SOLUTION: A boat is rowed 10 miles downstream in two hours, then rowed the same distance upstream in 10/3 hours. Find the rate of the boat in still water and the rate of the current.

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: A boat is rowed 10 miles downstream in two hours, then rowed the same distance upstream in 10/3 hours. Find the rate of the boat in still water and the rate of the current.       Log On


   



Question 1120780: A boat is rowed 10 miles downstream in two hours, then rowed the same distance
upstream in 10/3 hours. Find the rate of the boat in still water and the rate of the current.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


A basic type of problem that is easily solved with basic algebra.

You are looking for two unknown numbers -- the rate of the boat and the speed of the current. So define variables to represent those two numbers.

let b = boat speed
let c = current speed

Then observe that the effective speed upstream will be b-c (the current is opposing the boat) and the effective downstream speed will be b+c (the current is aiding the boat). So

upstream speed = b-c
downstream speed = b+c

Then distance is rate times time, so write equations that say the upstream and downstream distances are each 10 miles.

%2810%2F3%29%28b-c%29+=+10 --> 10%28b-c%29+=+30 --> {{10b-10c = 30}}}
2%28b%2Bc%29+=+10 --> 2b+2c = 10}}}

Multiply the second equation by 5 and add the two equations to eliminate c:
10b-10c+=+30
10b%2B10c+=+50

Add these two equations to solve for b; then use that value in either of the original equations to solve for c.

Note that there are many different ways to set up the problem for solving. Here is a completely different approach that seems to make the work easier.

The 10 miles upstream take 10/3 hours, so the effective upstream rate b-c is 10/(10/3) = 3mph.
The 10 miles downstream take 2 hours, so the effective downstream rate is b+c = 5mph.

Now the problem can be solved very easily by inspection: b+c=5 and b-c+3; clearly b=4 and c=1.