SSC CHSL · Quantitative Aptitude
Time, Work, Speed & Distance
Work efficiency problems combined with time, speed and distance including relative speed concepts.
Eight concepts, and one idea underneath all of them: you may add rates, and you may never add times. Work per day and distance per hour are the same kind of quantity, which is why one chapter holds both — and why the traps are the same traps in two costumes.
- SSC CHSL
- Medium level
- 8 concepts
- 107 practice questions
1Work is a rate, and rates add
If a job takes n days, then one day's work is 1/n of it. That reciprocal is the only conversion in the chapter, and everything else follows: two people working together do 1/a + 1/b in a day, so the job takes 1/(1/a + 1/b) = ab/(a + b) days.
Times do not add and they do not average. Twelve days and eighteen days together give 7.2, which is neither 30 nor 15 — and it must be less than the smaller of the two, because adding a second worker cannot make a job slower. That inequality is the fastest check there is on an answer of this kind.
Figure. Each line's steepness is a rate, and working together stacks the two steepnesses into one — the joint line's slope is exactly the sum of the other two. That is what "rates add" means graphically. What you cannot do is add or average the times, which are the horizontal distances at which each line would reach the finish.
How it works
- Turn every time into a raten days becomes 1/n of the job per day. Do this before anything else in the question.
- Add or subtract the ratesHelpers add; a leak or a dropout subtracts. The combined rate is a single number.
- Invert at the very endTime is one divided by the combined rate — the last step, never an intermediate one.
Two workers, one job
A can finish a job in 12 days and B in 18. How long do they take working together?
- A's rate = 1/12, B's rate = 1/18per day
- Together: 1/12 + 1/18 = 3/36 + 2/365/36 per day
- Time = 1 ÷ (5/36)36/5 = 7.2 days
- Formula check: ab/(a + b) = (12 × 18)/307.2 days
Pro tip. The ab/(a + b) formula is the two-worker case only. Three workers need the rates added properly — there is no abc/(a + b + c), and writing one is the most common way this shortcut backfires. Notice too that 7.2 is closer to 12 than to 18: the faster worker contributes more, so the combined time always leans towards the shorter of the two.
A alone takes 12 days over a job and B alone takes 18. Working together they take
- 7.2 days
- 15 days
- 30 days
Rates add, so 1/12 + 1/18 = 5/36 and the job takes 36/5 = 7.2 days. 15 averages the two times and 30 adds them; both fail the same sanity check, since two people together cannot take longer than the faster one alone. Any answer above 12 here is wrong before it is computed.
2The LCM method
Fractions are what make work problems slow, and they are avoidable. Instead of calling the job 1, call it the LCM of every time given in the question. Each worker's daily output then becomes a whole number, and the rest of the problem is integer arithmetic.
The choice is free because the job's size never appears in the answer — you divide by it again at the end. What you gain is that combinations, leaks, dropouts and part-days all stay in whole units until the final division.
Figure. Call the job the LCM of the three pair-times — 60 units — and every daily output is an integer: A+B=5, B+C=4, A+C=3. Adding the three pair bars double-counts each worker, so A+B+C=6 units a day and the job takes 60\div 6=10 days. No fraction appears until that final division.
How it works
- Size the job as the LCMTake the LCM of every alone-time or pair duration in the question. That whole number is the job in units, so every daily output becomes an integer.
- Combine in whole unitsRead each person's (or pair's) daily units from job ÷ duration. Helpers, leaks and dropouts stay in integers through the ledger.
- Divide once at the finishTime is job ÷ combined daily units. The job size cancels — you chose it only to keep the arithmetic clean.
A takes 20 days, B takes 30, C takes 60. On the left the job is 1; on the right it is the LCM, 60.
| Step | Job = 1 | Job = 60 units |
|---|---|---|
| A's day | 1/20 | 3 units |
| B's day | 1/30 | 2 units |
| C's day | 1/60 | 1 unit |
| All three | 1/20 + 1/30 + 1/60 = 6/60 | 6 units |
| Time | 1 ÷ (1/10) | 60 ÷ 6 = 10 days |
Three pairs, three unknowns
A and B together finish a job in 12 days, B and C in 15, and A and C in 20. How long do all three take together?
- Total work = LCM(12, 15, 20)60 units
- A+B = 60/12 = 5, B+C = 60/15 = 4, A+C = 60/20 = 3units per day
- Add all three pairs: 2(A + B + C) = 12A + B + C = 6
- Time = 60 ÷ 610 days
Pro tip. The individual times drop out for free once you have the total of 6. A alone is 6 − 4 = 2 units a day, so 30 days; B is 6 − 3 = 3 units, so 20 days; C is 6 − 5 = 1 unit, so 60 days. Each is the total rate minus the pair that excludes that worker, and a follow-up question asking for any one of them costs a single subtraction.
A and B together take 12 days over a job that A alone would take 20 days over. B alone would take
- 30 days
- 8 days
- 32 days
With the job at LCM(12, 20) = 60 units, A+B is 5 a day and A is 3, so B is 2 and takes 60/2 = 30 days. Subtracting the times gives 20 − 12 = 8, which would make B faster than the pair working together — impossible, and the whole reason the LCM method is worth the extra line.
3Men, days and hours
When the number of workers, the days and the hours per day all change at once, the invariant is the total labour: M₁D₁H₁/W₁ = M₂D₂H₂/W₂. If the job is the same on both sides the W's cancel, and the equation says that men × days × hours is a fixed quantity.
Think of it as an area. Workers up the side, days along the bottom, and the rectangle's area is the job. Squeeze the rectangle narrower and it must grow taller to keep the same area — which is inverse variation, drawn.
Figure. This is the same wall with the daily hours left alone, so only men and days trade off. The two rectangles are drawn to scale and hold the same area — 36 × 18 and 27 × 24 are both 648. Taking a quarter of the men off the height forces the width out by a third. Shortening the working day as well, as the worked example does, is a third dimension the picture cannot show, and it stretches the width further still, to 32.
How it works
- Multiply everything on the known sideMen × days × hours gives the job in man-hours. One number, computed once.
- Divide by the new daily capacityNew men × new hours is how much gets done each day under the new arrangement.
- Scale for a different jobIf the second job is twice as big, multiply the answer by two — that is what the W's are for.
Fewer men, shorter days
36 men working 8 hours a day build a wall in 18 days. How long would 27 men working 6 hours a day take over the same wall?
- Job = 36 × 8 × 185184 man-hours
- New daily capacity = 27 × 6162 man-hours a day
- Days = 5184 ÷ 16232 days
- Check: 27 × 6 × 325184 ✓
Pro tip. Both changes push the same way here — fewer men and shorter days — so the answer must be more than 18, and it is nearly double. When the two changes pull against each other, that direction check is even more valuable: work out which factor is larger before computing, and you will catch an inverted fraction without redoing the sum.
10 men working 6 hours a day finish a job in 18 days. 12 men working 9 hours a day will finish it in
- 10 days
- 15 days
- 12 days
The job is 10 × 6 × 18 = 1080 man-hours and the new capacity is 12 × 9 = 108 a day, so 10 days. 15 days is what you get from 10 × 18 = 180 man-days divided by 12 — the answer that ignores the change in daily hours, which is exactly what this question is checking.
5Speed, distance, time — and the units
Speed = distance / time, and the two other forms follow. The arithmetic is never the difficulty; the units are. Almost every question mixes km/h with metres and seconds, and the conversion is 5/18 going from km/h to m/s and 18/5 coming back.
Remember which way round by size rather than by memory: 18 km/h is a jogging pace and 5 m/s is the same jogging pace, so a km/h figure must come down when it becomes m/s. Anything that turns 90 km/h into a number in the hundreds has gone the wrong way.
Figure. Left cluster: 5\,\mathrm{m/s} and 18\,\mathrm{km/h} are the same jogging pace written in two units — equal meaning, unequal numbers. Right cluster: the conversion factors themselves, \times 18/5 going up to km/h and \times 5/18 coming down. Anything that turns 90\,\mathrm{km/h} into a number in the hundreds has used the up factor the wrong way.
How it works
- Pick one system and convert everythingMetres and seconds if any length is in metres; kilometres and hours otherwise.
- Multiply by 5/18 or 18/5km/h to m/s is × 5/18. The other way is × 18/5. Sanity-check the direction by size.
- Convert the answer back if askedA time found in seconds may be wanted in minutes; read the last line of the question.
Convert before you divide. The jogging-pace check: 5 m/s is 18 km/h, so km/h figures shrink by ×5/18 when you move to m/s.
| From | To | Factor | Check |
|---|---|---|---|
| km/h | m/s | × 5/18 | 72 km/h → 20 m/s |
| m/s | km/h | × 18/5 | 5 m/s → 18 km/h |
| minutes | seconds | × 60 | 2 min → 120 s |
| km | metres | × 1000 | 4.5 km → 4500 m |
One speed, three units
A man covers 600 metres in 2 minutes. Find his speed in m/s and in km/h, and the time he needs for 4.5 km.
- 600 m in 120 s: 600 ÷ 1205 m/s
- 5 × 18/518 km/h
- 4.5 km = 4500 m at 5 m/s900 s
- 900 ÷ 6015 minutes
Pro tip. 5 m/s and 18 km/h are the same speed, and that pair is worth carrying as a reference point — every other conversion can be scaled off it. 10 m/s is 36 km/h, 20 m/s is 72 km/h, 25 m/s is 90 km/h. Recognising a round m/s value in a question usually means the examiner chose the km/h figure to make it come out round.
A train travels at 90 km/h. Its speed in metres per second is
- 25
- 324
- 18
90 × 5/18 = 25 m/s. 324 is 90 × 18/5 — the conversion applied backwards, and it fails on sight, since no train covers 324 metres in a second. 18 is 90/5, the 18 dropped from the fraction.
6Average speed is not an average of speeds
Average speed is total distance over total time, always. For two equal distances covered at x and y, that works out to the harmonic mean 2xy/(x + y) — never the arithmetic mean (x + y)/2, because more time is spent on the slower leg and the slow speed therefore counts for more.
The arithmetic mean is right in the other case: equal times at x and y do give (x + y)/2. So the question is not which formula to memorise but which quantity the problem holds equal — the distances or the durations.
Figure. The outward leg is the steep segment and the return the shallow one — steeper means faster. The dashed line is a car driven at a constant 48 km/h, and it starts and finishes at exactly the same two points, which is the whole definition of an average speed. The reason 48 is not 50 is visible in the widths: the shallow segment occupies three of the five hours, so it drags the straight line down towards its own slope.
How it works
- Read what is equal"There and back" or "each half of the journey" means equal distances. "For two hours, then for three" means times.
- Total the distance and the time separatelyAssume a convenient distance if none is given — the answer does not depend on it.
- Divide once, at the endTotal distance ÷ total time. Never average two speeds and hope.
There at 60, back at 40
A car goes from P to Q at 60 km/h and returns at 40 km/h. Find its average speed for the whole journey.
- Take the one-way distance as 120 km; out: 120/602 hours
- Back: 120/403 hours
- 240 km in 5 hours48 km/h
- Formula: 2xy/(x + y) = 2 × 60 × 40 / 10048 km/h
Pro tip. The 120 was chosen because it divides by both speeds; any distance gives the same 48, so pick the LCM of the two speeds and skip the fractions. And note where the answer sits: 48 is below the midpoint 50 and closer to the slower speed, which is true of every harmonic mean and is enough to eliminate two options before you compute anything.
A car covers the first half of a journey at 40 km/h and the second half at 60 km/h. Its average speed is
- 48 km/h
- 50 km/h
- 52 km/h
Equal distances, so the harmonic mean: 2 × 40 × 60/100 = 48 km/h. 50 is the arithmetic mean, which would be right only if the two speeds were held for equal times rather than over equal distances. The answer must be below 50 because more of the journey's time is spent at 40.
7Relative speed
Sit on one of the two moving objects and ask how fast the other seems to approach. Moving towards each other, the speeds add; moving the same way, they subtract. Every overtaking, meeting and crossing problem is that one sentence plus a distance.
The difference between the two cases is enormous in practice. Two trains at 72 and 54 km/h close at 126 km/h head-on and at only 18 km/h in the same direction — seven times slower, so seven times longer to pass each other, for exactly the same pair of trains.
Figure. Arrow lengths are drawn to the two speeds, 72 and 54. On the left the arrows point into each other and their lengths add; on the right they point the same way and only the difference in their lengths counts. Hold the two right-hand arrows against each other and what is left over is a stub one third the length of the shorter one, which is why same-direction problems always take so much longer than they look like they should.
How it works
- Add or subtract the speedsTowards each other: u + v. Same direction: u − v, and it is the faster minus the slower.
- Work out the distance to be closedFor objects with length, it is the sum of the two lengths; for points, the gap between them.
- Divide, in consistent unitsConvert the relative speed to m/s if the lengths are in metres, then divide.
Sit on one object: the other approaches at the sum when you meet head-on, and at the difference when you travel the same way.
| Direction | Relative speed | Distance to clear | Time |
|---|---|---|---|
| Opposite | u + v | sum of lengths | distance ÷ relative speed |
| Same way | |u − v| | sum of lengths | distance ÷ relative speed |
Two trains passing, both ways
Trains of length 120 m and 180 m travel at 72 km/h and 54 km/h. How long does each take to pass the other, running in opposite directions and then in the same direction?
- Opposite: 72 + 54 = 126 km/h, and 126 × 5/1835 m/s
- Distance to clear = 120 + 180300 m
- 300 ÷ 358.57 s
- Same direction: 72 − 54 = 18 km/h = 5 m/s, so 300 ÷ 560 s
Pro tip. The ratio of the two answers, 60 : 8.57, is exactly 7 — and so is 126 : 18. The distance never changed, so the times are in the inverse ratio of the relative speeds, and once you have one answer the other is a single multiplication. That also explains why an overtaking train seems to take forever while an oncoming one is gone in a blink.
A train 150 m long overtakes a man running at 6 km/h in the same direction, taking 10 seconds to pass him completely. The train's speed is
- 60 km/h
- 54 km/h
- 48 km/h
The train covers its own 150 m relative to the man in 10 s, so the relative speed is 15 m/s = 54 km/h. That is the difference of the two speeds, so the train does 54 + 6 = 60 km/h. Stopping at 54 answers the relative speed rather than the train's, and subtracting instead of adding gives 48.
8What distance does a train actually cover?
A train passing a pole covers its own length, because the pole has none. A train crossing a platform covers its own length plus the platform's, because the rear must clear the far end before the crossing is complete. Every train question turns on which of those two distances is wanted.
The pairing is what makes the questions solvable in reverse. Give a candidate the pole time and the platform time and the difference between them is the platform's length divided by the speed — two facts, two unknowns, no algebra worth the name.
Figure. The same train is drawn twice, at the instant the crossing begins and the instant it ends. Between those two positions the train has moved its own length plus the platform's, which is the long bracket. Shrink the platform to a point and the two positions collapse onto each other with only the train's length between them — that is the pole case.
How it works
- Decide what has lengthA pole, a man, a signal: no length. A platform, a bridge, a tunnel, another train: length counts.
- Add the lengths that countDistance = train + object. For a pole that is just the train.
- Divide by the speed in m/sConvert first. A km/h figure divided into metres gives a meaningless number.
A pole and then a platform
A 240 m train runs at 54 km/h. How long does it take to pass a pole, and to cross a 360 m platform?
- 54 × 5/1815 m/s
- Pole: the train covers its own 240 m, so 240 ÷ 1516 s
- Platform: 240 + 360600 m
- 600 ÷ 1540 s
Pro tip. The gap between the two times is 24 s, and 24 × 15 = 360 m — the platform's length exactly. That is not a coincidence but the whole structure of the question: the extra time is the time to cover the extra distance. Given any two of platform length, speed and the pair of times, the third comes out of that one line.
A train crosses a pole in 12 seconds and a 180 m platform in 24 seconds. Its length is
- 180 m
- 240 m
- 120 m
The extra 12 seconds buy exactly the platform's 180 m, so the speed is 15 m/s; and in the 12 seconds against the pole the train covers its own length, 12 × 15 = 180 m. The train happens to be the same length as the platform, which is what makes the other options look reasonable — but the two 12s in the question force it.
Notes
- Work as Rate: If a person completes a job in n days, their one-day work is \frac{1}{n}; combined rates simply add, so A (\frac{1}{a}) and B (\frac{1}{b}) together finish in \frac{ab}{a+b} days.
- LCM Method for Work: Take total work as the LCM of the given times to make each person's daily work a whole number — this eliminates fractions in multi-worker problems.
- Speed-Distance-Time: \text{Speed} = \frac{\text{Distance}}{\text{Time}}; convert km/h to m/s by multiplying by \frac{5}{18} and m/s to km/h by \frac{18}{5}.
- Relative Speed: For objects moving in opposite directions add speeds; in the same direction subtract them — this underlies overtaking and meeting problems.
- Average Speed Trap: For equal distances at speeds x and y, average speed is the harmonic mean \frac{2xy}{x+y}, not the arithmetic mean \frac{x+y}{2}.
Formulas
- Combined time for A and B: \frac{ab}{a+b} days
- Work done = Rate \times Time; Total work via LCM of individual times
- Speed = \frac{\text{Distance}}{\text{Time}}; km/h to m/s multiply by \frac{5}{18}
- Average speed for equal distances: \frac{2xy}{x+y}
- Men-days: \frac{M_1 D_1 H_1}{W_1} = \frac{M_2 D_2 H_2}{W_2}
- Relative speed: opposite directions = u+v, same direction = u-v
Exam traps & shortcuts
- Set total work as the LCM of all given completion times; each worker's efficiency becomes an integer and the answer drops out by simple division.
- For equal-distance two-leg journeys use harmonic mean \frac{2xy}{x+y} directly rather than assuming a distance.
- When two people work in alternate days/turns, compute work per full cycle and multiply up to just below the target, then finish the remainder.
Reference tables
Multiply km/h by 5/18 for m/s and m/s by 18/5 for km/h. These five pairs cover most of what an examiner will choose, because they are the values that make the answer come out whole.
| km/h | m/s | Covers 1 km in |
|---|---|---|
| 18 | 5 | 3 min 20 s |
| 36 | 10 | 1 min 40 s |
| 54 | 15 | 1 min 6.7 s |
| 72 | 20 | 50 s |
| 90 | 25 | 40 s |
Every line here should be reconstructible from the concept above it, not merely recalled.
| Quantity | Relation | Watch for |
|---|---|---|
| One day's work | 1/n of the job | The only conversion in the chapter |
| Two together | ab/(a + b) days | Two workers only; three need rates added |
| LCM method | Job = LCM of the given times | Every rate becomes a whole number |
| Men, days, hours | M₁D₁H₁/W₁ = M₂D₂H₂/W₂ | The W's only matter if the job changes |
| Alternate days | Work per cycle, whole cycles, then the tail | The answer depends on who starts |
| Speed | Distance / Time | One unit system throughout |
| km/h to m/s | × 5/18 | The number must get smaller |
| Average speed, equal distances | 2xy/(x + y) | Below the arithmetic mean, always |
| Average speed, equal times | (x + y)/2 | This is the case where averaging is right |
| Relative speed | u + v opposite, u − v same direction | Same direction is far slower than it looks |
| Crossing a pole | own length ÷ speed | A pole, a man and a signal have no length |
| Crossing a platform | (train + platform) ÷ speed | The difference of the two times gives the platform |
Recap
Read only this the night before.
- Rates
- n days means 1/n a day. Add rates, never times. The combined time is always less than the faster worker's own.
- LCM
- Call the job the LCM of the given times and every rate is a whole number. Add the three pair-rates and halve to get all three together.
- Man-hours
- Men × days × hours is fixed. Picture it as a rectangle of constant area: narrower means taller.
- Alternating
- Size one cycle, fit whole cycles under the target, finish the tail at one person's rate. Who starts changes the answer.
- Average speed
- Total distance over total time. Equal distances give 2xy/(x + y), below the mean and nearer the slower speed. Equal times give the ordinary average.
- Trains
- Add speeds head-on, subtract them going the same way. A pole costs the train's own length; a platform costs both lengths.
Practise Time, Work, Speed & Distance
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- 107 exam-style questions on this topic, with explanations
- A 8-question practice set that ends the chapter
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