Q 116 - Intermediate Algebra

Question

The Jones family took a $12$-mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of canoe in still water and the rate of current.

Step-by-Step Solution

Answer

The rate of canoe in still water is $5$ mph.

The rate of current is $1$ mph.

Step 1. Given information.

Consider the given question,

Distance traveled by the Jones family one way is $12$ miles.

Time taken by the Jones family to ride down the river is $2$ hours and up the river is $3$ hours.

Assume $x$ to be the rate of the canoe in still water and $y$ to be the rate of the current.

Step 2. Form the equations.

Consider the given question,

When the family travels $12$ miles downstream in $2$ hours, then the current will help the canoe.

So, $$\begin{gathered} 2\left(x+y\right)=12 \\ x+y=6\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{i}\right) \end{gathered}$$.

When the travels $12$ miles upstream in $3$ hours, then the current will slow the canoe.

So, $$\begin{gathered} 3\left(x-y\right)=12 \\ x-y=4\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{ii}\right) \end{gathered}$$

Step 3. Solve the equations.

Add equations (i) and (ii),

$$\begin{gathered} \left(x+y\right)+\left(x-y\right)=6+4 \\ x+x=10 \\ 2x=10 \\ x=5 \end{gathered}$$

Substitute $x=5$ in equation (ii),

$$\begin{gathered} 5-y=4 \\ y=1 \end{gathered}$$

Step 4. Check the answers.

Substitute the values in equation (i),

$$\begin{gathered} 5+1=6 \\ 6=6 \end{gathered}$$

This is true.

Substitute the values in equation (ii),

$$\begin{gathered} 5-1=4 \\ 4=4 \end{gathered}$$

This is true.

The rate of canoe in still water is $5$ mph.

The rate of current is $1$ mph.