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Difference between revisions of "1954 AHSME Problems/Problem 36"
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== Solution 2 ==
 
== Solution 2 ==
 
Let the distance the boat traveled be <math>d</math>.
 
Let the distance the boat traveled be <math>d</math>.
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Then since <math>rate=\frac{distance}{time}</math>, the average speed of the round trip is <math>\frac{2d}{\frac{d}{10}+\frac{d}{20}}</math>, or <math>\frac{40}{3}</math>. Note that this means that the round trip speed is the same no matter how long the trip. The speed of the boat in still water is <math>15</math> mph, so the ratio is <math>\frac{\frac{40}{3}}{15}</math>, or <math>\frac{8}{9} \implies </math>\boxed{(\textbf{C})}$
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Then since <math>rate=\frac{distance}{time}</math>, the average speed of the round trip is <math>\frac{2d}{\frac{d}{10}+\frac{d}{20}}</math>, or <math>\frac{40}{3}</math>. Note that this means that the round trip speed is the same no matter how long the trip. The speed of the boat in still water is <math>15</math> mph, so the ratio is <math>\frac{\frac{40}{3}}{15}</math>, or <math>\frac{8}{9} \implies \boxed{(\textbf{C})}</math>

Revision as of 17:17, 15 April 2017

Problem 36

A boat has a speed of $15$ mph in still water. In a stream that has a current of $5$ mph it travels a certain distance downstream and returns. The ratio of the average speed for the round trip to the speed in still water is:

$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{1}{1}\qquad\textbf{(C)}\ \frac{8}{9}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{9}{8}$

Solution 1

WLOG, let the distance the boat travel be $1$ mile. Then the boat takes $\frac{1}{15+5}=\frac{1}{20}=3$ minutes, to travel down a mile, then to travel back up the river, the boat travels $15-5=10$ miles per hour, taking $\frac{1}{10}=\frac{1}{10}=6$ minutes to travel up the river. This gives an average speed of $\frac{20(3)+10(6)}{3+6}=\frac{120}{9}=\frac{40}{3}$ which as a ratio to $15$ is $\frac{\frac{5\cdot 8}{3}}{3\cdot 5}\implies\frac{8}{9}$ $\boxed{(\textbf{C})}$

Solution 2

Let the distance the boat traveled be $d$. Then since $rate=\frac{distance}{time}$, the average speed of the round trip is $\frac{2d}{\frac{d}{10}+\frac{d}{20}}$, or $\frac{40}{3}$. Note that this means that the round trip speed is the same no matter how long the trip. The speed of the boat in still water is $15$ mph, so the ratio is $\frac{\frac{40}{3}}{15}$, or $\frac{8}{9} \implies \boxed{(\textbf{C})}$

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