They turn around and paddle downstream for 4 hours. The next morning, they pack up and get back on the river to canoe to their original starting point 28 miles upstream and arrive at 2 pm. Assume the river always flows at a constant rate of 2 miles per hour, and the two friends always paddle at a constant rate. What time did they leave for their return trip?
My approach:
When going up they have the stream against them, so they go at their rate $(r) - 2m/h$. On the other hand, when going downstream they gain 2m/h.
$D=r*t$
$(r+2)*4-(r-2)*3 = 28$
$r=14$
The original point is upstream so they will paddle at $12m/h$
$28 = 12*t$
$t= \frac{7}{3}$ $\rightarrow$ $t=2h 20m$. Since they arrived at $2$ pm, they left at $11:40$ am.
What do you think?
Your math is correct. Here's an illustration I made to go with it: