Boats and Streams Model 2: Finding Time and Distance

Boats and Streams · 25 min

Aptitude · Boats and Streams · Model 2

Find the distance once, then use it twice

Once a real distance or a real time appears in the question, the ratios are no longer enough. Convert the boat and the stream into two travelling speeds, use the leg you were given to pin the distance, and the other leg follows — or skip the distance entirely with the time ratio.

Turn a time into a distance and back
Turn b and s into D and U before you touch a distance. The two legs share one distance, and that is what makes the second leg solvable.

01 The idea

The distance is the bridge between the two legs

Model 1 lived entirely in ratios and never needed a kilometre. Model 2 starts the moment the question hands you a genuine distance or a genuine time, and it wants a genuine distance or time back. The formulas do not change — you still have D = b + s and U = b − s — but now they feed straight into distance = speed × time.

The one habit worth building is to convert first. Write D and U down before anything else, and then forget the river exists. A boat at 20 km/h on a 12 km/h current is not a river problem at all after the first line; it is a vehicle that does 32 km/h one way and 8 km/h the other. Every mistake in this model comes from carrying b into a division where D belonged.

The second habit is to notice what the two legs share. They do not share a speed and they do not share a time, but they share the distance — the boat goes to a point and comes back to where it started. So a speed and a time in one direction fix the distance, and the distance then fixes everything in the other direction. One number does the whole crossing.

And because the distance is shared, you can often avoid computing it. Time is inversely proportional to speed over a fixed distance, so if D : U is 4 : 1 then the times are 1 : 4, and a 16-hour trip down means a 64-hour trip back — without ever mentioning the 512 kilometres in between. Use the long route when the distance is the answer and the short route when it is not.

Convert to D and U, use the given leg to find the shared distance, then divide again. Or flip the speed ratio and scale the time directly.
The shared distanceThe one quantity both legs have in common. It is what turns two separate speed-time-distance problems into a single solvable question, and it is why “the same distance” in the wording is load-bearing rather than decorative.
Inverse time scalingOver a fixed distance, tup / tdown = D / U. So the upstream time is the downstream time multiplied by the speed ratio — no distance required.
Fractional hours2 hours 45 minutes is 11/4 hours, not 2.45. Convert minutes to a fraction of an hour on the first line and keep it as a fraction; decimals here are where the arithmetic starts to slip.

02 Worked example

Twenty on a twelve, sixteen hours down

One boat runs the whole lesson. A boat moves at 20 km/h in still water while the current flows at 12 km/h. It takes 16 hours to travel a certain distance downstream. How much time will it take to cover that same distance upstream?

1
Convert to the two travelling speedsDo this before anything else. After this line the river never appears again.D = 20 + 12 = 32 km/h  ·  U = 20 − 12 = 8 km/h
2
The downstream leg gives the distanceA speed and a time in the same direction, so multiply. Note how far it is — the current is doing most of the work.distance = 32 × 16 = 512 km
3
Same 512 km, at 8 km/hThe distance is shared, so the return leg is one division away.512 / 8 = 64 hours
4
The route that skips the 512D : U is 32 : 8 = 4 : 1, so the times must be 1 : 4. The downstream time is one part.16 × 4 = 64 hours, and the distance never appeared
5
Check it is the right way roundUpstream must be the slower leg, and the round trip must average below the still-water speed of 20.64 > 16 ✓  ·  1,024 km / 80 h = 12.8 km/h, below 20 ✓

The two routes cost very different amounts of writing for the same answer. If the question had asked how far the point was, you would need the 512 km; because it asked for a time, the flip does it in one multiplication. The general rule is worth stating plainly: compute the distance only when the distance is what you were asked for, or when the two legs are different lengths so the flip does not apply.

03 The method

Two routes, and how the given arrives

The method is short. Most of the difficulty in exam versions is in the first line, where the speeds are handed to you as a percentage, a ratio, or a distance over an awkward time.

Long route: D = b + s, U = b − s, distance = D × tdown, tup = distance / U. On the lesson boat: 32, 8, 512 km, 64 hours.
Short route: tup = tdown × D / U, valid only when the two legs are the same length. With D : U = 4 : 1 the times are 1 : 4, so 16 hours becomes 64. If the question gives two different distances — 90 km down and 87.5 km up — the flip does not apply and you must work in speeds.
How the speeds arriveFirst lineThen
b and s outrightadd and subtractD = 32, U = 8
A distance and a timedivide for the speed90 km in 4 h → D = 22.5
b : s given as 17 : 3D = 20 parts, U = 14 partsD = 40 → 1 part = 2, U = 28
“b is 250% higher than s”b : s = 350 : 100 = 7 : 2D = 9 parts, U = 5 parts
2 hours 45 minuteswrite it as 11/4 hoursnever 2.45
Equal legs, time wantedflip the speed ratio16 h × 4 = 64 h
Unequal legsthe flip is invalidfind both speeds instead

05 Cheat sheet

Model 2 on one page

Both routes and the conversions that feed them. The example column is the lesson boat: b = 20, s = 12, 16 hours downstream.

StepRuleOn b 20, s 12, 16 h down
Travelling speedsD = b + s, U = b − s32 and 8 km/h
Shared distanceD × t(down)32 × 16 = 512 km
Return legdistance / U512 / 8 = 64 h
Same thing, no distancet(down) × D/U16 × 4 = 64 h
Round tript(down) + t(up)80 hours
Round-trip average(b² − s²)/b(400 − 144)/20 = 12.8
Using b in a divisionnever — use D or Unot 512/20
Ratios still need one real number“b : s = 17 : 3, and 110 km downstream took 2 hours 45 minutes.” D is 20 parts and U is 14. The real downstream speed is 110 ÷ 11/4 = 40 km/h, so a part is 2 and U = 28. Then 98 km upstream takes 98/28 = 3.5 hours.
Percentages hide the ratio“250% higher than the stream” means b : s = 350 : 100 = 7 : 2, so D is 9 parts and U is 5. If 90 km downstream took 4 hours then D = 22.5, a part is 2.5, and U = 12.5 — so 87.5 km upstream takes exactly 7 hours.
The flip needs equal legsInverse time scaling comes from holding the distance fixed. The moment the question gives different distances in the two directions the flip is simply false, and you have no choice but to find both speeds and divide twice.

06 Where & why

Where this shows up

Model 2 is the workhorse of the chapter in bank and staff-selection papers, because it lets an examiner bolt a percentage or a fractional time onto arithmetic that is otherwise easy.

TCS NQT · Infosys
One leg to the other

b, s and a downstream time, asking for the upstream time. Two lines with the flip. The mark is for not using b as a travelling speed.

Bank PO · IBPS Clerk
Fractional times and a ratio

“110 km in 2 hours 45 minutes, with b : s = 17 : 3.” Convert the time to 11/4 first, then price the parts. Decimalising 2 hours 45 minutes as 2.45 is a standard trap answer.

SSC CGL
Percentage-masked speeds

“250% higher”, “25% of”, “80% more”. Decode into parts on line one and the question collapses into this model.

Flight and delivery planning
Out-and-back timings

Any round trip through a helping-then-hindering medium: a drone against the wind, a courier route with a gradient, a ferry across a tide. The out leg times the route and the back leg follows from the speed ratio.

What carries beyond boats is the discipline of converting to the quantity the formula actually wants before substituting. Half the errors in speed-time-distance questions of every kind are a value used in a formula that was built for a different value.

07 Interview questions

What gets asked

Ten, from why you convert first through to the case where the shortcut everyone reaches for is not available.

Why convert to D and U before doing any distance arithmetic?
Because distance = speed × time needs the speed the bank sees, not the rower’s own speed. A boat at 20 km/h on a 12 km/h current covers 32 km in an hour downstream and 8 km upstream, and neither of those is 20. Using b in the division is the single commonest error in this model.
A boat does 20 km/h in still water on a 12 km/h current and takes 16 hours downstream. How long back?
64 hours. D = 32 and U = 8, so the distance is 32 × 16 = 512 km and the return leg is 512/8 = 64 hours. Or skip the distance: D : U = 4 : 1, so the times are 1 : 4 and 16 × 4 = 64.
What is the shortcut that avoids the distance?
Over a fixed distance, time is inversely proportional to speed, so t(up) / t(down) = D / U. Take the speed ratio, flip it into a time ratio, and scale the time you were given. It replaces a multiplication and a division with a single multiplication.
When should you compute the distance anyway?
When the distance is what the question asked for, or when the two legs are different lengths so the flip does not apply. Also when you want the round-trip average speed, since that needs the total distance and the total time rather than a ratio.
How do you handle “110 km downstream in 2 hours 45 minutes”?
Convert the time to 11/4 hours on the first line, then divide: 110 ÷ 11/4 = 110 × 4/11 = 40 km/h. Writing 2 hours 45 minutes as 2.45 gives 44.9 km/h, which is wrong and is usually one of the options.
“A boat’s speed in still water is 250% higher than the stream’s.” What ratio?
b : s = 350 : 100 = 7 : 2, because “higher than” adds to the original 100%. That makes D = 9 parts and U = 5 parts. If 90 km downstream took 4 hours then D = 22.5, so a part is 2.5 and U = 12.5 km/h.
Given b : s = 17 : 3 and 110 km downstream in 11/4 hours, how long for 98 km upstream?
3.5 hours. D is 17 + 3 = 20 parts and U is 17 − 3 = 14. The real downstream speed is 40 km/h, so a part is 2 km/h and U = 28 km/h. Then 98/28 = 3.5. Notice the ratio alone was not enough — the 110 km and the time were what priced the parts.
The two distances are different. Does the time flip still work?
No. The inverse relation between time and speed holds only when the distance is held fixed, so with 90 km one way and 87.5 km the other you must find both speeds and divide each distance by its own. The flip is a shortcut for a special case, not a law of the chapter.
What tells you whether a question is Model 1 or Model 2?
What it gives and what it wants. Model 1 gives you a time ratio plus one absolute speed and wants speeds — pure ratio work, no distance. Model 2 gives a real distance or time and wants a real distance or time, so speed-time-distance has to be used at least once. Many exam questions do the Model 1 chain first and then a Model 2 division.
When would you actually use this?
Any out-and-back journey through something that moves: a delivery van with a strong prevailing wind, a boat crossing a tidal channel, a plane flying a route in both directions. Timing the outbound leg and scaling by the speed ratio is a genuinely practical way to estimate the return without re-measuring the distance.

08 Practice problems

Six on times and distances

Convert to D and U on the first line every time, and convert minutes into fractions of an hour before you divide anything. One of these deliberately breaks the time-flip shortcut.

Ratio plus a fractional time

Easy
The speed of a boat in still water and the speed of the stream are in the ratio 7 : 5. If the boat covers 100 km downstream in 1 hour and 40 minutes, how long will it take to travel 45 km upstream?
Follow-up
The ratio gives you the parts and the 100 km gives you their price, but the two distances are not the same, so the time flip is unavailable and you need the actual upstream speed.
Show the hint
1 hour 40 minutes is 5/3 hours. D is 12 parts and U is only 2.

A bigger ratio, the same shape

Easy
A boat covers 110 km downstream in 2 hours and 45 minutes. If the ratio of its speed in still water to the speed of the stream is 17 : 3, how long will it take to travel 98 km upstream?
Follow-up
The parts are large and close together, so the upstream speed is not far below the downstream one — a good check on whether you have subtracted rather than divided. Convert the mixed time first.
Show the hint
2 hours 45 minutes is 11/4 hours. D is 20 parts and U is 14.

The percentage mask

Medium
A boat’s speed in still water is 250% higher than that of the stream. It takes 4 hours to travel 90 km downstream. Find the time required by the boat to cover 87.5 km upstream.
Follow-up
“250% higher” is not 250% of, and getting that wrong changes every number after it. The two distances are also deliberately close together, so a wrong ratio still produces a plausible looking answer.
Show the hint
Higher than adds to the original: b : s = 350 : 100. Then 90 km in 4 hours prices the parts.

The stream, given with a downstream speed

Medium
A boat takes 6 hours to cover 96 km downstream on a river flowing at 4 km/h. How long will the same boat take to cover 84 km upstream?
Follow-up
You are given a downstream speed and the current, so the still-water speed has to be recovered before the upstream speed is available — two subtractions in a row, and it is easy to do only one of them.
Show the hint
96 km in 6 hours gives D. Then b = D − s, and U = b − s, which is D − 2s.

Minutes going in and coming out

Medium
A boat whose still-water speed is 15 km/h travels down a river flowing at 3 km/h for 2 hours 30 minutes, then turns around and returns to its starting point. How long does the return leg take, and what is the total journey time?
Follow-up
Mixed units on both ends: the given time is in hours and minutes and the answer wants to be too. The flip does apply here, so there are two routes — try both and see which is shorter to write.
Show the hint
D = 18 and U = 12, so the speed ratio is 3 : 2 and the time ratio is 2 : 3.

Doubling the current is not symmetrical

Hard
A boat covers a certain distance downstream in 2 hours and returns over the same distance in 3 hours. Suppose the current were twice as fast, with the boat unchanged. Find the new time for each leg, and explain why the downstream time and the upstream time do not change by the same factor.
Follow-up
The first part is the Model 1 chain followed by a Model 2 division, with the distance chosen freely because only ratios are given. The second part is the real question: adding s to one speed and subtracting it from the other is symmetrical in speed but never in time, because the same absolute change costs proportionally more on the slower leg. That asymmetry is why the round trip gets slower overall even though one leg got faster — the same fact behind the harmonic mean.
Show the hint
Times 2 : 3 give D : U = 3 : 2 and so b : s = 5 : 1. Take s = 1 and b = 5, pick the distance that makes both legs whole, then rebuild D and U with the stream at 2.