Time, Speed & Distance and Time & Work at CAT Depth

Arithmetic word problems supply three to five of the twenty-two QA questions, and Time–Speed–Distance with Time & Work is the largest slice of them. At +3/−1 for an MCQ and no negative marking on a TITA item, these are the questions where a candidate who reads carefully earns marks that a faster but careless candidate loses. The unifying idea is a single equation, quantity = rate × time: distance = speed × time for motion and work = efficiency × time for labour. Every result in this chapter — the harmonic mean for average speed, the LCM method for work, the count of meeting points on a circular track, the 65 5/11 minutes between coincidences of clock hands — is that one equation applied to a fixed quantity. Read each worked example for the ratio it uses, not the numbers, because CAT changes the numbers and keeps the ratio.

Proportionality, Units and Average Speed

Distance = speed × time carries three proportionalities. With distance fixed, speed and time are inversely proportional: a speed ratio of 4 : 5 forces a time ratio of 5 : 4. With time fixed, distance is proportional to speed. With speed fixed, distance is proportional to time. Most CAT questions state one fixed quantity in words ("the same route", "they started together") and expect you to name the proportionality without being told.

Conversions and the two average-speed cases
SituationRuleCheck
km/h → m/sMultiply by 5/1872 km/h = 72 × 5/18 = 20 m/s
m/s → km/hMultiply by 18/515 m/s = 15 × 18/5 = 54 km/h
Equal distances at speeds a and bAverage = 2ab/(a + b), the harmonic mean60 and 40 → 2 × 60 × 40/100 = 48
Equal times at speeds a and bAverage = (a + b)/2, the arithmetic mean60 and 40 → 50
Three equal distances at a, b, cAverage = 3abc/(ab + bc + ca)20, 30, 60 → 3 × 36000/(600 + 1800 + 1200) = 30
  • Worked example (late and early). Walking at 4 km/h a student reaches school 10 minutes late; at 5 km/h she reaches 5 minutes early. How far is the school? The two journeys differ by 15 minutes = 1/4 hour over the same distance d, so d/4 − d/5 = 1/4. The left side is d/20, so d = 20/4 = 5 km. Check: 5/4 h = 75 min and 5/5 h = 60 min, a difference of 15 min ✓
  • Worked example (why the harmonic mean). A car goes 120 km at 60 km/h and returns at 40 km/h. Outward time = 120/60 = 2 h; return time = 120/40 = 3 h; total distance 240 km in 5 h, so the average speed is 240/5 = 48 km/h. The formula 2ab/(a + b) = 4800/100 = 48 agrees. The arithmetic mean 50 is wrong because the car spends more time at the slower speed, which drags the average down.
  • Worked example (percentage change in speed). If speed rises by 25% it becomes 5/4 of the original, so over the same distance the time becomes 4/5 of the original — a fall of 20%, not 25%. Always convert the percentage to a fraction, invert it, and read the new percentage off the inverted fraction.
⚠️ The arithmetic-mean trap
Whenever two speeds cover the same distance, their average is the harmonic mean 2ab/(a + b), which is always less than (a + b)/2. The options will contain the arithmetic mean, and it is there precisely because half the room will pick it. Use the arithmetic mean only when the two speeds are held for the same time.

Relative Speed — Trains, Boats and Streams

Two bodies moving in opposite directions approach each other at the sum of their speeds; in the same direction the faster one gains on the slower at the difference. Everything about trains, boats, circular tracks and clock hands is this one sentence. For trains, the distance to be covered is the length that must pass a fixed point: a train crossing a pole covers its own length; crossing a platform or bridge it covers its own length plus the platform's; two trains crossing each other cover the sum of both lengths at their relative speed.

  • Worked example (pole and platform). A 150 m train runs at 54 km/h = 54 × 5/18 = 15 m/s. It passes a pole in 150/15 = 10 s. It passes a 300 m platform in (150 + 300)/15 = 450/15 = 30 s. A candidate who forgets to add the train's length gets 20 s, which will be among the options.
  • Worked example (two trains). Trains of 120 m and 180 m run at 54 km/h and 90 km/h. Opposite directions: relative speed 144 km/h = 40 m/s, distance 300 m, time 300/40 = 7.5 s. Same direction: relative speed 36 km/h = 10 m/s, time 300/10 = 30 s. Same lengths, same speeds, four times the crossing time — the direction word is the whole question.
  • A train passing a person walking. A man walking at 6 km/h in the direction of a 200 m train at 60 km/h is overtaken at relative speed 54 km/h = 15 m/s, in 200/15 = 13⅓ s. The man has no length; only the train's counts.
  • Boats and streams. With boat speed b in still water and stream speed s: downstream speed D = b + s, upstream speed U = b − s. Hence b = (D + U)/2 and s = (D − U)/2. If a boat does 12 km/h downstream and 8 km/h upstream, b = 10 and s = 2.
  • Worked example (round trip). With b = 10 and s = 2, a 24 km round trip takes 24/12 + 24/8 = 2 + 3 = 5 h. Dividing the total 48 km by the still-water speed gives 4.8 h, which is wrong: the stream never cancels out, because the boat spends longer fighting it than riding it. The round-trip average speed is 48/5 = 9.6 km/h = (b² − s²)/b, always less than b.
  • Time ratios give speed ratios. If a boat takes twice as long upstream as downstream over the same distance, then U : D = 1 : 2, so (b − s) : (b + s) = 1 : 2, giving b + s = 2b − 2s and b = 3s. The still-water speed is three times the stream — no distance needed.
🧠 Convert before you compute
When lengths are in metres and speeds in km/h, convert the speed to m/s with 5/18 as the very first line. The CAT setters choose speeds like 54, 72, 90 and 144 km/h precisely because they become the integers 15, 20, 25 and 40 m/s — if your conversion produces a messy decimal, you have probably misread a number.

Circular Tracks, Races, Escalators and Clocks

On a circular track of length L, two runners meet for the first time when the faster has gained exactly one lap on the slower (same direction) or when together they have covered one lap (opposite directions). They are together at the starting point when each has completed a whole number of laps — the LCM of their individual lap times. And the number of distinct meeting points depends only on the ratio of their speeds in lowest terms a : b: it is a − b in the same direction and a + b in opposite directions.

Two runners on a 600 m track: A at 5 m/s, B at 3 m/s, starting together
QuestionSame directionOpposite directions
First meeting anywhereL/(5 − 3) = 600/2 = 300 sL/(5 + 3) = 600/8 = 75 s
First meeting at the startLCM(600/5, 600/3) = LCM(120, 200) = 600 sSame: 600 s — direction does not matter
Distinct meeting points5 − 3 = 2 (at 300 m and at the start)5 + 3 = 8, spaced 75 m apart
Meetings before they are back at the start together600/300 = 2600/75 = 8
  • Races and head-starts. "A beats B by 10 m in a 100 m race" means that when A runs 100, B runs 90, so their speeds are 10 : 9. "A gives B a start of 20 m and they finish together" means A runs 100 while B runs 80: speeds 5 : 4. Worked example. In 100 m, A beats B by 10 m; in 200 m, B beats C by 20 m. By how much does A beat C in 100 m? A : B = 100 : 90 = 10 : 9 and B : C = 200 : 180 = 10 : 9, so A : C = 100 : 81. When A finishes 100 m, C has run 81 m — A wins by 19 m, not 20, because the two 10% margins compound.
  • Escalators — the steps model. Let the escalator have N visible steps and move at e steps per second. A person walking down at p steps per second for t seconds has covered N = pt + et = (p + e)t steps if the escalator moves with him, and N = (p − e)t if it moves against him. Worked example. Walking down a down-moving escalator, a man takes 30 steps in 30 s. Walking three times as fast he takes 60 steps. How many steps does the escalator have? First trip: N = 30 + 30e. Second trip: 3 steps/s for 60 steps takes 20 s, so N = 60 + 20e. Equating: 30 + 30e = 60 + 20e, so 10e = 30, e = 3 steps/s and N = 30 + 90 = 120 steps. Check with the second trip: 60 + 20 × 3 = 120 ✓
  • Clocks are a circular track. The minute hand moves 6° per minute and the hour hand 0.5° per minute, so the minute hand gains 5.5° per minute. The hands coincide every 360/5.5 = 720/11 = 65 5/11 minutes, which is why they coincide only 11 times in 12 hours (22 in a day), not 12. They are at right angles every 180/11 minutes — 22 times in 12 hours. Worked example. When between 3 and 4 o'clock do the hands coincide? At 3:00 the hour hand is 90° ahead; the gap closes at 5.5°/min, so the time is 90/5.5 = 180/11 = 16 4/11 minutes past 3. And the angle at 4:20 is |4 × 30 + 20 × 0.5 − 20 × 6| = |130 − 120| = 10°.
🎯 Why these four topics are one topic
A circular track, a race, an escalator and a clock face all ask the same question: how fast does one thing gain on another, and how much gap must be closed? Once you write "gap ÷ relative speed" the problem is arithmetic. Candidates who memorise a separate formula for each lose the connection and, under time pressure, the formula.

Time & Work — the LCM Method, Wages and Pipes

Take the total work as the LCM of the individual times; each worker's efficiency is then a whole number of units per day, and every question becomes addition. If A takes 12 days and B takes 18, call the work 36 units: A does 3 units a day, B does 2. Together they do 5 a day and finish in 36/5 = 7.2 days. The alternative, adding 1/12 + 1/18 = 5/36 and inverting, is the same calculation with fractions and more places to slip.

  • Worked example (one leaves). A and B above work together for 4 days, then A leaves. In 4 days they finish 4 × 5 = 20 units; the remaining 16 units are B's alone at 2 per day, taking 16/2 = 8 more days. Total 12 days.
  • Efficiency ratios. "A is twice as efficient as B" means A does 2 units when B does 1. If together they finish in 12 days, then 3 units/day × 12 = 36 units of work, so B alone takes 36/1 = 36 days and A alone 36/2 = 18 days. Efficiency and time are inversely proportional: efficiency 2 : 1 means time 1 : 2.
  • Wages follow work done, not days available. A can finish a job in 10 days and B in 15; they do it together for ₹1500. Work = 30 units; A does 3/day, B does 2/day, so A : B = 3 : 2 and A gets 3/5 × 1500 = ₹900, B gets ₹600. Splitting in the ratio of days, 10 : 15, gives A ₹600 — exactly inverted, because the slower worker cannot earn more for the same job.
  • Alternate days — who starts matters. A does a job in 10 days, B in 20; they work on alternate days. Work = 20 units, A = 2/day, B = 1/day, so each two-day cycle does 3 units. Six cycles (12 days) do 18 units, leaving 2. If A started, day 13 is A's and he does the 2 units: 13 days. If B started, day 13 is B's (1 unit, leaving 1) and day 14 is A's, who needs half a day for the last unit: 13.5 days.
Pipes and cisterns: filling is positive, emptying is negative
PipeAloneUnits per hour (tank = 24 units)
A (inlet)Fills in 6 h+4
B (inlet)Fills in 8 h+3
C (leak)Empties in 12 h−2
All three open24/5 = 4.8 h = 4 h 48 min4 + 3 − 2 = +5
  • Worked example (finding the leak). A pipe fills a tank in 8 h, but with a leak open it takes 10 h. How long would the leak alone take to empty a full tank? Tank = 40 units; the pipe gives +5/h, the pipe with leak gives 40/10 = +4/h, so the leak is −1/h and empties the tank in 40 h. The leak is small precisely because it slowed the filling by only 2 hours.
  • A tank that is partly full. If the tank above is already one-quarter full (6 of 24 units) when A, B and C are all opened, only 18 units remain, taking 18/5 = 3.6 h = 3 h 36 min. Always subtract what is already there before dividing.

Man-Days and the Chain Rule

When the number of workers, the hours per day and the size of the job all change at once, use the chain rule: M₁D₁H₁/W₁ = M₂D₂H₂/W₂, where M is the number of workers, D the days, H the hours per day and W the work. It is nothing more than "work = number of workers × time × efficiency" written for two situations with the same efficiency per worker. The quantity M × D × H is measured in man-hours, and a job needs a fixed number of them however you arrange the labour.

  • Worked example (chain rule). 12 men working 8 hours a day build a 240 m wall in 10 days. How many days will 16 men working 6 hours a day take to build a 360 m wall? Left side: 12 × 8 × 10/240 = 960/240 = 4 man-hours per metre. Right side: 16 × 6 × D/360 = 96D/360. Equate: 96D/360 = 4, so D = 4 × 360/96 = 1440/96 = 15 days. Sanity check by ratios: more men (×12/16) and a longer wall (×360/240) and fewer hours (×8/6): 10 × 12/16 × 360/240 × 8/6 = 10 × 0.75 × 1.5 × 4/3 = 15 ✓
  • Men and women — convert to one unit. 3 men or 5 women can do a job in 12 days. How long will 6 men and 10 women take? 3 men = 5 women in output, so 6 men = 10 women, and 6 men + 10 women = 20 women = 4 groups of 5 women. Four such groups finish in 12/4 = 3 days.
  • Workers joining midway — count man-days. A group can finish a job in 30 days. After 10 days, 5 more workers join and the job is finished in 15 more days. How large was the original group? With N workers the job is 30N man-days. Done: 10N (first phase) + 15(N + 5) (second phase) = 25N + 75. So 30N = 25N + 75, 5N = 75, N = 15. Check: 15 × 30 = 450 man-days; 10 × 15 + 15 × 20 = 150 + 300 = 450 ✓
ℹ️ Which method for which question
Use the LCM method when named individuals have different rates (A, B, pipes, leaks). Use man-days when the workers are identical and only their number and hours change. Use efficiency ratios when the question says "twice as efficient" or "takes 20% less time". Every Time & Work question in CAT is one of these three, sometimes two in sequence.

Key takeaways

  • For equal distances the average speed is the harmonic mean 2ab/(a + b); the arithmetic mean applies only to equal times. Convert km/h to m/s with 5/18 before touching lengths in metres.
  • Relative speed is the sum for opposite directions and the difference for the same direction; a train covers its own length past a pole and the sum of lengths past a platform or another train.
  • On a circular track the runners first meet after L/(relative speed), meet at the start after the LCM of lap times, and meet at a − b (same direction) or a + b (opposite) distinct points for speeds in the ratio a : b.
  • Clock hands are a relative-speed problem at 5.5° per minute: they coincide every 65 5/11 minutes, 11 times in 12 hours.
  • Take the work as the LCM of the times so efficiencies are integers; wages split in the ratio of efficiencies, not days; a leak is a negative pipe; and when workers and hours both change, M₁D₁H₁/W₁ = M₂D₂H₂/W₂.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A man drives from town P to town Q at 60 km/h and returns along the same road at 40 km/h. His average speed for the whole journey is:

    1. 50 km/h
    2. 45 km/h
    3. 24 km/h
    4. 48 km/h
    Show answer

    Answer: D — 48 km/h

    The two legs cover equal distances, so the average is the harmonic mean 2ab/(a + b) = 2 × 60 × 40/(60 + 40) = 4800/100 = 48 km/h. Check with a concrete distance: 120 km each way takes 2 h out and 3 h back, so 240 km in 5 h = 48 km/h. The value 50 is the arithmetic mean of the speeds, which would be right only if the two speeds were held for equal times; the value 24 divides the one-way distance by the round-trip time (120/5), forgetting that the 5 hours covered 240 km.

  2. Two trains, 120 m and 180 m long, run on parallel tracks at 54 km/h and 90 km/h in opposite directions. The time they take to cross each other completely is:

    1. 7.5 s
    2. 12 s
    3. 30 s
    4. 3.75 s
    Show answer

    Answer: A — 7.5 s

    Opposite directions means the relative speed is the sum, 54 + 90 = 144 km/h = 144 × 5/18 = 40 m/s. The distance to be covered is the sum of both lengths, 120 + 180 = 300 m, so the time is 300/40 = 7.5 s. The value 30 s uses the difference of speeds, 36 km/h = 10 m/s, which is the same-direction case; 12 s divides 300 m by the faster train's speed alone, 25 m/s; and 3.75 s uses only half the combined length, as if just one train's length had to pass.

  3. A and B start together from the same point on a circular track 600 m long and run in the same direction at 5 m/s and 3 m/s respectively. At how many distinct points on the track will they meet?

    1. 8
    2. 2
    3. 1
    4. 15
    Show answer

    Answer: B — 2

    For speeds in the lowest-terms ratio a : b = 5 : 3 running the same way, the number of distinct meeting points is a − b = 2. Verify: they first meet after 600/(5 − 3) = 300 s, when A has run 1500 m = 2.5 laps, so the meeting is at the 300 m mark; the second meeting at 600 s has A at 3000 m = 5 laps, back at the start. From there the pattern repeats, so only two points are ever used. The value 8 is a + b, the count for opposite directions; the value 1 assumes they meet only at the starting point; and 15 is the product of the speeds, which counts nothing.

  4. Walking down a descending escalator, a man takes 30 steps and reaches the bottom in 30 seconds. Walking three times as fast, he takes 60 steps to reach the bottom. The number of steps visible on the stationary escalator is:

    1. 60
    2. 90
    3. 120
    4. 150
    Show answer

    Answer: C — 120

    Let the escalator move e steps per second and have N steps. First trip: 1 step/s for 30 s, so N = 30 + 30e. Second trip: 3 steps/s for 60 steps takes 60/3 = 20 s, so N = 60 + 20e. Equating, 30 + 30e = 60 + 20e gives e = 3 and N = 30 + 90 = 120. Check: 60 + 20 × 3 = 120 ✓. The value 90 simply adds the two step counts; 60 takes the larger step count as the total, ignoring the escalator's contribution; and 150 keeps 30 s for the second trip (60 + 30 × 3) instead of the 20 s the faster walk actually took.

  5. A can complete a piece of work in 12 days and B can complete the same work in 18 days. Working together, they will complete it in:

    1. 7.2 days
    2. 15 days
    3. 30 days
    4. 6 days
    Show answer

    Answer: A — 7.2 days

    Take the work as LCM(12, 18) = 36 units. A does 36/12 = 3 units a day and B does 36/18 = 2 units a day, so together they do 5 units a day and finish in 36/5 = 7.2 days. The value 15 is the average of the two times, and 30 is their sum — neither has any meaning, since two people together must finish faster than the faster one alone, which rules out any answer above 12. The value 6 halves A's 12 days, as if B were exactly as fast as A — but B is slower, so the joint time must exceed 6.

  6. A alone can finish a job in 10 days and B alone in 15 days. They finish it together and are paid ₹1500 in total. B's share of the payment is:

    1. ₹900
    2. ₹750
    3. ₹600
    4. ₹500
    Show answer

    Answer: C — ₹600

    Wages are shared in the ratio of work done, which is the ratio of efficiencies. With the job as 30 units, A does 3 units a day and B does 2, so A : B = 3 : 2 and B receives 2/5 × 1500 = ₹600. The value ₹900 is A's share — it is what you get for B if you split in the ratio of days, 10 : 15 = 2 : 3, which rewards the slower worker and is exactly inverted. The value ₹750 is an equal split, and ₹500 is one-third, from treating 15 : 10 as 2 : 1.

  7. Pipe A fills a tank in 6 hours and pipe B fills it in 8 hours, while a leak at the bottom can empty the full tank in 12 hours. If both pipes and the leak are open together, the empty tank fills in:

    1. 3 h 26 min
    2. 4 h 48 min
    3. 4 h 30 min
    4. 8 h
    Show answer

    Answer: B — 4 h 48 min

    Let the tank be LCM(6, 8, 12) = 24 units. A adds 4 units an hour, B adds 3, and the leak removes 2, so the net rate is 4 + 3 − 2 = 5 units an hour and the tank fills in 24/5 = 4.8 h = 4 h 48 min. The value 3 h 26 min ignores the leak altogether, 24/7 ≈ 3.43 h; the value 8 h comes from subtracting B instead of the leak, 4 − 3 + 2 = 3 units an hour; and 4 h 30 min is a rounding of 4.8 h read as 4 hours 30 minutes rather than 0.8 × 60 = 48 minutes.

  8. 12 men working 8 hours a day build a 240 m wall in 10 days. The number of days that 16 men working 6 hours a day will take to build a 360 m wall of the same kind is:

    1. 10 days
    2. 11.25 days
    3. 20 days
    4. 15 days
    Show answer

    Answer: D — 15 days

    Apply M₁D₁H₁/W₁ = M₂D₂H₂/W₂: 12 × 8 × 10/240 = 16 × 6 × D/360. The left side is 960/240 = 4, so 96D/360 = 4 and D = 1440/96 = 15 days. As ratios from the original 10 days: × 12/16 for more men, × 8/6 for fewer hours, × 360/240 for the longer wall gives 10 × 0.75 × 4/3 × 1.5 = 15 ✓. The value 10 forgets that the wall is longer (960 = 96D gives D = 10); 11.25 ignores the change in hours per day (12 × 10/240 = 16 × D/360); and 20 keeps 12 men on the right side instead of 16, forgetting that the workforce grew.