Difference between revisions of "1971 AHSME Problems/Problem 18"
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== Solution == | == Solution == | ||
− | <math>\boxed{\textbf{(D) }2:1}</math>. | + | Let the boat travel over still water at rate <math>r</math>. Then, it travels downstream at rate <math>r+3</math> and upstream at rate <math>r-3</math>. The distance travelled downstream can be expressed as <math>(r+3)t</math>, where <math>t</math> is the time taken to travel downstream. Because the total time taken is <math>1</math> hour, the distance travelled upstream can be represented as <math>(r-3)(1-t)</math>. Because both distances are the same (<math>4</math> miles), we can equate the two expressions to solve for <math>t</math> in terms of <math>r</math>: |
+ | \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, <math>(r+3)t</math>, is <math>4</math> miles, we can substitute the above expression for <math>t</math> to solve for <math>r</math>: | ||
+ | \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 <math>r>0</math>, <math>r=9</math>. Thus, the ratio of the downstream rate <math>r+3</math> to the upstream rate <math>r-3</math> is <math>\tfrac{9+3}{9-3}=\tfrac{12}6=\boxed{\textbf{(D) }2:1}</math>. | ||
== See Also == | == See Also == |
Latest revision as of 09:33, 5 August 2024
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
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