A motorboat 5 hours to travel 100km upstream. The return trip 2 hours going downstream. What is the rate in still water and the rate of the current?

Question

i don't know what to do

Answer 1

Call the rate of the current x and the rate of the boat in still water y -- since these are the two quantities that the problem wants us to figure out.  Let's see what kinds of equations we can come up with.

 

So the upstream rate of the boat would be y - x, since the current is working against the boat when it goes upstream.  So after 5 hours, the distance traveled upstream would be 5(y-x) . which is 100 km.  So we have one equation:  5(y-x) = 100

 

Let's use the same logic going downstream.  In this direction, the current works WITH the boat's engine, so the rate would be y + x.  So after 2 hours, the distance would be 2(y+x), which is also 100 km.  So now we have a second equation: 2(y+x) = 100.

 

So there are two equations, with two unknowns:

 

5(y-x) = 100

2(y+x) = 100

 

Let's solve:

 

First expand the equations:

 

5(y-x)  = 5y - 5x = 100

2(y+x) = 2y + 2x =100

 

There are a number of ways to solve these, but one easy way is to multiply both sides of the second equation by 2.5:

 

5y + 5x = 250

 

Add this to the first equation and the x's cancel out:

 

10y = 350

 

So y is 35.

 

Substitute y back into one of the original equations.  The arithmetic is easier in the second one, so:

 

2*35 + 2x = 100

70 + 2x = 100

2x = 30

x = 15

 

Go back to the original definitions of x and y to interpret the results.  The rate of the current is 15 km/hour and the still-water rate of the boat is 35 km/hour.

 

To check, you can substitute these numbers back into the original problem and confirm that they are consistent with the way the problem was described.