Difference between revisions of "1951 AHSME Problems/Problem 30"

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==Problem==
 
==Problem==
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If two poles <math>20''</math> and <math>80''</math> high are <math>100''</math> apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
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<math> \textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these} </math>
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==Solution==
 
==Solution==
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The two pole formula says this height is the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is <math>\frac1{\frac1{20}+\frac1{80}}</math>, or <math>\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}</math>.
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== See Also ==
 
== See Also ==
 
{{AHSME 50p box|year=1951|num-b=29|num-a=31}}  
 
{{AHSME 50p box|year=1951|num-b=29|num-a=31}}  

Revision as of 15:34, 19 April 2014

Problem

If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:

$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$

Solution

The two pole formula says this height is the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is $\frac1{\frac1{20}+\frac1{80}}$, or $\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 31
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All AHSME Problems and Solutions

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