Traveling against the current a ship went 220 km in 3 hours & 20 minutes. The return trip with the current took 35 minutes less. Find (a) the ship's rate in still water and (b) the rate of the current. Please explain.

Question

sirm3d · Answer

let s = ship's rate in still water
     c = rate of the current

effective rate against the current is $s-c$
effective rate with the current is $s+c$

distance=rate x time
against the current: $220=(s-c)\times 200 \text{ (minutes)}$
with the current: $220=(s+c)\times 165 \text{ (35 minutes shorter/faster)}$