Time, Speed, Distance and Work
CSAT time, speed and work with worked examples — unit conversion, average and relative speed, trains, boats and streams, the LCM method for work, pipes and man-days.
SyllabusCSAT · Basic numeracy (numbers and their relations, orders of magnitude, etc.) — Class X level
Prerequisites: Time, Speed and Work — overview · Ratio, Proportion and Averages
The Hook: The Commute That Did Not Average 50
You drive 60 km to work at 60 km/h and return on the same road in heavy traffic at 40 km/h. What is your average speed for the round trip? Almost everyone says 50 km/h. The answer is 48 km/h — the trip out took 1 hour, the trip back took 1.5 hours, and $120 \div 2.5 = 48$.
You spend more time at the slower speed, so it pulls the average down. CSAT builds whole questions around this kind of intuition failure. The fix is a habit: always go back to total distance ÷ total time.
The Core Concept
1. The basic relation and units
$$\boxed{\text{Distance} = \text{Speed} \times \text{Time}}$$| Conversion | Multiply by | Example |
|---|---|---|
| km/h → m/s | $\dfrac{5}{18}$ | $72$ km/h $= 72 \times \dfrac{5}{18} = 20$ m/s |
| m/s → km/h | $\dfrac{18}{5}$ | $15$ m/s $= 54$ km/h |
Constant-distance rule: for a fixed distance, time is inversely proportional to speed. If speed becomes $\tfrac{3}{4}$ of usual, time becomes $\tfrac{4}{3}$ of usual.
2. Average speed
$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$$For equal distances at speeds $x$ and $y$:
$$\boxed{\text{Average speed} = \frac{2xy}{x + y}}$$Hook check: $\dfrac{2 \times 60 \times 40}{100} = 48$ km/h ✓. For equal times at two speeds, the average is the simple mean $\dfrac{x + y}{2}$.
3. Relative speed
| Direction | Relative speed |
|---|---|
| Opposite (towards or away from each other) | $S_1 + S_2$ |
| Same direction | $\lvert S_1 - S_2 \rvert$ |
Meeting time for two objects starting $D$ apart and moving towards each other: $\dfrac{D}{S_1 + S_2}$.
4. Trains
| A train of length $L$ crosses… | Distance covered |
|---|---|
| A pole, a person or a signal post | $L$ |
| A platform or bridge of length $P$ | $L + P$ |
| Another train of length $L_2$ | $L + L_2$ (at relative speed) |
Worked example 1. A 240 m train runs at 54 km/h ($= 15$ m/s).
- Time to pass a pole $= \dfrac{240}{15} = 16$ s.
- Time to cross a 360 m platform $= \dfrac{240 + 360}{15} = \dfrac{600}{15} = 40$ s.
Worked example 2. Trains of 150 m and 200 m run at 50 km/h and 40 km/h on parallel tracks.
- Opposite directions: relative speed $= 90$ km/h $= 25$ m/s; time to cross $= \dfrac{350}{25} = 14$ s.
- Same direction: relative speed $= 10$ km/h $= \dfrac{25}{9}$ m/s; time $= 350 \times \dfrac{9}{25} = 126$ s.
5. Boats and streams
With boat speed $b$ in still water and stream speed $s$:
| Quantity | Formula |
|---|---|
| Downstream speed $D$ | $b + s$ |
| Upstream speed $U$ | $b - s$ |
| Boat speed | $b = \dfrac{D + U}{2}$ |
| Stream speed | $s = \dfrac{D - U}{2}$ |
Worked example. Downstream 18 km/h, upstream 12 km/h → boat $= 15$ km/h, stream $= 3$ km/h.
6. Time and work — the LCM method
Instead of fractions, assume the total work is the LCM of the days, then find each worker’s units per day.
graph LR
A["Days taken by each worker"] --> B["Total work = LCM of the days"]
B --> C["Efficiency = total work / days"]
C --> D["Add efficiencies for workers together; subtract for pipes that empty"]
D --> E["Time = work remaining / combined efficiency"]Worked example 1. A finishes a job in 12 days, B in 18 days. Total work $= \text{LCM}(12, 18) = 36$ units. A does $3$/day, B does $2$/day. Together: $\dfrac{36}{3 + 2} = 7.2$ days.
Worked example 2. A and B together take 10 days; A alone takes 15 days. How long does B take alone? Total $= \text{LCM}(10, 15) = 30$. A + B $= 3$/day, A $= 2$/day → B $= 1$/day → 30 days.
Efficiency language. “A is twice as efficient as B” means A does 2 units for B’s 1 — and therefore takes half the time.
7. Pipes and cisterns
A filling pipe does positive work; an emptying pipe (or a leak) does negative work.
Worked example. Pipe A fills a tank in 20 hours, pipe B in 30 hours; pipe C empties it in 15 hours. All three are opened together. Total $= \text{LCM}(20, 30, 15) = 60$. Per hour: $+3$, $+2$, $-4$ → net $+1$. Time $= 60$ hours.
8. Man-days
When the work is the same:
$$\boxed{M_1 D_1 H_1 = M_2 D_2 H_2}$$(and in general $\dfrac{M_1 D_1 H_1}{W_1} = \dfrac{M_2 D_2 H_2}{W_2}$ for different amounts of work).
Worked example. 15 workers finish a job in 24 days. How many days will 20 workers take? $15 \times 24 = 20 \times D_2$ → $D_2 = 18$ days.
Memory Anchors
- “5/18 to slow down” — km/h to m/s makes the number smaller.
- “Opposite? Add. Same? Subtract.” — relative speed.
- “Cross the train, add the lengths.”
- “Down adds, up subtracts” — boats.
- “LCM is the job” — total work = LCM of the days.
- “Harmonic for equal distance, arithmetic for equal time.”
| Confusable pair | Rule |
|---|---|
| Equal distances vs equal times | $\dfrac{2xy}{x+y}$ vs $\dfrac{x+y}{2}$ |
| Efficiency vs time | Twice as efficient → half the time |
| Pole vs platform | $L$ vs $L + P$ |
For Prelims
- $1$ km/h $= \dfrac{5}{18}$ m/s; $36$ km/h $= 10$ m/s; $54$ km/h $= 15$ m/s; $72$ km/h $= 20$ m/s; $90$ km/h $= 25$ m/s.
- Average speed over equal distances is always less than the simple mean of the two speeds (unless the speeds are equal).
- A train crossing a pole covers only its own length.
- In work problems, adding workers adds efficiencies (work per day), never days.
- An emptying pipe’s efficiency is subtracted.
Speed Strategy
- Convert units first. Mixing km/h with metres is the most common source of wrong answers. Learn the 36–54–72–90 → 10–15–20–25 ladder.
- Use LCM units, not fractions, for every work and pipe question.
- Use ratios for constant-distance questions. Speed ratio $3 : 4$ → time ratio $4 : 3$; the difference of 1 part equals the given late/early time.
- Test options when the question asks for a distance or a time and the options are clean numbers.
- What to skip: multi-stage questions where workers join and leave several times can be done, but leave them for your second pass.
Common Traps
Practice
Prelims (CSAT)
A man travels from his village to a town at 30 km/h and returns along the same road at 20 km/h. The total travel time is 5 hours. How far is the town from the village?
Solution
Average speed $= \dfrac{2 \times 30 \times 20}{30 + 20} = 24$ km/h. Total distance $= 24 \times 5 = 120$ km; one way $= 60$ km.
Check: $\dfrac{60}{30} + \dfrac{60}{20} = 2 + 3 = 5$ hours ✓.
Answer: B
A train 300 m long passes a pole in 15 seconds. How long will it take to cross a bridge 450 m long?
Solution
Speed $= \dfrac{300}{15} = 20$ m/s. To cross the bridge it covers $300 + 450 = 750$ m: time $= \dfrac{750}{20} = 37.5$ s.
Answer: C
A can finish a job in 20 days and B in 30 days. They work together for 6 days, after which A leaves. In how many more days will B finish the remaining work?
Solution
Total work $= \text{LCM}(20, 30) = 60$ units. A $= 3$/day, B $= 2$/day.
In 6 days together: $6 \times 5 = 30$ units. Remaining $= 30$ units; B alone at 2/day → $15$ days.
Answer: A
A boat’s speed in still water is 10 km/h and the speed of the stream is 2 km/h. How long will the boat take to go 48 km downstream and return to the starting point?
Solution
Downstream $= 12$ km/h → $\dfrac{48}{12} = 4$ h. Upstream $= 8$ km/h → $\dfrac{48}{8} = 6$ h. Total $= 10$ hours.
(9.6 h is the trap — it assumes the boat moves at 10 km/h both ways.)
Answer: D
12 workers, working 8 hours a day, complete a job in 10 days. How many days will 16 workers take to complete the same job if they work 6 hours a day?
Solution
$M_1 D_1 H_1 = M_2 D_2 H_2$: $12 \times 10 \times 8 = 16 \times D_2 \times 6$ → $960 = 96 D_2$ → $D_2 = 10$ days.
Answer: B
Quick Revision
| Situation | Formula / method |
|---|---|
| Units | km/h × 5/18 = m/s |
| Average speed | Total distance ÷ total time; $\dfrac{2xy}{x+y}$ for equal distances |
| Relative speed | Opposite: add; same: subtract |
| Train crosses pole / platform | $L$ / $L + P$ |
| Boats | $D = b + s$, $U = b - s$, $b = \dfrac{D + U}{2}$, $s = \dfrac{D - U}{2}$ |
| Work | Total = LCM of days; add efficiencies |
| Pipes | Emptying pipe = negative efficiency |
| Man-days | $M_1 D_1 H_1 = M_2 D_2 H_2$ |
| Constant distance | Time ∝ 1/speed |
Key Takeaways
- Rate × time = amount — the same structure underlies distance and work.
- Always return to total distance ÷ total time for average speed.
- Relative speed handles every meeting, overtaking and train-crossing question.
- The LCM method removes fractions from work and pipe questions.
- Convert units before calculating.
Related Topics
- Time, Speed and Work — quick revision
- Time, Speed and Work — practice questions
- Ratio, Proportion and Averages for CSAT
- Number System: Divisibility, HCF–LCM and Remainders
- CSAT mock tests