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Difference between revisions of "1971 AHSME Problems/Problem 18"

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== Solution ==
 
== Solution ==
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<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 08:33, 5 August 2024

Problem

The current in a river is flowing steadily at $3$ miles per hour. A motor boat which travels at a constant rate in still water goes downstream $4$ 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

$\textbf{(A) }4:3\qquad \textbf{(B) }3:2\qquad \textbf{(C) }5:3\qquad \textbf{(D) }2:1\qquad \textbf{(E) }5:2$

Solution

Let the boat travel over still water at rate $r$. Then, it travels downstream at rate $r+3$ and upstream at rate $r-3$. The distance travelled downstream can be expressed as $(r+3)t$, where $t$ is the time taken to travel downstream. Because the total time taken is $1$ hour, the distance travelled upstream can be represented as $(r-3)(1-t)$. Because both distances are the same ($4$ miles), we can equate the two expressions to solve for $t$ in terms of $r$: \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, $(r+3)t$, is $4$ miles, we can substitute the above expression for $t$ to solve for $r$: \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 $r>0$, $r=9$. Thus, the ratio of the downstream rate $r+3$ to the upstream rate $r-3$ is $\tfrac{9+3}{9-3}=\tfrac{12}6=\boxed{\textbf{(D) }2:1}$.

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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