When the boat travels downstream, the current helps it move faster because it works in the direction of travel. When the boat moves upstream, the opposite happens. So, let \( s \) be the speed of the boat in still water and \( c \) be the speed of the current, both in miles per hour.

Remember that \( rate * time = distance \). Let’s use this formula to write what happens when the boat is going downstream and upstream. When the boat goes downstream, the current goes in the same direction as the boat and speeds it up, so the rate is given by \( s + c \). When the boat is going upstream, the current goes against the boat’s movement and slows it down, so the rate is \( s - c \). Thus,

Upstream: \( (s + c)(4 h. ) = 120 mi. \)

Downstream: \( (s - c)(12 h. ) = 120 mi. \)

This is now a system of equations. Let’s first divide both sides of each equation by the time to get everything in miles per hour:

\( \left \{ {{s + c = 30 mph} \atop {s - c = 10 mph}} \right. \)

We can then add the two equations and solve for the speed of the boat in still water first:

\( (s + c) + (s - c) = 30 mph + 10 mph \)

\( 2s = 40 mph \)

\( \bf s = 20 mph \)

Finally, we can find the speed of the current by plugging the speed of the boat back into one of the previous equations:

\( s + c = 30 mph \)

\( 20 mph + c = 30 mph \)

\( \bf c = 10 mph \)