Core formulas
Let B = speed of the boat in still water and S = speed of the stream.
| Quantity | Formula |
|---|---|
| Downstream speed D | B + S |
| Upstream speed U | B - S |
| Boat speed | B = (D + U)/2 |
| Stream speed | S = (D - U)/2 |
| Round trip time over distance d | d/(B + S) + d/(B - S) |
| Average speed for a round trip | (B² - S²)/B |
| Same distance, time upstream = k x time downstream | (B + S)/(B - S) = k |
Target: 40 seconds.
Worked example
Downstream speed is 16 km/h and upstream speed is 10 km/h. Then B = (16 + 10)/2 = 13 km/h and S = (16 - 10)/2 = 3 km/h.
Worked example 2 (round trip)
B = 12 km/h, S = 4 km/h. Time to go 32 km downstream and return?
32/16 + 32/8 = 2 + 4 = 6 hours. Check with the average-speed formula: (144 - 16)/12 = 10.67 km/h and 64/10.67 = 6 hours.
Worked example 3 (ratio)
A man rows 4 km downstream in the time he rows 3 km upstream. Then D : U = 4 : 3. If the round trip of 48 km each way takes 14 hours, then 48/4k + 48/3k = 14, k = 2, so D = 8 and U = 6 km/h, stream = 1 km/h.
Quadratic-type questions (Mains)
"A boat takes 1 hour more upstream than downstream to cover 24 km; boat speed is 18 km/h." Write 24/(18 - S) - 24/(18 + S) = 1, giving S² + 48S - 324 = 0 and S = 6. Tip: instead of solving, test the options - put S = 6: 24/12 - 24/24 = 1. Done in 10 seconds.
Common traps
- Swapping downstream and upstream: downstream is always faster.
- Taking the average of downstream and upstream speeds as the average speed of the round trip (that gives B, not the average speed).
- The man's speed "in still water" is B, not D.
- Escalator problems use the same logic: walking with the escalator adds speeds.
Worked example 4 (time ratio)
A man rows at 9 km/h in still water and takes twice as long to go upstream as to go the same distance downstream. (9 + S)/(9 - S) = 2, so 9 + S = 18 - 2S and S = 3 km/h.
General result: if upstream time is k times downstream time, then B : S = (k + 1) : (k - 1). For k = 2 this gives 3 : 1, matching 9 and 3.