Question

A boat can travel 30 miles against a current in the same time that it can travel 90 miles with the current. The rate of the current is 3 mph. Find the rate of the boat in Stillwater.

Answer 1

Solution:

Let r represent the rate(speed) of the boat.

Given that the rate of the current is 3 mph, this implies that

$\begin{gathered} speed\text{ of the boat with current =\lparen r+3\rparen mph} \\ speed\text{ of the boat against current = \lparen r-3\rparen mph} \end{gathered}$

Recall that

$\begin{gathered} speed(rate)\text{ = }\frac{d}{t} \\ where \\ d\Rightarrow distance\text{ covered/traveled} \\ t\Rightarrow time\text{ taken} \end{gathered}$

Thus, the time taken to travel 30 miles against the current is evaluated as

$\begin{gathered} r-3=\frac{30}{t} \\ cross-multiply, \\ t(r-3)=30 \\ \Rightarrow t=\frac{30}{r-3}\text{ ---- equation 1} \end{gathered}$

In a similar manner, the time taken to travel 90 miles with the current is evaluated as

$\begin{gathered} r+3=\frac{90}{t} \\ cross-multiply, \\ t(r+3)=90 \\ \Rightarrow t=\frac{90}{r+3}\text{ ----- equation 2} \end{gathered}$

Given that the boat can cover these distances at the same time, this implies that we equate the time in equations 1 and 2.

Thus,

$\frac{30}{r-3}=\frac{90}{r+3}$

To solve for r, we cross-multiply

$30(r+3)=90(r-3)$

open parentheses,

$30r+90=90r-270$

collect like terms,

$\begin{gathered} 30r-90r=-270-90 \\ \Rightarrow-60r=-360 \end{gathered}$

divide both sides by the coefficient of r.

the coefficient of r is -60.

thus,

$\begin{gathered} -\frac{60r}{-60}=-\frac{360}{-60} \\ \Rightarrow r=6\text{ mph} \end{gathered}$

Hence, the rate of the boat in still water is

$6\text{ mph}$