Q. 416 - Intermediate Algebra

Question

Greg is paddling his canoe

upstream, against the current, to

a fishing spot 10 miles away. If he

paddles upstream for 2.5 hours

and his return trip takes 1.25

hours, find the speed of the current

and his paddling speed in still

water.

Step-by-Step Solution

Answer

The speed of boat is $6 mi/hr$ and the speed of the stream is $2 mi/hr$.

Step 1. Given information.

Fishing spot $=10miles$.

Time to go upstream $=2.5hours$

Time to go downstream $=1.25hour$

Step 2. Write the concept.

Let us assume speed of boat is $x mi/hr$ and speed of stream is $y mi/hr$.

To go upstream speed of boat will be $x-y mi/hr$ while to go downstream speed of boat will be $x+y mi/hr.$

Step 3. Calculation.

Given equation:

$$\begin{gathered} Dis\tan ce=Speed \times Time \\ 10 = (x-y)\times 2.5 ..........\left(1\right) \\ 10= (x+y)\times 1.25 ..........\left(2\right) \end{gathered}$$

Simplifying equation (1) we get,

$$\begin{gathered} x-y=4 ...........\left(3\right) \\ x+y=8 ...........\left(4\right) \end{gathered}$$

Adding equation $\left(3\right)$ and $\left(4\right)$ we get,

Solve for $x$,

$$\begin{gathered} 2x=12 \\ x=6 \end{gathered}$$

Put value of x in equation $\left(1\right)$ we get,

$$\begin{gathered} y=6-4 \\ y=2 \end{gathered}$$