Aptitude · Boats and Streams · Model 1
Flip the times, add and subtract, anchor once
Most of these questions never give you a distance, and never need to. A pair of times over the same stretch of river fixes the ratio of every speed in the problem, and one real number anywhere turns that ratio into kilometres per hour.
Walk the ratio chain →01 The idea
A ratio needs no distance
A boat goes down a river in 4 hours and comes back up the same stretch in 6. You are not told how far it went, and you do not need to be. Because the distance is identical both ways, the speeds must be in the exactly opposite ratio to the times: 6 : 4 downstream to upstream, which is 3 : 2. The distance cancels out of the comparison entirely.
That flip is the whole first step, every time. It follows from distance = speed × time with the distance held fixed — if one leg takes twice as long, it must have been done at half the speed. And the slower leg is always the one going upstream, so the upstream number always lands on the small side.
Getting from D : U to b : s takes one more line and no new idea. Since b = (D + U)/2 and s = (D − U)/2, in parts the boat is worth the sum of the two numbers and the stream is worth their difference. The two halvings cancel against each other, so there is no division by 2 in the parts step — putting one in is the most frequent slip in this model.
At that point you have four ratios and no speeds. One absolute number fixes everything: the stream is 6 km/h, or the boat is 12, or a downstream leg was 27 km/h. Divide it by its own part count to get the value of one part, then multiply out. Everything before that line is dimensionless; everything after it is in kilometres per hour.
02 Worked example
Twenty-four down, thirty-two back, a six km/h river
One question runs the whole lesson. X swims a certain distance downstream in 24 hours and takes 32 hours to swim back upstream. The stream flows at 6 km/h. Find X’s speed in still water.
Two things are worth noticing. First, the answer came out of a ratio, so the distance was never required — and indeed the implied 1,152 km makes this a very strange swim, which the ratio cheerfully ignores. Second, the alternative route is real algebra: set the distance as an unknown, write D = d/24 and U = d/32, put (D − U)/2 = 6 and solve for d. It gives the same 42 km/h after four lines of fractions. The ratio does it in two.
03 The method
The chain, and every way the given is hidden
The method is three lines. What varies is the disguise on the two inputs — the time ratio and the one real number.
| How the question hides it | What to do first | Example |
|---|---|---|
| Two times, same distance | flip straight away | 24 : 32 → D : U = 4 : 3 |
| Different distances | scale one leg to match | 75% in 7.5 h → 100% in 10 h |
| Equal times, different distances | distance ratio is speed ratio | 36 and 24 in 2 h → 3 : 2 |
| “Upstream takes 4 times as long” | read it as times 1 : 4 | D : U = 4 : 1, b : s = 5 : 3 |
| “Current is 25% of the boat” | b : s = 4 : 1 directly | U = 3 parts, D = 5 parts |
| “b is 250% higher than s” | 100 + 250 = 350, so 7 : 2 | D = 9 parts, U = 5 parts |
| Halving inside the parts step | never needed | 7 : 1, not 3.5 : 1 |
05 Cheat sheet
Model 1 on one page
The chain, then the conversions that turn each style of wording into its first line. The example column is the lesson question: 24 hours down, 32 up, stream 6 km/h.
| Step | Rule | On 24 h down, 32 h up, s = 6 |
|---|---|---|
| Flip the times | D : U = t(up) : t(down) | 32 : 24 = 4 : 3 |
| Boat in parts | D + U | 4 + 3 = 7 parts |
| Stream in parts | D − U | 4 − 3 = 1 part |
| Anchor | given / its part count | 6 / 1 = 6 km/h a part |
| The four speeds | parts × one part | b 42, s 6, D 48, U 36 |
| Check | D × t(down) = U × t(up) | 1,152 km both ways |
| Dividing parts by 2 | never | 7 : 1, not 3.5 : 0.5 |
06 Where & why
Where this shows up
Model 1 is the highest-frequency shape in the chapter because it can be written a dozen ways and still be the same three lines.
Straight down the chain with small numbers. The only way to lose the mark is to flip the times the wrong way, and the sanity check — upstream must be the slow one — costs nothing.
“The current is 25% of the boat’s speed” is b : s = 4 : 1 before you write anything. Decode the percentage into parts first and the rest of the question is arithmetic.
One leg given as a fraction or percentage of the other. Equalise, then flip. Candidates who flip first get a clean-looking wrong answer, which is why this version is set.
Two travel times against and with a wind fix the ratio of aircraft speed to wind speed. A single known airspeed then prices the wind — identical arithmetic, different vocabulary.
07 Interview questions
What gets asked
Ten, from why the flip is legitimate through to the wordings that hide the ratio in a percentage.
Why can you just flip the two times to get the speed ratio?
A swimmer goes downstream in 24 hours and back in 32. The stream is 6 km/h. Find his still-water speed.
Why is there no division by 2 when you go from D : U to b : s?
What if the two legs are not the same distance?
The two times are equal but the distances differ. What then?
How do you handle “the speed of the current is 25% of the boat’s speed”?
And “the boat’s speed is 250% higher than the stream’s”?
Do you ever need the distance in this model?
When is the ratio route better than setting up algebra?
When would you actually use this?
08 Practice problems
Six on ratios and anchors
Write the time ratio, the speed ratio and b : s as three separate lines every time. Two of these hide the ratio in a percentage and one hides it in a mixed fraction.