\pi r^3&=\pi r^2h/3\\
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Then the ratio of the cone's height to its radius is <math>h:r=\boxed{\textbf{(B)}\ 3:1}</math>
 
Then the ratio of the cone's height to its radius is <math>h:r=\boxed{\textbf{(B)}\ 3:1}</math>
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==Video Solution by WhyMath==
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https://youtu.be/OQEfY3PweeA
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2003|ab=B|num-b=16|num-a=18}}
 
{{AMC10 box|year=2003|ab=B|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 15:43, 29 June 2021

Problem

An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies $75\%$ of the volume of the frozen ice cream. What is the ratio of the cone's height to its radius? (Note: a cone with radius $r$ and height $h$ has volume $\pi r^2 h / 3$ and a sphere with radius $r$ has volume $4 \pi r^3 / 3$.)

$\textbf{(A) } 2:1 \qquad\textbf{(B) } 3:1 \qquad\textbf{(C) } 4:1 \qquad\textbf{(D) } 16:3 \qquad\textbf{(E) } 6:1$

Solution

Let $r$ be the radius of both the cone and the sphere of ice cream, and let $h$ be the height of the cone. The cone can hold a maximum of $\pi r^2h/3$ melted ice cream. The volume of the frozen ice cream is $4\pi r^3/3.$ The volume of the melted ice cream is $75 \%$ of that, which is $\pi r^3.$ Since the melted ice cream fills the cone exactly,

\begin{align*}\pi r^3&=\pi r^2h/3\\ r&=h/3\\ 3&=h/r\end{align*}

Then the ratio of the cone's height to its radius is $h:r=\boxed{\textbf{(B)}\ 3:1}$

Video Solution by WhyMath

https://youtu.be/OQEfY3PweeA

~savannahsolver

See Also

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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