Difference between revisions of "1950 AHSME Problems/Problem 50"

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\textbf{(D)}\ 2\text{:}45\text{ p.m.} \qquad
 
\textbf{(D)}\ 2\text{:}45\text{ p.m.} \qquad
 
\textbf{(E)}\ 5\text{:}30\text{ p.m.}</math>
 
\textbf{(E)}\ 5\text{:}30\text{ p.m.}</math>
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==Solution==
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{{solution}}
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==See Also==
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{{AHSME 50p box|year=1950|num-b=49|after=Last Question}}
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[[Category:Introductory Algebra Problems]]

Revision as of 07:43, 29 April 2012

Problem

A privateer discovers a merchantman $10$ miles to leeward at 11:45 a.m. and with a good breeze bears down upon her at $11$ mph, while the merchantman can only make $8$ mph in her attempt to escape. After a two hour chase, the top sail of the privateer is carried away; she can now make only $17$ miles while the merchantman makes $15$. The privateer will overtake the merchantman at:

$\textbf{(A)}\ 3\text{:}45\text{ p.m.} \qquad \textbf{(B)}\ 3\text{:}30\text{ p.m.} \qquad \textbf{(C)}\ 5\text{:}00\text{ p.m.} \qquad \textbf{(D)}\ 2\text{:}45\text{ p.m.} \qquad \textbf{(E)}\ 5\text{:}30\text{ p.m.}$

Solution

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See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 49
Followed by
Last Question
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All AHSME Problems and Solutions