Aptitude · Boats and Streams · Model 4
The round trip averages less than the boat
Give the total time for a there-and-back journey and one formula returns the distance. Give the boat and the river and another returns the average speed — and it is never the still-water speed, which is the single most reliable mistake in this chapter.
Watch the average fall below the boat →01 The idea
Same distance, two speeds, unequal clocks
In a round trip the boat rows out to a point and comes back to where it started, so the distance is the same in both directions and only the speed changes. That is the whole setup, and everything interesting about it follows from one observation: because the two speeds differ, the two times differ, and the slower leg eats more of the clock.
Which makes average speed behave in a way most people do not expect. Average speed is total distance divided by total time. It is never the average of two speeds unless you spent equal time at each — and here you never do. So averaging D and U is wrong, and it is wrong in a particularly awkward way, because (D + U)/2 is exactly b. The wrong answer is the still-water speed, which looks like a clean result and is always in the option list.
The right answer is the harmonic mean of the two speeds, 2DU/(D + U). Substitute D = b + s and U = b − s and it collapses beautifully: the denominator becomes 2b, the twos cancel, and you are left with (b² − s²)/b. Written as b − s²/b, the shortfall is impossible to miss — any current at all costs you speed, and the cost grows with the square of the current.
The other half of this model is the distance formula. If you are told only the total time T for the whole journey, then one-way distance = T × DU/(D + U). That is just total distance = average speed × total time, with the two-way distance halved back down to one way — which is where the missing 2 in the formula goes.
02 Worked example
Ten on a two, five hours there and back
One boat runs the whole lesson. A rower does 10 km/h in still water on a river flowing at 2 km/h. He rows to a point and returns to his starting place, and the whole journey takes exactly 5 hours. How far away is the point, and what was his average speed?
Look hard at those two times: 2 hours out and 3 hours back. The rower was at 12 km/h for only 40% of the trip and at 8 km/h for 60% of it, so an honest average has to sit below the midpoint — 9.6, not 10. That is the whole argument, and it works for every boat and every river. Write it as 10 − 4/10 and you can even see the size of the penalty: s²/b, which for a 4 km/h current on the same boat would be 16/10 = 1.6, four times as much for twice the stream.
03 The method
Two formulas and the two things they get confused with
One formula for the distance, one for the average speed, and both of them derived from the same single fact that the two legs share a distance.
| Given | Wanted | One line |
|---|---|---|
| b, s, total time T | one-way distance | T × DU/(D + U) |
| b, s, total time T | total distance | twice that |
| b and s only | average speed | (b² − s²)/b |
| b, s, one-way distance | total time | d/D + d/U |
| s, distance, time gap | b | d/U − d/D = gap, solve for b |
| s, distance, total time | b | a quadratic — or test the options |
| Averaging D and U | gives b, always wrong | 10, not 9.6 |
05 Cheat sheet
Model 4 on one page
Both formulas, both traps, and the harmonic-mean result that makes this model worth learning properly. The example column is the lesson trip: b = 10, s = 2, 5 hours in all.
| Item | Formula | On b 10, s 2, T 5 h |
|---|---|---|
| Travelling speeds | b + s and b − s | 12 and 8 km/h |
| Average speed | 2DU/(D + U) | 192/20 = 9.6 km/h |
| Same, from b and s | (b² − s²)/b | 96/10 = 9.6 km/h |
| Speed lost to the river | s²/b | 4/10 = 0.4 km/h |
| One-way distance | T × DU/(D + U) | 5 × 4.8 = 24 km |
| Total distance | twice the one-way | 48 km |
| Averaging the two speeds | returns b, always wrong | 10, not 9.6 |
06 Where & why
Where this shows up
Round trips are set constantly in the higher-difficulty slots, because the average-speed trap lets an examiner put a wrong answer in the options that most candidates will reach for.
One formula. The only difficulty is remembering that the answer is one way and the question may want both ways.
Almost always with b in the options, sometimes as the first option. The harmonic mean is the point of the question, not an incidental detail.
“One hour more coming back.” Write d/U − d/D = gap, combine over the common denominator b² − s², and it becomes a one-line equation rather than a quadratic.
The harmonic mean is why a commute that is fast one way and slow the other averages worse than the midpoint, and why halving one leg’s time helps the overall average far less than it feels like it should.
07 Interview questions
What gets asked
Ten, and the third one is the question that most reliably separates a memorised formula from an understood one.
What is the round-trip distance formula?
A boat does 10 km/h in still water on a 2 km/h river and the round trip takes 5 hours. How far away is the point?
What was its average speed for that trip — and why is it not 10?
Give the general formula for the average speed and show where it comes from.
A 25 km/h boat on a 5 km/h river. Average speed for a round trip?
If the boat is exactly three times the stream, what is the ratio of average speed to still-water speed?
A round trip takes 10 hours with b = 9 and s = 3. What is the total distance covered?
A boat covers 24 km each way and the return leg takes 1 hour longer. The stream is 2 km/h. Find b.
How do you tell a Model 3 question from a Model 4 one?
When would you actually use this?
08 Practice problems
Six on the round trip
For each one, decide first whether the question wants the one-way distance, the total distance, or a speed — then solve. Two of these have a correct number as a wrong answer.