Difference between revisions of "1954 AHSME Problems/Problem 27"

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== Solution ==
 
== Solution ==
 
Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have <math>\frac{1}{3} \pi r^2 h= \frac{2}{3} \pi r^3 \implies h=2r\implies \frac{h}{r}=2</math> <math>\boxed{(\textbf{D})}</math>
 
Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have <math>\frac{1}{3} \pi r^2 h= \frac{2}{3} \pi r^3 \implies h=2r\implies \frac{h}{r}=2</math> <math>\boxed{(\textbf{D})}</math>
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==See Also==
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{{AHSME 50p box|year=1954|num-b=26|num-a=28}}
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{{MAA Notice}}

Latest revision as of 01:30, 28 February 2020

Problem 27

A right circular cone has for its base a circle having the same radius as a given sphere. The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:

$\textbf{(A)}\ \frac{1}{1}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{2}{1}\qquad\textbf{(E)}\ \sqrt{\frac{5}{4}}$

Solution

Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have $\frac{1}{3} \pi r^2 h= \frac{2}{3} \pi r^3 \implies h=2r\implies \frac{h}{r}=2$ $\boxed{(\textbf{D})}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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All AHSME Problems and Solutions


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