Given:
The Jones family took a 20-mile canoe ride down the Indian River [with the current] in 4 hours and The next day, the return trip back up the river [against the current] took 10 hours.
To find:
the rate of the canoe in still water and the rate of the current.
Solution:
Let x be the rate canoe in still water and y be the rate of current.
So, the rate of canoe with the current will be (x + y). So, by the given condition
[tex]\begin{gathered} 4(x+y)=20 \\ x+y=5\ldots(a) \end{gathered}[/tex]The rate of canoe against the current will be (x - y). So, by the given condition:
[tex]\begin{gathered} 10(x-y)=20 \\ x-y=2\ldots(b) \end{gathered}[/tex]Add (a) and (b), to get:
[tex]\begin{gathered} x+y+x-y=5+2 \\ 2x=7 \\ x=\frac{7}{2} \\ x=3.5 \end{gathered}[/tex]Substitute x = 3.5 in (a),
[tex]\begin{gathered} 3.5+y=5 \\ y=5-3.5 \\ y=1.5 \end{gathered}[/tex]Thus, the rate of canoe in still water is 3.5 mph and the rate of current is 1.5 mph.
