1971 AHSME Problems/Problem 18
Problem
The current in a river is flowing steadily at 
miles per hour. A motor boat which travels at a constant rate in still
water goes downstream 
miles and then returns to its starting point. The trip takes one hour, excluding the time spent
in turning the boat around. The ratio of the downstream to the upstream rate is
Solution
Let the boat travel over still water at rate 
. Then, it travels downstream at rate 
and upstream at rate 
. The distance travelled downstream can be expressed as 
, where 
is the time taken to travel downstream. Because the total time taken is 
hour, the distance travelled upstream can be represented as 
. Because both distances are the same ( miles), we can equate the two expressions to solve for in terms of :
\begin{align*}
(r+3)t &= (r-3)(1-t) \\
rt+3t &= r-rt-3+3t \\
2rt &= r-3 \\
t &= \frac{r-3}{2r} \\
\end{align*}
Recalling that the distance travelled downstream, , is miles, we can substitute the above expression for to solve for :
\begin{align*}
(r+3)t &= 4 \\
(r+3)\frac{r-3}{2r} &= 4 \\
r^2-9 &= 8r \\
r^2-8r-9 &= 0 \\
(r-9)(r+1) &= 0 \\
\end{align*}
Because 
, 
. Thus, the ratio of the downstream rate to the upstream rate is 
.
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
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