Question

A motorboat takes 5 hours to travel 150km going upstream. The return trip takes 3 hours going downstream. What is the rate of the boat in still water and what is the rate of the current?

Answer 1

=10km/h

Explanation:

Let motor boat speed be represented by x and current y

The speed upstream = Motor boats speed - rate of current

=x-y

The net speed down stream = Motor boats speed + rate of current

=x+y

Let us find the speed upstream =distance/ time taken

=150km/5hrs

=30km/h

Speed down stream= 150km/3h

=50 km/h

The problem forms simultaneous equations.

x-y=30

x+y=50

Using elimination method we solve the equations.

Add the two equations to eliminate y.

2x=80

x=40

Current, y= 50-x

=10km/h

Answer 2

1)
`40\ (km)/(h)`

2)
`10\ (km)/(h)`

Explanation:

Let' call "b" the speed of the motorboat and "c" the speed of the current.

We know that:

`V=(d)/(t)`

Where "V" is the speed, "d" is distance and "t" is time.

Then:

`d=V*t`

We know that distance traveled upstream is 150 km and the time is 5 hours. Then, we set up the folllowing equation:

`5(b-c)=150` (Remember that in the trip upstream the speed of the river is opposite to the motorboat)

For the return trip:

`3(b+c)=150`

By solving the system of equations, we get:

- Make both equations equal to each other and solve for "c".

`5(b-c)=3(b+c)\\\\5b-5c=3b+3c\\\\5b-3b=3c+5c\\\\2b=8c\\\\c=(b)/(4)`

- Substitute "c" into the any original equation and solve for "b":

`5(b-(b)/(4))=150\\\\(3)/(4)b=30\\\\b=40\ (km)/(h)`

- Substitute "b" into
`c=(b)/(4)`:

`c=(40)/(4)\\\\c=10\ (km)/(h)`