In this guide
Everything in this topic comes from one line: distance = speed × time. What changes from question to question is the setup. Two things move towards or away from each other, a train has its own length, a boat is helped or held back by a stream. Get the setup right and the arithmetic is short.
These questions are also where units trip people most. A train's length is in metres, its speed in km/h, and the answer in seconds. Converting before you calculate saves more marks here than any formula.
Units: why 5/18
- km/h to m/s: multiply by 5/18.
- m/s to km/h: multiply by 18/5.
Why: 1 km/h is 1,000 metres in 3,600 seconds, which is 1,000/3,600 = 5/18 m/s. So 54 km/h = 54 × 5/18 = 15 m/s, and 72 km/h = 20 m/s. Multiples of 18 convert cleanly, which is a hint that you should convert.
Speed, time and ratios
For a fixed distance, speed and time are inversely proportional. If speed becomes 3/4 of usual, time becomes 4/3 of usual.
For a fixed time, distance is proportional to speed. For a fixed speed, distance is proportional to time.
Many questions become one-line ratio questions once you spot which quantity is fixed.
Average speed
Average speed = total distance ÷ total time. It is not the average of the speeds.
For two equal distances at speeds x and y, average speed = 2xy/(x + y).
Why: the slower leg takes longer, so it carries more weight in the total time. Ride 36 km out at 12 km/h (3 hours) and back at 18 km/h (2 hours): 72 km in 5 hours is 14.4 km/h. The formula is just this calculation done once in general: total distance 2d, total time d/x + d/y.
Relative speed
| Moving | Relative speed |
|---|---|
| Towards each other (opposite directions) | Sum of speeds |
| Same direction | Difference of speeds |
Why: if you sit on one object, the other appears to move at the relative speed. Moving towards each other, both close the gap; moving the same way, only the difference closes it.
Trains
| The train crosses… | Distance covered |
|---|---|
| A pole, a person or a signal | The train's own length |
| A platform, bridge or tunnel | Train length + platform length |
| Another train | Sum of the two lengths, at relative speed |
Why: the crossing starts when the engine reaches the object and ends when the last coach clears it. The engine therefore travels the train's length plus the object's length.
Boats and streams
With boat speed b in still water and stream speed s:
- Downstream speed = b + s.
- Upstream speed = b − s.
- b = (down + up)/2 and s = (down − up)/2.
Why: downstream the current adds to the boat's speed; upstream it subtracts. Adding the two speeds cancels s, and subtracting cancels b.
Seven worked questions
Q1. A cyclist rides to a market at 12 km/h and returns at 18 km/h. Find the average speed.
2 × 12 × 18/(12 + 18) = 432/30 = 14.4 km/h.
Q2. Walking at 5 km/h, a student reaches school 6 minutes late. At 6 km/h, the student is 4 minutes early. Find the distance.
The two times differ by 10 minutes = 1/6 hour. d/5 − d/6 = 1/6, so d/30 = 1/6 and d = 5 km.
Q3. A 150 m train moving at 72 km/h crosses a 250 m platform. How long does it take?
72 km/h = 20 m/s. Distance = 150 + 250 = 400 m. Time = 20 seconds.
Q4. A train crosses a pole in 12 seconds and a 180 m platform in 24 seconds. Find its length.
The extra 12 seconds cover the platform's 180 m, so speed = 15 m/s. Length = 15 × 12 = 180 m.
Q5. Two trains, 150 m and 100 m long, run in the same direction at 60 km/h and 42 km/h. How long does the faster take to pass the slower?
Relative speed = 18 km/h = 5 m/s. Distance = 250 m. Time = 50 seconds.
Q6. A boat covers 30 km downstream in 2 hours and the same distance upstream in 3 hours. Find the boat's speed in still water and the stream's speed.
Down = 15 km/h and up = 10 km/h. Boat = (15 + 10)/2 = 12.5 km/h. Stream = (15 − 10)/2 = 2.5 km/h.
Q7. Walking at 3/4 of the usual speed, a person reaches office 20 minutes late. Find the usual time.
Time becomes 4/3 of usual, so the extra 1/3 of the usual time is 20 minutes. Usual time = 60 minutes.
Common mistakes
| Mistake | Fix |
|---|---|
| Averaging two speeds directly | Total distance ÷ total time, or 2xy/(x + y) |
| Mixing km/h with metres and seconds | Convert with 5/18 first |
| Forgetting the train's own length | Distance = train + object |
| Adding speeds for the same direction | Same direction: subtract |
| Swapping upstream and downstream | Downstream is faster: b + s |
Practice
- Convert 90 km/h into m/s.
- A car travels to a town at 40 km/h and returns at 60 km/h. Find the average speed.
- A 200 m train at 36 km/h crosses a bridge in 30 seconds. Find the length of the bridge.
- Two trains of 120 m and 80 m run in opposite directions at 50 km/h and 40 km/h. How long do they take to cross each other?
- A boat's speed in still water is 10 km/h and the stream flows at 2 km/h. How long will it take to go 24 km upstream and come back?
- Two cyclists start 60 km apart and ride towards each other at 12 km/h and 18 km/h. When do they meet?
- A car covers a distance in 6 hours at 50 km/h. At what speed must it travel to cover the same distance in 5 hours?
- A 240 m train passes a person running at 6 km/h in the same direction in 16 seconds. Find the train's speed.
Answers:
- 25 m/s. 90 × 5/18.
- 48 km/h. 2 × 40 × 60/100.
- 100 m. 36 km/h = 10 m/s, so 300 m in 30 seconds, minus the train's 200 m.
- 8 seconds. Relative speed 90 km/h = 25 m/s, and 200 m ÷ 25.
- 5 hours. 3 hours upstream at 8 km/h and 2 hours downstream at 12 km/h.
- 2 hours. 60 km at a relative speed of 30 km/h.
- 60 km/h. The distance is 300 km, and 300 ÷ 5 = 60.
- 60 km/h. Relative speed = 240 ÷ 16 = 15 m/s = 54 km/h. Add the runner's 6 km/h.
What to do next
- Learn km/h to m/s for 18, 36, 54, 72, 90 and 108 so you convert without writing.
- Solve 10 train questions, sketching which lengths are covered before calculating.
- Solve 10 boats-and-streams questions using only the two average formulas.
- Revise time and work, which uses the same rate × time idea, and averages if average speed still feels slippery.
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Staff Selection Commission website .
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