Aptitude · Boats and Streams · Model 5
Solve for the travelling speeds, not the boat
Two journeys, each mixing some upstream distance with some downstream distance, each with its own total time. In b and s the algebra is horrible. Treat 1/U and 1/D as the unknowns and it becomes a pair of ordinary simultaneous equations.
Eliminate one speed at a time →01 The idea
The substitution that makes it linear
This is the hardest-looking model in the chapter and the most mechanical once set up. A trip of 8 km upstream and 12 km downstream taking 3 hours says 8/(b−s) + 12/(b+s) = 3. Two such trips give two of those, and trying to solve them for b and s directly means clearing denominators that contain both unknowns — which produces a mess with squared terms in it.
The fix is to stop asking for b and s. Set x = 1/U and y = 1/D and each trip becomes u₁x + d₁y = t₁ — linear, with constant coefficients, exactly the shape you can eliminate a variable from in one line. Solve for x and y, invert them to get U and D, and only then use b = (D + U)/2 and s = (D − U)/2.
Elimination works the way it always does. Scale the two equations so the upstream distances match — use the LCM — and subtract, which removes the 1/U term completely and leaves a single equation in 1/D. The source material also teaches a faster guess: take the HCF of the two downstream distances and try it as D. That works surprisingly often because examiners choose friendly numbers, but it is a guess, and elimination is not.
There is one thing to check before any of this. If u₁d₂ = u₂d₁, the second trip is the first one multiplied by a constant and tells you nothing new — the system has no unique solution, and any single answer offered for it is one of infinitely many. That test is two multiplications and it is worth doing first, because published question sets do contain this mistake.
02 Worked example
Eight and twelve, then twelve and twenty-four
One pair of trips runs the whole lesson. A rower goes 8 km upstream and 12 km downstream in 3 hours. On another day he goes 12 km upstream and 24 km downstream in 5 hours. Find his speed in still water.
Compare the two routes honestly. In b and s the first trip is 8/(b−s) + 12/(b+s) = 3, and clearing that denominator gives you b² − s² terms to carry through a second equation of the same kind. In 1/U and 1/D it took one LCM and one subtraction. The source also offers a quicker guess — the HCF of the downstream distances 12 and 24 is 12, so try D = 12, and it works — which is excellent in a timed paper and useless when the numbers are not chosen to cooperate. Know the guess, trust the elimination.
03 The method
The method, the guess, and the test
Three things to carry: the substitution that makes it easy, the shortcut that makes it fast, and the check that stops you solving a broken question.
| Move | Why | On the lesson pair |
|---|---|---|
| Substitute 1/U and 1/D | makes both equations linear | 8x + 12y = 3, 12x + 24y = 5 |
| Test the determinant | catches a rescaled second trip | 192 − 144 = 48 ✓ |
| Scale to LCM of upstream | cancels one unknown exactly | ×3 and ×2 to 24/U |
| Subtract | one equation, one unknown | 12/D = 1, so D = 12 |
| Back-substitute | recovers the other speed | U = 8/2 = 4 |
| Convert last | b and s from D and U | b = 8, s = 4 |
| Solving in b and s first | creates squared terms | avoid entirely |
05 Cheat sheet
Model 5 on one page
The whole method in six rows, plus the two things that go wrong. The example column is the lesson pair: 8 up and 12 down in 3 hours, 12 up and 24 down in 5 hours.
| Step | Rule | On the lesson pair |
|---|---|---|
| Unknowns | x = 1/U, y = 1/D | 8x + 12y = 3, 12x + 24y = 5 |
| Independence | u1·d2 − u2·d1 ≠ 0 | 192 − 144 = 48 |
| Eliminate | scale to LCM of u1, u2 | ×3 and ×2, both 24/U |
| Downstream | subtract, then invert | 12/D = 1, D = 12 |
| Upstream | back-substitute in trip 1 | 3 − 1 = 2 h, U = 4 |
| Boat and stream | (D+U)/2 and (D−U)/2 | b = 8, s = 4 |
| Determinant zero | no unique answer exists | 30/45/9 with 40/60/12 |
06 Where & why
Where this shows up
This is a Bank PO Mains and SSC Tier-2 shape almost exclusively. Its value beyond the exam is that it is a genuine two-variable elimination dressed as a word problem.
Usually with the HCF guess available. The candidates who finish it are the ones who wrote both equations in reciprocals rather than in b and s.
“24 up and 30 down in 7 hours, 36 up and 20 down in 8 hours.” The HCF of 30 and 20 happens to work here, but the elimination is barely slower and cannot be caught out.
The determinant test is exactly the question data-sufficiency items ask. Two statements that scale into one another are one statement, and spotting that is the whole mark.
Two delivery runs each mixing uphill and downhill, two transfers each mixing fast and slow links: any pair of totals over two unknown rates is this model, reciprocals and all.
07 Interview questions
What gets asked
Ten, from the substitution through to the published question that has no answer.
Why substitute 1/U and 1/D instead of solving for b and s?
A boat does 8 km upstream and 12 downstream in 3 hours, and 12 up and 24 down in 5 hours. Find b.
What is the HCF hit-and-trial shortcut, and when should you trust it?
Describe the elimination method properly.
What is the determinant test and why does it matter?
The set says 30 km up and 45 down in 9 hours, and 40 up and 60 down in 12 hours, answer 5 km/h. Is that right?
Given that, how would you fix the question?
24 km up and 30 down in 7 hours, 36 up and 20 down in 8 hours. Find the rowing speed.
How is this different from the earlier models?
When would you actually use this?
08 Practice problems
Six on two-trip systems
Test the determinant before you solve, every time. One of these is a published question with no unique answer, and recognising that is the whole point of it.