Aptitude · Boats and Streams

Add going with it, subtract coming back

In every other speed question the road holds still. Here it moves, so the same boat has two different speeds depending on which way it points. Four short formulas cover the whole chapter, and one of them is the trap that catches most students.

Set the boat and the river →

Downstream adds, upstream subtracts. The boat is the average of the two travelling speeds and the stream is half their gap.

01 The idea

When the road itself is moving

Ride a bicycle at 10 km/h on a still day and you move at 10 km/h. Now let a 5 km/h wind push from behind: you are pedalling at 10, the air is carrying you at 5, and the ground goes past at 15. Turn round into that same wind and you are down to 5. Nothing about your legs changed — the medium changed sides. A river does exactly this, and an escalator does it too.

That gives four quantities and only four. b is the speed of the boat in still water, which is the rower’s own power. s is the speed of the stream. D is the downstream speed, going with the current, and U is the upstream speed, going against it. Every question in the chapter hands you two of them and asks for the others.

The relations are as short as they look: D = b + s and U = b − s going one way, and b = (D + U)/2 with s = (D − U)/2 coming back. The second pair is just the first pair rearranged: add the two equations and s cancels, subtract them and b cancels.

One consequence is worth keeping in your head for the rest of the module. The gap between the two travelling speeds is D − U = 2s, never s, because the current is added on one leg and subtracted on the other. A 12 km/h boat on a 4 km/h river runs at 16 one way and 8 the other, and 16 − 8 = 8 = 2 × 4.

Two of b, s, D and U always give you the other two. Add for the boat, subtract for the stream, and remember the halving.

b — still water speedWhat the boat would do on a dead-flat lake. It is the rower’s own contribution and it never changes when the boat turns around. Phrases like “the speed of the boat” and “his rowing speed” all mean this.

D and U — travelling speedsWhat the bank sees: b + s with the current and b − s against it. “Along the current”, “with the flow” and “downstream” all mean D.

The 2s gapD − U = 2s always. It is the fastest route from a pair of travelling speeds to the current, and it is why a stated difference between the two directions is halved rather than used as it stands.

02 Worked example

Twelve on a four, over forty-eight kilometres

One boat runs the whole lesson. A boat rows at 12 km/h in still water on a river that flows at 4 km/h. It travels 48 km downstream and then 48 km back up. Find the two travelling speeds, the total time, and the average speed for the whole trip.

Downstream, the speeds addThe river is helping, so its 4 km/h sits on top of the rower’s 12.D = 12 + 4 = 16 km/h

Upstream, the stream is subtractedThe same rower, the same river, pointing the other way.U = 12 − 4 = 8 km/h · gap 16 − 8 = 8 = 2s

Time each leg separatelyDistance over speed, twice. The 48 km is the same both ways; only the speed moved.48/16 = 3 h down · 48/8 = 6 h up · total 9 h

Average speed is total distance over total time96 km covered in 9 hours. Note that you spent 6 of those 9 hours at the slow speed, so the slow leg dominates.96 / 9 = 10⅔ km/h — not 12

The same answer straight from b and sSubstituting into 2DU/(D + U) collapses the denominator to 2b, so the distance drops out entirely and the average depends only on the boat and the river.(12² − 4²)/12 = (144 − 16)/12 = 128/12 = 10⅔ km/h

The tempting move is to average 16 and 8, which gives 12 — and 12 is exactly the still-water speed, which makes the wrong answer feel right. It is wrong because average speed weights by time, and the two legs took 3 hours and 6 hours, not 4½ each. The honest figure is 10⅔ km/h, and the shortfall of 1⅓ km/h is exactly s²/b = 16/12. Any current at all costs you speed, both ways round.

03 The method

Four formulas and three traps

The formulas take one line. The traps are what the marks are actually for, and the source material names the first two explicitly.

D = b + s · U = b − s · b = (D + U)/2 · s = (D − U)/2. And underneath all four, distance = speed × time — used with D or U, never with b.

Trap 1: b must be greater than s, or there is no upstream journey to speak of. Trap 2: read the English carefully — “in still water” is b but “along the current” is D, and they are different numbers. Trap 3, the expensive one: the average speed of a round trip is not b. It is (b² − s²)/b, always strictly less.

You are given	You want	One line
b and s	D and U	add, then subtract
D and U	b and s	half the sum, half the difference
b and U	D	s = b − U, then D = b + s
D − U	s	halve it — the gap is 2s
b, s and a distance	each leg’s time	distance / D and distance / U
b and s	round-trip average	(b² − s²)/b, below b
D and U	round-trip average	never (D + U)/2 — that is b

05 Cheat sheet

Boats and streams on one page

The four formulas, the three traps and the one derived result worth memorising. The example column is the lesson boat: b = 12, s = 4, 48 km each way.

Quantity	Formula	On b 12, s 4, 48 km
Downstream speed	`D = b + s`	16 km/h
Upstream speed	`U = b − s`	8 km/h
Boat from D and U	`b = (D + U)/2`	(16 + 8)/2 = 12
Stream from D and U	`s = (D − U)/2`	(16 − 8)/2 = 4
Times	`distance / D and / U`	3 h and 6 h, total 9 h
Round-trip average	`(b² − s²)/b`	128/12 = 10⅔ km/h
Averaging D and U	`gives b, and is wrong`	12, not 10⅔

The gap is 2s, not s“A boat goes 8 km/h faster downstream than upstream” means 2s = 8, so the current is 4 km/h. The same halving turns up as a distance: if 3 hours downstream covers 12 km more than 3 hours upstream, then D − U = 4 and s = 2.

The average speed loss is s²/bWrite (b² − s²)/b as b − s²/b and the penalty is visible. A 4 km/h current costs a 12 km/h boat 16/12 = 1⅓ km/h; an 8 km/h current on the same boat would cost 64/12 = 5⅓, four times as much for twice the stream.

Units come in disguisedSpeeds arrive in m/s as often as km/h. Multiply m/s by 18/5 to get km/h, and do the conversion at the end rather than the start — D = 15 m/s and U = 5 m/s give b = 10 m/s, which is 10 × 18/5 = 36 km/h.

06 Where & why

Where this shows up

The setting is quaint and the arithmetic is not. Boats and streams is really relative speed with two frames of reference, which is why every exam sets it.

TCS NQT · Infosys

Two of the four, find the rest

The whole question is one addition or one halving. Speed matters more than method here, so learn to spot which two of b, s, D and U you were handed.

SSC CGL · Bank PO

The 2s gap, hidden as a distance

“Covers 12 km more downstream than upstream in the same 3 hours.” Divide by the time to get D − U, then halve. Candidates who forget the halving get exactly double.

CAT · XAT

Round-trip average speed

Set precisely because averaging the two speeds returns the still-water speed, which looks like a clean answer and is one of the options. Total distance over total time, every time.

Aircraft, escalators, walkways

Any moving medium

A plane with a tailwind, a walker on a travelator, a swimmer in a current — identical arithmetic. Headwind and tailwind times over the same route give you the wind speed by halving the difference in speeds.

The five models ahead are all this page with one extra sentence: find the speeds and their ratios, then time and distance, then the “upstream takes n times as long” trick, then the round trip, and finally two mixed journeys solved together. Nothing new gets added to the four formulas — only new ways of hiding them.

07 Interview questions

What gets asked

Ten, starting from the definitions and ending on the average-speed question that separates people who have thought about it from people who have not.

What are the four formulas of this chapter?

Downstream speed D = b + s, upstream speed U = b − s, and read backwards, b = (D + U)/2 and s = (D − U)/2. The second pair is the first pair rearranged — add the two equations and the stream cancels, subtract them and the boat cancels.

Why do the speeds add downstream and subtract upstream?

Because the water is moving too. Downstream the current carries the boat forward on top of whatever the rower does, so the bank sees both contributions. Upstream part of the rowing goes into holding position against the flow, so what the bank sees is the rowing minus the current.

A boat rows 18 km/h downstream and 10 km/h upstream. Find b and s.

b = (18 + 10)/2 = 14 km/h and s = (18 − 10)/2 = 4 km/h. The boat is the average of the two travelling speeds because the current is added once and subtracted once; the stream is half the gap for the same reason.

Why is the difference between the two speeds 2s and not s?

Because the current appears twice with opposite signs: D = b + s and U = b − s, so D − U = 2s. Any statement about how much faster the boat is one way than the other is therefore worth exactly half as much current as it first looks.

A boat does 15 km/h in still water and only 11 km/h upstream. What is its downstream speed?

19 km/h. The 4 km/h that upstream lost is the current, so s = 4 and D = 15 + 4 = 19. This is the “hidden missing link” shape: you are never given the number you need directly, but two of the four always fix the other two.

Why must the boat be faster than the stream?

Otherwise b − s is zero or negative and there is no upstream journey — the boat is held still or swept backwards. Every well-formed question in this chapter has b greater than s, so a negative upstream speed always means a misread, usually of which number is the stream.

“Speed of the boat along the current” — which letter is that?

D, the downstream speed, not b. The phrases that mean b are “in still water”, “in calm water” and “his rowing speed”. “Along the current”, “with the flow” and “aided by the stream” all mean D. Mixing the two up costs the whole question rather than a step.

A 12 km/h boat covers 48 km down a 4 km/h river and 48 km back. What is its average speed?

10⅔ km/h, not 12. The legs are at 16 and 8 km/h, so they take 3 hours and 6 hours — 96 km in 9 hours. Averaging 16 and 8 gives 12, which is the still-water speed and the trap: average speed weights by time, and you spent twice as long crawling as flying.

Is that always true, or just for those numbers?

Hard

A boat rowing at 12 km/h in still water goes 48 km down a river flowing at 4 km/h and returns. Find its average speed for the whole journey. Then show that for any positive current the round-trip average is strictly less than the still-water speed, and say what the shortfall depends on.

Follow-up

The first half is arithmetic; the second half asks you to prove something about every case at once, which means the 48 km has to disappear from the working. This is the harmonic mean of two speeds meeting the arithmetic mean, and the inequality between them is a result you will meet again in mixtures and in averages.

Show the hint

Average = total distance / total time. Then write the general version as 2DU/(D + U), substitute D = b + s and U = b − s, and look at what is left after b.