Difference between revisions of "1954 AHSME Problems/Problem 36"
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Let the distance the boat traveled be <math>d</math>. | Let the distance the boat traveled be <math>d</math>. | ||
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> | 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> | ||
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+ | ==See Also== | ||
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+ | {{AHSME 50p box|year=1954|num-b=35|num-a=37}} | ||
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+ | {{MAA Notice}} |
Latest revision as of 17:09, 22 April 2020
Contents
[hide]Problem 36
A boat has a speed of mph in still water. In a stream that has a current of 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:
Solution 1
WLOG, let the distance the boat travel be mile. Then the boat takes minutes, to travel down a mile, then to travel back up the river, the boat travels miles per hour, taking minutes to travel up the river. This gives an average speed of which as a ratio to is
Solution 2
Let the distance the boat traveled be . Then since , the average speed of the round trip is , or . 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 mph, so the ratio is , or
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 35 |
Followed by Problem 37 | |
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