A boat crew rowed 20 miles downstream, with the current, in 2hours. The return trip upstream, against the current, covered the same distance, but took 5 hours. Dine the crew's rowing rate in still water and the rate of the current? Still water _ MPH and Current _ MPH.

I know I already posted this question I didn't realize that I left out a few things. oops sorry!

Letting V = the rowing speed and C = the current speed:

20/(V+C) = 2 and
20/(V-C) = 5
Then, 2(V+C) = 5(V-C)
From which we get V = 7C/3
Substituting,
20/(7C/3+C) = 2 from which
C = 3MPH and V = 7MPH

No problem! To determine the crew's rowing rate in still water and the rate of the current, we can use a simple formula:

Rate = Distance / Time

Let's name the rowing rate in still water as R (in mph) and the rate of the current as C (in mph).

Given that the crew rowed 20 miles downstream in 2 hours, the effective rate would be R + C (since the current helps). Therefore, we get the equation:

(R + C) = 20 / 2

Simplifying this equation, we have:

R + C = 10

Similarly, when rowing upstream, the effective rate would be R - C (since the current opposes them). Using the same formula, we get:

(R - C) = 20 / 5

Simplifying this equation, we have:

R - C = 4

Now we have a system of two equations:

R + C = 10
R - C = 4

To solve this system, we can use the method of elimination. By adding these two equations, the 'C' terms cancel out:

(R + C) + (R - C) = 10 + 4

Which simplifies to:

2R = 14

Dividing both sides of the equation by 2:

R = 7

Now, substituting the value of 'R' into one of the original equations (R + C = 10), we can solve for 'C':

7 + C = 10

Subtracting 7 from both sides:

C = 3

Therefore, the crew's rowing rate in still water is 7 mph, and the rate of the current is 3 mph.