Paul Websters motorboat can travel 15 mph in still water. Traveling with the current of a river, the boat can travel 20 miles in the same time it takes to go 10 miles againt the current. Find the speed of the current.

(15+Vc)t = 20

Eq1: 15t + Vc*t = 20

(15-Vc)t = 10
Eq2: 15t - Vct = 10
Add Eq1 and Eq2:
15t + Vct = 20
15t _ Vct = 10

30t = 30
t = 1 h.

15*1 + Vc*1 = 20
Vc = 20-15 = 5 mi/h. = Velocity of current.

To find the speed of the current, we can set up a basic equation. Let's denote the speed of the current as 'c'.

Let's assume the boat is traveling against the current. In this case, its effective speed will be the difference between its speed in still water (15 mph) and the speed of the current (c mph). Therefore, the effective speed in this scenario is (15 - c) mph.

If the boat travels a distance of 10 miles against the current, the time it takes can be represented as:

Time = Distance / Speed
Time = 10 / (15 - c)

Now let's consider the situation where the boat is traveling with the current. In this case, the boat's effective speed will be the sum of its speed in still water (15 mph) and the speed of the current (c mph). Therefore, the effective speed in this scenario is (15 + c) mph.

If the boat travels a distance of 20 miles with the current, the time it takes can be represented as:

Time = Distance / Speed
Time = 20 / (15 + c)

According to the given information, these times are equal. Therefore, we can set up the equation:

10 / (15 - c) = 20 / (15 + c)

To solve this equation, we can cross-multiply and simplify:

10(15 + c) = 20(15 - c)
150 + 10c = 300 - 20c
30c = 150
c = 150 / 30
c = 5

Therefore, the speed of the current is 5 mph.