Downstream the current helps: b + c. Upstream it fights: b − c. Two trips give two equations, and the pair unlocks both speeds.
The problem
A boat travels 10 km/h in still water; the current runs at 2 km/h. How long does it take to go 36 km downstream?
Downstream the current pushes: 10 + 2 = 12 km/h, so 36 ÷ 12 = 3 hours. B divides by the upstream speed 8 — the current fights only on the way back. C ignores the current entirely.
downstream: b + c
boat speed plus the water's push
upstream: b − c
the same boat, minus the fight
b is the boat's speed in still water — what the engine produces. The river adds or subtracts its own speed c depending on direction. Neither trip ever happens at b itself.
A boat covers 24 km downstream in 2 hours and returns in 3 hours.
b + c = 24 ÷ 2 = 12
the downstream speed
b − c = 24 ÷ 3 = 8
the upstream speed
add: 2b = 20 → b = 10, c = 2
adding kills c; subtracting kills b
b = down + up2, c = down − up2
boat is the average, current is half the gap
b = 202 = 10, c = 42 = 2 ✓
same numbers, one line
Swap "boat" for "plane" and "current" for "wind" and nothing changes: with the wind is p + w, against it p − w. The trap to dodge everywhere: a round trip's average speed is NOT b — the against-leg takes longer and drags the average below it (see the average-speed lesson).
A walkway: a walker covers 40 m in 10 s going with it (4 m/s) and in 40 s going against it (1 m/s). The walker’s own speed is (4 + 1) ÷ 2 = 2.5 m/s and the walkway moves at (4 − 1) ÷ 2 = 1.5 m/s.
A round trip’s average, worked: b = 8 km/h and c = 4 km/h over 24 km each way. Down takes 24 ÷ 12 = 2 h and up takes 24 ÷ 4 = 6 h, so the average is 48 ÷ 8 = 6 km/h — below b = 8.
Time = distance ÷ speed for each leg, then add.
3512 + c + 3512 − c = 6
down + up
35 × 24 = 6(144 − c²)
over (12 + c)(12 − c)
840 = 864 − 6c²
c² = 4, so c = 2 km/h
a speed is positive
Check: 35 ÷ 14 + 35 ÷ 10 = 2.5 + 3.5 = 6 h ✓. On a multiple-choice paper, trying each choice this way is often faster than the algebra.
Still b + c and b − c; here b changes or the units do.
Times in hours and minutes: a plane flies 450 km against the wind in 1 h 30 min and returns with the wind in 1 h 15 min.
against: 450 ÷ 1.5 = 300
1 h 30 min = 1.5 h
with: 450 ÷ 1.25 = 360
1 h 15 min = 1.25 h
plane (360 + 300) ÷ 2 = 330 km/h; wind (360 − 300) ÷ 2 = 30 km/h
average and half the gap
Convert the minutes first; 1 h 15 min as “1.15 h” spoils both speeds.
The engine stops: a boat goes 14 km/h in still water. It travels upstream for 2 hours against a 2 km/h current, then its engine fails. How long does it take to drift back to the start?
upstream: (14 − 2) × 2 = 24 km
b − c for 2 hours
drifting: b = 0, so speed = c = 2 km/h
only the river moves it
24 ÷ 2 = 12 hours
back downstream
Using b + c = 16 km/h for the drift gives 1.5 hours, as if the engine were still running. With no engine, b is 0.
A current too strong: a rowing boat makes 4 km/h in still water on a river flowing at 5 km/h.
upstream speed: 4 − 5 = −1 km/h
a negative speed
it cannot go upstream
it is carried back 1 km each hour
Any “how long upstream” question then has no answer. If a solved b comes out at or below c, the set-up is wrong, or the trip is impossible.
Same time means distance ÷ speed is equal both ways.
40b + 3 = 24b − 3
equal times
40b − 120 = 24b + 72
cross-multiply
16b = 192, so b = 12 km/h
check: 40 ÷ 15 = 24 ÷ 9 = 83 h ✓
If a solved b is not bigger than c, the boat could not go upstream at all — recheck the set-up.
Boat, wind and walkway items all reduce to b + c and b − c. With both trip times given, take half the sum and half the difference; with only a total time or equal times, write time = distance ÷ speed for each leg — and check by timing both legs.
The problem
A plane flies 550 km/h with the wind and 450 km/h against it. How fast is the wind?
The wind is half the gap: 550 − 4502 = 50 km/h — it got added on one leg and subtracted on the other, so the 100 km/h difference is TWO winds. A forgets the halving; C is the plane's own speed, the average.
