Boats and streams is a classic aptitude topic built on two simple speeds. Enter the boat's speed in still water and the stream's speed, or the downstream and upstream speeds, to get every speed and travel time.
Choose what you know.
Enter the two speeds and the distance.
Read the speeds, times and average speed.
Downstream: b + s
Upstream: b − s
Boat: (down + up) ÷ 2
Boats and streams questions describe a boat moving in a river whose current helps it going downstream and holds it back going upstream. From the boat's speed in still water and the stream's speed you get the two effective speeds, and from those, travel times and the average speed for a round trip. This calculator also works backwards, from downstream and upstream speeds to the boat and stream speeds.
The chapter is regular in SSC, banking and railway aptitude tests, and it is an easy scoring topic once the two basic relations are clear. The same reasoning applies to swimmers in a river, aircraft with a tailwind or headwind, and walking on an airport travelator.
Choose what you know, enter the two speeds and the one-way distance, and read all the results together.
1. Let b be the boat's speed in still water and s the stream's speed, both in km/h.
2. Downstream speed D = b + s.
3. Upstream speed U = b − s. If U is 0 or negative, the boat cannot move upstream.
4. Reverse: b = (D + U) ÷ 2 and s = (D − U) ÷ 2.
5. Travel times: distance ÷ D downstream and distance ÷ U upstream.
6. Round-trip average speed = 2 × distance ÷ (time down + time up), which also equals (b² − s²) ÷ b.
Relative to the land, the boat's velocity is its velocity relative to the water plus the water's velocity relative to the land. Going with the current both point the same way, so the speeds add. Going against it they point opposite ways, so the stream's speed subtracts. Adding and subtracting D and U then isolates each part: D + U = 2b and D − U = 2s. This is why the reverse formulas are simple halves.
For equal distances d each way, total time is d/(b + s) + d/(b − s) = 2db ÷ (b² − s²). Average speed = 2d ÷ total time = (b² − s²) ÷ b = b − s²/b. So the average is always below the still-water speed whenever there is a current, and falls further as the current grows. The boat spends longer going upstream than it saves going downstream. This is the harmonic mean of D and U, not their arithmetic mean.
These questions assume a uniform current across the river, a constant rowing effort and no time lost turning around. Real rivers flow faster in the middle than near the banks, and wind adds another effect. Exam questions sometimes state that a man takes twice as long to row upstream as downstream; then D = 2U, which gives b = 3s. Setting up such ratios as equations in b and s is the quickest route.
A ferry on the Hooghly covers its route at 18 km/h going downstream and 12 km/h coming back upstream. The one-way distance between the two ghats is 30 km.
Boat and stream from the two speeds: Boat = (18 + 12) ÷ 2 = 15 km/h Stream = (18 − 12) ÷ 2 = 3 km/h
Travel times: Down: 30 ÷ 18 = 1.6667 h Up: 30 ÷ 12 = 2.5 h
Average speed for the round trip: 2 × 30 ÷ 4.1667 = 14.4 km/h
Answer: Round-trip time 4.1667 hours; Downstream / upstream speed 18 / 12 km/h; Boat / stream speed 15 / 3 km/h
Taking the round-trip average speed as (D + U) ÷ 2, which equals the still-water speed and is too high.
Swapping the downstream and upstream speeds in the reverse formulas.
Ignoring that the boat cannot go upstream if the stream is as fast as or faster than the boat.
Mixing km/h with m/s or distances in metres.
Scoring the boats and streams questions in aptitude exams.
Estimating ferry or boat travel times on rivers.
Understanding headwind and tailwind effects on flights.
Explaining vector addition of velocities in simple terms.
Why isn't average speed (down + up) ÷ 2?
The boat spends longer going upstream, so the average is total distance ÷ total time, which is lower.
What if the stream is faster than the boat?
The boat cannot make progress upstream.