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1950 AHSME Problems/Problem 50

Revision as of 17:03, 18 July 2015 by 1=2 (talk | contribs) (Added category: Rate Problems)

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

Assume that the two boats are traveling along the positive real number line, with the merchantman starting at the number $10$ and the privateer starting at the number $0$. After two hours the merchantman is at $26$ while the privateer is at $22$. When the top sail of the privateer is carried away, the speed of the merchantman is unaffected; he still travels at $8$ miles per hour. Therefore the privateer travels at $17\cdot \frac{8}{15}=\frac{136}{15}=9\frac{1}{15}$ miles per hour. The remaining time $t$ in hours it takes for the privateer to catch up to the merchantman satisfied the equation

\[26+8t=22+\frac{136}{15}t\]

Simplification yields the equation $\frac{16}{15}t=4$, which shows that $t=\frac{15}{4}=3\frac{3}{4}$. It therefore takes a total of $5\frac{3}{4}$ hours for the privateer to catch the merchantman, so this will happen at $\boxed{\textbf{(E)}\ 5\text{:}30\text{ p.m.}}$

See Also

1950 AHSC (Problems • Answer Key • Resources)
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