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SSC CHSL · Quantitative Aptitude

Boats, Streams, Trains & Races

Relative speed problems involving boats in water, trains passing objects and races.

Six concepts under one relative-speed idea: a current adds or subtracts, a train clears a length, and a race is finished when one runner still has metres left. Boats own the still-water/stream split; trains and races reuse the same add-or-subtract speed, with a distance that is not always the race length printed on the question.

  • SSC CHSL
  • Medium level
  • 6 concepts
  • 42 practice questions

1Downstream adds the stream; upstream subtracts it

Call the boat's speed in still water b and the stream's speed s. With the current the water helps, so downstream speed is b+s; against it the water fights, so upstream speed is b-s. The stream never changes which way the boat points — it only changes how fast the ground goes by.

Swap b and s in your head and the whole chapter breaks: a stream faster than the boat means the boat cannot make way upstream at all. So b>s is not a nicety, and the upstream speed b-s is the quantity that goes to zero first when the current strengthens.

Figure. The stream bar is the same length whether you add it or subtract it. Downstream is still water plus that length; upstream is still water minus it. The four bars are to a common scale in km/h.

How it works

  1. Name b and sStill-water speed is the boat alone; stream speed is the water alone. Neither is a journey speed until you add or subtract.
  2. Add with, subtract againstDownstream = b+s. Upstream = b-s. Same arithmetic for a swimmer in a current.
  3. Divide into the distanceTime = distance ÷ the speed that applies on that leg. A round trip needs two divisions, never one average of b.

Out with the current, back against it

A boat's still-water speed is 12 km/h and the stream is 3 km/h. How long does the boat take to cover 45 km downstream and return the same 45 km upstream?

  • Downstream speed = 12 + 315 km/h
  • Upstream speed = 12 − 39 km/h
  • Time down = 45 ÷ 153 h
  • Time up = 45 ÷ 9; round trip = 3 + 58 h

Pro tip. The round-trip average speed is not b. Distance both ways is 90 km in 8 h, so 11.25 km/h — below the still-water 12, because more time was spent on the slow leg. Averaging 15 and 9 as (15+9)/2 = 12 recovers b, not the journey rate.

A boat's still-water speed is 10 km/h and the stream is 2 km/h. Its upstream speed is
  1. 12 km/h
  2. 8 km/h
  3. 5 km/h

Upstream subtracts: 10-2=8 km/h. 12 is the downstream speed, and 5 is half the still-water figure — a common panic guess when b-s has not been written down.

2Still water is the average of down and up

If you are handed a downstream speed and an upstream speed, the boat and the stream fall out in one pair of lines: b=\frac{\text{down}+\text{up}}{2} and s=\frac{\text{down}-\text{up}}{2}. Adding cancels the stream; subtracting cancels the boat.

The trap is to treat either given speed as b. Downstream is already inflated by the current, so calling the downstream figure the boat speed overstates the still-water value by exactly the stream — and then every upstream time you predict comes out too short.

Figure. The dashed rule sits at the average of the two journey speeds — that height is b. Half the gap between the bars is s. Bars and rule share one km/h scale.

How to recover b and s

  1. Find the two journey speedsIf the question gives times and a common distance, divide: down = d/t_{\text{down}}, up = d/t_{\text{up}}.
  2. Average for the boatb = (\text{down}+\text{up})/2. The stream terms cancel.
  3. Halve the gap for the streams = (\text{down}-\text{up})/2. The boat terms cancel.

Boat speed and stream speed

A boat covers 28 km downstream in 2 hours and the same distance upstream in 4 hours. Find the speed of the boat in still water.

  • Downstream speed = 28 ÷ 214 km/h
  • Upstream speed = 28 ÷ 47 km/h
  • Still-water speed b = (14 + 7) ÷ 210.5 km/h
  • Stream speed s = (14 − 7) ÷ 23.5 km/h

Pro tip. Boat speed is the average of downstream and upstream speeds; stream speed is half their difference. Asking only for b does not excuse skipping s in the check: 10.5+3.5=14 and 10.5-3.5=7 must rebuild the two journey speeds you started from.

Downstream speed is 18 km/h and upstream speed is 10 km/h. The boat's still-water speed is
  1. 18 km/h
  2. 14 km/h
  3. 8 km/h

(18+10)/2 = 14 km/h. Taking the downstream figure as b leaves the 18; halving the difference alone gives the stream s=4, not the boat.

3A train clears its own length — plus the object's, if it has one

A train passing a pole or a standing person covers a distance equal to its own length: the object has none, so the crossing ends when the rear clears the same point the front reached. A train crossing a platform, bridge or tunnel covers train length plus object length, because the rear must clear the far end.

The whole question is which lengths count. Forget the platform and you understate the distance; add a pole's "length" you invented and you overstate it. Sibling topic `quant_time_work_speed` owns the same idea under trains, poles and platforms — this concept keeps the boats-and-trains source example that lives here.

Figure. The third bar is the sum of the first two — that sum is the distance the train must cover before the platform crossing is complete. Lengths are to a common metre scale.

How it works

  1. Decide what has lengthPole, man, lamp-post: length zero. Platform, bridge, tunnel, another train: length counts.
  2. Add the lengths that countDistance to clear = train + object. For a pole that is just the train.
  3. Convert speed, then dividePut the speed in m/s before dividing into metres. Time = distance ÷ speed.

Train crossing a platform

A 120 m long train running at 54 km/h crosses a platform of length 180 m. Find the time taken.

  • Speed = 54 × 5/1815 m/s
  • Distance to clear = 120 + 180300 m
  • Time = 300 ÷ 1520 s
  • Pole check: 120 ÷ 158 s

Pro tip. Crossing a platform means the train's own length plus the platform length — never forget to add both. The pole time is the platform time with the platform term deleted; their difference, 12 s, times 15 m/s rebuilds the platform's 180 m.

A 100 m train at 54 km/h crosses a 200 m platform. The time taken is
  1. 10 s
  2. 20 s
  3. 6\tfrac{2}{3} s

Speed =54\times 5/18=15 m/s; distance =100+200=300 m; time =300/15=20 s. 10 s uses only one of the lengths; 6\tfrac{2}{3} s divides 100 m by 15 and forgets the platform.

4Two trains: add speeds head-on, subtract when overtaking

Two trains pass each other over the sum of their lengths. Running opposite ways, their relative speed is the sum of the speeds; running the same way, the faster overtakes at the difference of the speeds. Same relative-speed rule as any other pair of moving objects — only the distance is both lengths added.

`quant_time_work_speed` already teaches relative speed with a two-train ledger. What this concept locks is the length sum: opposite or same direction changes only the denominator, never which lengths you add.

Figure. Relative speeds in km/h for the worked example's pair. Same distance (250 m) divided by these two rates gives 10 s and 50 s — the bars are in the ratio 5:1, and so are the times reversed. Scale is km/h.

How it works

  1. Add the lengthsDistance to clear = L_1 + L_2 in both the opposite and the same-direction cases.
  2. Add or subtract the speedsOpposite: u+v. Same direction: |u-v|.
  3. Convert, then divideRelative speed into m/s, then time = (L_1+L_2) ÷ relative speed.

Opposite and same direction

Trains of lengths 150 m and 100 m run at 54 km/h and 36 km/h. How long do they take to pass each other running in opposite directions, and then in the same direction?

  • Opposite: 54 + 36 = 90 km/h; 90 × 5/1825 m/s
  • Distance to clear = 150 + 100250 m
  • Opposite time = 250 ÷ 2510 s
  • Same direction: 54 − 36 = 18 km/h = 5 m/s; 250 ÷ 550 s

Pro tip. The same-direction time is five times the opposite time here because the relative speed shrank by the same factor, 25 to 5. The distance never changed — if your two answers do not sit in the ratio of the two relative speeds, a length was dropped in one of the cases.

Two 100 m trains run at 60 km/h and 40 km/h in the same direction. The time for the faster to overtake the slower is
  1. 10 s
  2. 20 s
  3. 36 s

Relative speed =20 km/h =20\times 5/18=\tfrac{50}{9} m/s; distance =200 m; time =200/(\tfrac{50}{9})=36 s. 10 s adds the speeds as if they were opposite; 20 s uses 10 m/s as though the conversion were \times 1/2.

5When A finishes, B is still short of the tape

'A beats B by x metres' in a race of D metres means that when A has finished the full race, B is still short of the tape. The speeds sit in the ratio of those distances in the same time: \frac{v_A}{v_B}=\frac{D}{D-x}. 'Beats by t seconds' means B still needs t more seconds to finish after A has already finished.

The wording names how far B still has to go, not how far B has already gone. Reading 'beats by 100 m' as B having run 100 m is the inversion that flips the ratio upside down.

Figure. The freeze-frame when A hits the tape. B's bar is 100 m short — that shortfall is the 'beats by 100 m', and the two bars in the same time are the speed ratio. Metre scale shared.

How it works

  1. Fix the moment A finishesA has run the full race D. B has run D-x if the beat is by x metres.
  2. Same time, two distancesIn that shared time, speeds are in the ratio D:(D-x).
  3. Translate a time beatIf A finishes in time T and beats B by t seconds, B's time for D is T+t.

Beaten by a hundred metres

In a 1000 m race, A beats B by 100 m. A finishes the race in 100 s. Find the speeds of A and B.

  • When A has run 1000 m, B has run 1000 − 100900 m
  • Shared time100 s
  • A's speed = 1000 ÷ 10010 m/s
  • B's speed = 900 ÷ 100; ratio check 1000/9009 m/s (= 10/9)

Pro tip. The ratio formula \frac{v_A}{v_B}=\frac{D}{D-x} is exactly the two distances in the ledger's first row. If a later part asks how many seconds A beats B by, give B the full 1000 m at 9 m/s: 1000/9 s, then subtract A's 100 s.

In a 500 m race, A beats B by 50 m. The ratio of A's speed to B's speed is
  1. 10 : 9
  2. 9 : 10
  3. 11 : 10

When A runs 500, B runs 450, so v_A:v_B=500:450=10:9. The upside-down 9:10 is the inversion trap; 11:10 would be a 50 m head start misread as an 11-part split.

6Lengths in metres force speeds into m/s

Train and race lengths arrive in metres and times in seconds; boat speeds usually stay in km/h. Mixing the two without converting is the slip that produces a "time" of several hundred seconds for a platform crossing that should be twenty. Multiply km/h by \frac{5}{18} to get m/s, or multiply m/s by \frac{18}{5} to go back.

Do the conversion before the division, not after. Dividing metres by a km/h figure does not give seconds — it gives a number with no unit worth trusting.

Figure. Platform and train lengths arrive in metres and times in seconds, so the speed must be in m/s. Multiply km/h by 5/18 (here 72\times 5/18=20); the reverse factor is 18/5.

How to keep units consistent

  1. List the units you were givenLengths in m and times in s demand m/s. Distances in km and times in h may stay in km/h.
  2. Convert the speed firstkm/h → m/s: multiply by 5/18. m/s → km/h: multiply by 18/5.
  3. Only then divideTime = distance ÷ speed, now in matching units.
Speeds worth converting without thinking
km/hm/skm/hm/s
1857220
36109025
541510830
90 km/h in m/s is
  1. 25 m/s
  2. 18 m/s
  3. 50 m/s

90\times 5/18=25 m/s. 18 is the factor itself mistaken for the answer; 50 is 90\times 5/9, the wrong fraction.

Notes

  • Downstream & Upstream: If boat speed in still water is b and stream speed is s, then downstream speed =b+s and upstream speed =b-s; hence b=\frac{\text{down}+\text{up}}{2} and s=\frac{\text{down}-\text{up}}{2}.
  • Train Crossing a Point: A train crossing a pole/man covers a distance equal to its own length; crossing a platform/bridge it covers (train length + platform length).
  • Two Trains Passing: Trains moving in opposite directions pass each other using the sum of speeds; in the same direction, the faster overtakes using the difference of speeds, over the combined length.
  • Races & Head Starts: 'A beats B by x metres' means when A finishes the race, B is x metres behind; 'beats by t seconds' means B needs t more seconds to finish.
  • Unit Care: Always convert km/h to m/s using \times\frac{5}{18} when lengths are in metres and times in seconds, a frequent slip in train problems.

Formulas

  • Downstream speed = b+s; Upstream speed = b-s
  • b = \frac{(b+s)+(b-s)}{2}; s = \frac{(b+s)-(b-s)}{2}
  • Train crossing platform: time = \frac{L_{train}+L_{platform}}{\text{speed}}
  • Two trains (opposite): time =\frac{L_1+L_2}{u+v}; (same direction): \frac{L_1+L_2}{u-v}
  • km/h to m/s: multiply by \frac{5}{18}
  • A beats B by distance d: \frac{\text{A's speed}}{\text{B's speed}} = \frac{D}{D-d}

Exam traps & shortcuts

  • Find still-water and stream speed at once using b=\tfrac12(\text{down}+\text{up}) and s=\tfrac12(\text{down}-\text{up}).
  • For a train crossing a platform, add lengths first, then divide by speed — a single distance figure avoids two-step errors.
  • Convert all speeds to m/s (\times\frac5{18}) at the very start of any train problem to keep units consistent.

Reference tables

Same distance to clear in both columns — only the relative speed changes. Read across one row to see why same-direction times blow up.

Opposite versus same direction
SituationDistance to clearRelative speedTime
Train past a poleTrain lengthTrain speedL/v
Train past a platformTrain + platformTrain speed(L_t+L_p)/v
Two trains, oppositeL_1+L_2u+v(L_1+L_2)/(u+v)
Two trains, same wayL_1+L_2|u-v|(L_1+L_2)/|u-v|
A beats B by x m in race DA ran D; B ran D-xRatio D:(D-x)Shared time when A finishes

Every line should be reconstructible from the concepts above, not merely recalled.

Formula sheet
QuantityRelationWatch for
Downstream / upstreamb+s / b-sNeed b>s to make way upstream
Still-water / streamb=(\text{down}+\text{up})/2; s=(\text{down}-\text{up})/2Do not call downstream speed b
Train past platformtime =(L_{\text{train}}+L_{\text{platform}})/vConvert v to m/s first
Two trains oppositetime =(L_1+L_2)/(u+v)Lengths still add
Two trains same waytime =(L_1+L_2)/|u-v|Same lengths, smaller speed
Race beat by distancev_A/v_B=D/(D-x)B has run D-x, not x
km/h to m/smultiply by 5/18Before dividing into metres

Recap

Read only this the night before.

Boat ± stream
Downstream b+s, upstream b-s. Still water is the average of the two journey speeds; stream is half their difference.
Round trip
More time on the slow leg, so average speed for the journey is below b. Do not average the two speeds and call it the round-trip rate.
Train lengths
Pole → train length only. Platform → train + platform. Two trains → sum of lengths either way.
Relative speed
Opposite: add. Same direction: subtract. The distance formula does not care which case you are in.
Races
Beats by x m means when A finishes D, B has run D-x. Speeds in the ratio D:(D-x).
Units
Metres and seconds need m/s: multiply km/h by 5/18 before you divide.

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