A boat can travel 38 mph in still water. If the boat can travel 100 miles with the current in the same time it can travel 80 miles against the current, find the rate of the current r in miles per hour.

speed of current --- x mph

time to go 100 miles with the current = 100/(38+x)
time to go 80 miles against the current = 80/(38-x)

so 100/(38+x) = 80/(38-x)
cross-multiply
3040 + 80x =3800 - 100x
180x =760
x = 760/180 = 38/9 mph
the current is appr 4.2 mph

Well, I must confess, I'm not a big fan of currents. They can be quite pushy, you know? But let me try to calculate this for you.

Let's call the rate of the current "r" (because who doesn't love a good letter 'r', right?). We know that the boat can travel 38 mph in still water, but when it's going with the current, it can travel 100 miles in the same time it takes to go 80 miles against the current. So, let's break it down.

When the boat is going with the current, its effective speed is 38 + r (you know, the boat's regular speed plus the push from the current). And when it's going against the current, its effective speed is 38 - r (because it's fighting against the current).

Now we introduce the magical element called time. We know that time is constant in the two scenarios, so we can set up an equation to solve for 'r':

(100 / (38 + r)) = (80 / (38 - r))

Now, let the math begin!

100(38 - r) = 80(38 + r)

And solving that, we get:

3,800 - 100r = 3,040 + 80r

Combining like terms:

180r = 760

Dividing by 180:

r = 4.22

So, my friend, the rate of the current is approximately 4.22 mph. I hope that helps, and remember, currents might pull you in the wrong direction, but at least you'll have an answer to this question!

To solve this problem, let's denote the rate of the boat in still water as B, and the rate of the current as r.

When the boat is moving with the current, its effective speed will be the sum of the boat's speed in still water and the current's speed. So the boat's effective speed with the current is B + r.

Similarly, when the boat is moving against the current, its effective speed will be the difference between the boat's speed in still water and the current's speed. So the boat's effective speed against the current is B - r.

According to the problem, the boat can travel 100 miles with the current in the same time it can travel 80 miles against the current. We can set up the following equation to represent this:

100 / (B + r) = 80 / (B - r)

To solve for r, we can cross multiply and solve for r:

100(B - r) = 80(B + r)

Applying the distributive property:

100B - 100r = 80B + 80r

Combining like terms:

100B - 80B = 100r + 80r

20B = 180r

Dividing both sides by 180:

r = (20B) / 180

Simplifying further:

r = B / 9

Therefore, the rate of the current is B/9 miles per hour.

To find the rate of the current, we can set up a system of equations based on the given information.

Let's assume the rate of the current is r mph.

When the boat is traveling with the current, its effective speed is the sum of its speed in still water (38 mph) and the rate of the current (r mph).

So, the effective speed of the boat when traveling with the current is 38 + r mph.

When the boat is traveling against the current, its effective speed is the difference between its speed in still water (38 mph) and the rate of the current (r mph).

So, the effective speed of the boat when traveling against the current is 38 - r mph.

Let's say the time taken to travel 100 miles with the current is t hours.

Therefore, the time taken to travel 80 miles against the current is also t hours.

Now, we can use the formula:

Speed = Distance / Time.

For traveling with the current:
38 + r = 100 / t.

For traveling against the current:
38 - r = 80 / t.

We have a system of two equations:

38 + r = 100 / t,
38 - r = 80 / t.

To solve this system, let's eliminate the variable t.

Multiply both sides of the first equation by t:
(38 + r) * t = 100.

Multiply both sides of the second equation by t:
(38 - r) * t = 80.

Expanding both equations, we get:
38t + rt = 100,
38t - rt = 80.

Add the two equations together to eliminate the variable rt:
(38t + rt) + (38t - rt) = 100 + 80,
76t = 180,
t = 180 / 76.

Now, substitute the value of t back into one of the original equations to find the rate of the current (r):

38 + r = 100 / (180 / 76),
38 + r = 76 / 18,
r = 76 / 18 - 38,
r = (76 - 18 * 38) / 18,
r = (76 - 684) / 18,
r = -608 / 18,
r ≈ -33.78.

Therefore, the rate of the current is approximately -33.78 mph.