Time and Work is the single most rewarding chapter in the quantitative section of judiciary and CLAT-PG screening papers: a handful of ideas — that work is the product of rate and time, that rates simply add up, and that efficiency is the reciprocal of time — generate dozens of question types, from a lone mason building a wall to three pipes quarrelling over a cistern. Once you stop chasing fractions and start counting units of work, the whole topic collapses into mental arithmetic. This chapter builds the concept from first principles, derives every standard formula, and walks through the recurring problem patterns — combined work, alternate days, leaving midway, pipes and cisterns, and wage distribution — that examiners recycle year after year.
The core relationship: Work = Rate x Time
Every problem in this chapter rests on one identity: Work = Rate × Time, usually written W = R × T. Here work is the total task (a wall, a field, a full tank), rate (also called efficiency) is how much of that task a worker finishes in one unit of time, and time is the number of units taken. Rearranging gives the two forms you will use constantly: R = W / T and T = W / R.
The chapter's golden insight is that rate is the reciprocal of time when the whole job is taken as one unit. If A finishes a job in 10 days, A's one-day work is 1/10 of the job; if B finishes it in 15 days, B's one-day work is 1/15. This reciprocal relationship — efficiency is inversely proportional to time taken — is the hinge on which the entire topic turns, and it is the same proportional reasoning you practised in percentage, ratio and proportion.
Because rates measure work per unit time, they are additive. When several workers labour together on the same task, their combined one-day work is the simple sum of their individual one-day works. That additivity — and nothing more exotic — is what lets us solve almost every "working together" question.
It is worth pausing on why the reciprocal works. "Work" in this chapter is a dimensionless whole: we do not care whether the task is a 200-metre wall or a hectare of harvest, only that it is one complete unit. If the unit is done in 10 days at a steady pace, then in one day exactly one-tenth is done, by definition of "steady pace". The assumption of constant rate is the silent premise of every problem; examiners rarely state it, but it is what makes the arithmetic linear. When a question deliberately breaks the assumption — say a worker tires and slows after lunch — it must spell that out, and you handle each phase separately. Absent such wording, treat every rate as fixed.
Working together: the fraction method
Suppose A can do a job in 12 days and B in 18 days. A's daily work is 1/12, B's is 1/18. Together their daily work is 1/12 + 1/18. Taking the LCM of 12 and 18, which is 36, this becomes 3/36 + 2/36 = 5/36. So together they complete 5/36 of the job each day, and the time to finish is the reciprocal, 36/5 = 7.2 days.
The general two-worker shortcut follows directly: if A takes a days and B takes b days, together they take ab / (a + b) days. For the example, (12 × 18) / (12 + 18) = 216 / 30 = 7.2 days, confirming the longhand. Note that the combined time is always shorter than the faster worker alone — a useful sanity check that catches sign and reciprocal errors in the exam hall.
A second classic uses the same shortcut in reverse. "A and B together finish a job in 8 days; A alone takes 12 days. How long does B take alone?" Together their daily work is 1/8; A's is 1/12; so B's daily work is 1/8 − 1/12. With LCM 24, that is 3/24 − 2/24 = 1/24, so B alone takes 24 days. The lesson is that subtraction of rates recovers a missing worker's solo time exactly as addition combines them — the operations are symmetric because rates live on a single additive number line.
The fraction method is transparent but the denominators grow ugly with three or more workers or with mixed fractions. That is exactly where the LCM unit method, covered next, earns its keep.
The LCM unit method: stop using fractions
Instead of treating the job as 1, assume the total work equals the LCM of the given times. Each worker's rate then becomes a whole number of units per day, and the messy fractions vanish. This is the technique every serious aspirant adopts because it converts the topic into integer arithmetic you can do in your head.
Reworking the earlier problem: A takes 12 days, B takes 18 days, so let total work = LCM(12, 18) = 36 units. A's efficiency is 36/12 = 3 units/day; B's is 36/18 = 2 units/day. Together they do 5 units/day, and 36 / 5 = 7.2 days — the same answer, reached without a single fraction. The LCM is a convenient fiction: any common multiple works, but the lowest keeps the numbers small.
For three workers — A in 20 days, B in 30 days, C in 60 days — take LCM(20, 30, 60) = 60 units. Efficiencies are 3, 2 and 1 unit/day, summing to 6 units/day, so together they finish 60/6 = 10 days. The method scales effortlessly and, because efficiencies emerge as clean integers, it doubles as the natural setup for wage-distribution questions. Aspirants who internalise the LCM method here find it equally indispensable in time, distance and speed, where the structurally identical Distance = Speed × Time governs every problem.
Efficiency, ratios and the inverse relationship
Efficiency is rate by another name, and the crucial fact is that the ratio of efficiencies is the inverse of the ratio of times. If A is twice as efficient as B, then A takes half the time B does. Formally, if times are in the ratio a : b, efficiencies are in the ratio b : a (equivalently 1/a : 1/b).
Examiners love to dress this up. "A is 25% more efficient than B; B takes 25 days alone — how long does A take?" If A is 25% more efficient, the efficiency ratio A : B is 125 : 100 = 5 : 4, so the time ratio is the inverse, 4 : 5. Since B's time corresponds to 5 parts = 25 days, one part = 5 days, and A's 4 parts = 20 days. Watch the direction of the percentage: "more efficient" means less time, a trap that mirrors the gain-versus-cost confusion in profit, loss and discount.
Comparisons such as "3 men do as much work as 5 women" are pure ratio statements: the efficiency of one man to one woman is 5 : 3. Convert every worker into a common currency of efficiency units and the rest is arithmetic.
The M1D1H1 work-equivalence rule
When a question varies the number of workers, the days, the hours per day, and the size of the job all at once, the unified tool is the work-equivalence (or "chain") rule. If M1 men working H1 hours a day for D1 days complete work W1, and M2 men working H2 hours a day for D2 days complete work W2, then:
(M1 × D1 × H1) / W1 = (M2 × D2 × H2) / W2
The product M × D × H is total man-hours, and man-hours per unit of work is constant for a fixed difficulty. Drop H if hours are not mentioned and drop W if the same job is done in both scenarios. Example: if 10 men build a 200-metre wall in 8 days working 6 hours a day, how many days do 12 men need to build a 300-metre wall working 8 hours a day? Set (10 × 8 × 6)/200 = (12 × D2 × 8)/300. The left side is 480/200 = 2.4; so 2.4 = 96 D2 / 300, giving D2 = 720/96 = 7.5 days.
The rule is just disguised direct and inverse proportion — work is directly proportional to men, days and hours, and the number of days is inversely proportional to the number of men. Keeping that intuition prevents you from inverting a ratio by accident.
Pattern: together, then someone leaves
A staple of judiciary papers: "A and B start together; after some days A leaves; B finishes the rest. Find the total time." The LCM method dispatches these cleanly. Take A in 16 days and B in 24 days, with A leaving 4 days before completion. Let total work = LCM(16, 24) = 48 units; efficiencies are 3 and 2 units/day.
In the last 4 days only B works, doing 2 × 4 = 8 units. The remaining 48 − 8 = 40 units are done by both together at 5 units/day, taking 40/5 = 8 days. Total time = 8 + 4 = 12 days. The reliable strategy is to peel off the solo stretch first, then divide the remainder by the combined rate.
A close cousin asks for how many days A actually worked, or what fraction of the job each did — answered immediately once you have efficiencies in integer units. In the worked example, A did 3 units/day for the 8 combined days = 24 units, exactly half of 48, while B did the other 24 units (16 in the joint stretch plus 8 solo); if a wage were attached, the pay would split equally. Notice how the unit method hands you the wage answer as a by-product of the timing answer.
The phrasing also varies: sometimes the slower worker leaves, sometimes the faster; sometimes a worker joins late rather than leaving early. Whatever the storyline, the same two-step discipline holds — isolate any stretch worked by a lone rate, then divide the remaining units by whatever combined rate is in force. Always state the total work in units at the top of your rough work; it anchors every subsequent step.
Pattern: working on alternate days
"A and B work on alternate days, A starting first" requires you to think in two-day cycles. With A in 9 days and B in 12 days, let total = LCM(9, 12) = 36 units; efficiencies 4 and 3 units/day. A two-day cycle (A then B) produces 4 + 3 = 7 units.
After 4 full cycles (8 days) the pair has done 28 units, leaving 8 units. Day 9 is A's turn: A does 4 units, leaving 4 units after 9 days. Day 10 is B's turn: B does 3 units/day, so B needs 4/3 of a day to finish the last 4 units. Total time = 9 + 4/3 = 31/3 ≈ 10.33 days. Crucially, who starts matters — start with the slower worker and the answer can differ, because the leftover at the end is handled by a different rate.
The disciplined approach: compute work per cycle, find how many complete cycles fit, then finish the remainder day by day in the correct order. Rushing straight to a formula is where most candidates lose the mark.
Pipes and cisterns: positive and negative work
Pipes and cisterns is Time and Work wearing a plumber's overalls. A pipe that fills a tank does positive work; a pipe that empties it (an outlet or leak) does negative work. The net rate when several pipes are open is the sum of inlet rates minus the sum of outlet rates, and the time to fill is the tank size divided by that net rate.
Example: pipe A fills a tank in 6 hours, pipe B in 8 hours, and outlet C empties it in 12 hours; all three are open. Take tank = LCM(6, 8, 12) = 24 units. Rates: A = +4, B = +3, C = −2 units/hour. Net = 4 + 3 − 2 = 5 units/hour, so the tank fills in 24/5 = 4.8 hours. If the net rate ever comes out negative or zero, the tank never fills — a result examiners use to test whether you are reading signs correctly.
A favourite twist: a tank is full and a leak empties it; an inlet that should fill it in x hours now takes longer because of the leak. Treat the leak as a negative rate and solve for it. Worked through: an inlet fills a tank in 5 hours, but with a leak at the bottom it takes 6 hours. Take tank = LCM(5, 6) = 30 units. The inlet alone is 30/5 = 6 units/hour; inlet-minus-leak is 30/6 = 5 units/hour; so the leak removes 6 − 5 = 1 unit/hour, meaning the leak alone would empty the full tank in 30/1 = 30 hours. The arithmetic is identical to worker problems; only the vocabulary changes.
One more layer that examiners enjoy: pipes opened at different times. "Pipe A (fills in 10 hours) is opened first; after 4 hours pipe B (fills in 15 hours) is opened too." Take tank = 30 units, rates 3 and 2 units/hour. In the first 4 hours A alone puts in 12 units, leaving 18; thereafter both run at 5 units/hour, taking 18/5 = 3.6 more hours. Total fill time = 4 + 3.6 = 7.6 hours. As with workers leaving midway, handle the solo stretch first, then the combined stretch.
Wages: dividing pay in the ratio of work done
The governing principle is that wages are divided in the ratio of work actually done, which — when everyone works the same number of days — reduces to the ratio of efficiencies. This follows from W = R × T: for equal T, work is proportional to rate.
If A, B and C can individually finish a job in 6, 8 and 12 days, and they complete it together for a total payment of ₹1,800, then efficiencies (taking LCM(6,8,12) = 24 units) are 4, 3 and 2 units/day. Working the same number of days together, their shares are in the ratio 4 : 3 : 2. The total is 9 parts = ₹1,800, so one part = ₹200, giving A = ₹800, B = ₹600, C = ₹400.
When workers put in different numbers of days, do not use efficiency alone — multiply each worker's rate by the days actually worked to get units done, and split the wages in that ratio. Confusing the two cases is the single most common wage-distribution error. The proportional split is the same machinery you met in ratio and proportion; only the quantity being shared has changed.
Scaling a workforce up or down
Many questions never name individual workers; they scale a crew and ask for the new time. The principle is inverse proportion: with a fixed job, the number of workers and the number of days multiply to a constant. If 15 workers finish a job in 24 days, the job is 15 × 24 = 360 worker-days. Twenty workers then finish it in 360/20 = 18 days, and if 5 workers walk off after the start the remaining-work logic from earlier sections applies to the balance.
A richer version interrupts the work: "40 men can build a bridge in 30 days. After 12 days, 20 more men join. When is it finished?" The job is 40 × 30 = 1200 man-days. In the first 12 days, 40 men do 480 man-days, leaving 720. From day 13 there are 60 men, who clear 720/60 = 12 more days. Total = 12 + 12 = 24 days, six days ahead of schedule. The reliable routine is to convert the whole job into man-days at the outset, subtract what is finished before the change, and divide the balance by the new headcount.
The mirror image — work falling behind because labourers leave or are absent — is solved identically; only the sign of the change flips. Keep the man-days total fixed and let the days float, and these problems become bookkeeping rather than algebra.
Mixed groups: men, women and children
Questions mixing men, women and children are equivalence puzzles. The bridge is a sentence such as "2 men = 3 women = 4 children (in work done)", which fixes the efficiency ratio. Convert everyone into the smallest unit. If 2 men equal 3 women, one man's efficiency is 3/2 that of a woman; if 3 women equal 4 children, scale all three onto a common base.
Take "6 men or 8 women can do a job in 10 days; how long do 3 men and 4 women take?" Total work in man-equivalents: 6 men × 10 days = 60 man-days, so the job is 60 man-days. Since 6 men = 8 women in rate, one woman = 6/8 = 3/4 of a man. The mixed team of 3 men and 4 women equals 3 + 4 × (3/4) = 3 + 3 = 6 men-equivalents, finishing 60/6 = 10 days. (Here the team happens to match the original 6-man crew.)
The discipline is always the same: pick one worker type as the unit, express everyone in that unit, and the problem becomes a one-line equivalence. The same conversion-to-a-common-unit habit underlies success across the whole quantitative section, which is why the aptitude and reasoning hub treats ratio fluency as the spine of the syllabus.
Fraction of work done and remaining work
Some questions never ask for total time; they ask what fraction is left. Here the LCM units shine because a fraction is just "units done over total units". If a job is 60 units and a worker does 36 units, the fraction completed is 36/60 = 3/5 and the fraction remaining is 2/5.
A typical phrasing: "A can do a job in 20 days. A works for 8 days, then leaves. What fraction is left, and how long will B (alone, 30 days for the whole job) take to finish it?" A's 8 days at 1/20 per day complete 8/20 = 2/5; so 3/5 remains. B's rate is 1/30 per day, so B needs (3/5) ÷ (1/30) = (3/5) × 30 = 18 days. Setting total = LCM(20, 30) = 60 units gives the same path with integers: A does 8 × 3 = 24 units, 36 remain, B at 2 units/day takes 18 days.
Always read whether the examiner wants the fraction done or remaining, and whether the finisher works alone or with help — the words "left", "remaining" and "alone" carry the marks.
Common traps and exam strategy
First, the reciprocal trap: a worker's time and a worker's one-day work are reciprocals, not equals — never add days, only add rates. Second, the efficiency-direction trap: "more efficient" means less time; flip the ratio. Third, the alternate-days trap: the starting worker changes the leftover at the end, so the answer is order-sensitive. Fourth, the wage trap: split by units of work actually done, not by efficiency, whenever the days worked differ.
Strategically, default to the LCM unit method for everything except the simplest two-worker cases, where ab/(a+b) is faster. Write the total work in units at the very top of your rough sheet and keep all efficiencies as integers. Sanity-check by remembering that the combined time of a team is always less than the time of its fastest member, and that a net-negative pipe system never fills.
Finally, pace yourself: a Time and Work question is typically a 45-to-60-second item once set up correctly. If you find yourself wrestling with denominators, you have abandoned the unit method — reset, take the LCM, and the numbers will fall into place.
Frequently asked questions
What is the basic formula for Time and Work problems?
The single governing identity is Work = Rate × Time. Since efficiency (rate) is the reciprocal of time for a one-unit job, a worker finishing a task in n days does 1/n of it per day. Rates are additive, so workers together complete the sum of their individual one-day works each day, and the time to finish is the reciprocal of that combined rate.
Why is the LCM method preferred over fractions?
Assuming the total work equals the LCM of the given times turns every worker's rate into a whole number of units per day, eliminating fractional arithmetic. It scales painlessly to three or more workers, to pipes and cisterns, and to wage problems, because efficiencies emerge as clean integers that you can add, subtract and divide mentally.
If two workers take a and b days alone, how long do they take together?
Together they take ab / (a + b) days. For A in 12 days and B in 18 days, that is (12 × 18) / (12 + 18) = 216/30 = 7.2 days. The combined time is always shorter than the faster worker alone, which is a quick way to catch an arithmetic slip.
How does the M1D1H1 work-equivalence rule work?
When men, days, hours per day and the size of the job all vary, use (M1 × D1 × H1) / W1 = (M2 × D2 × H2) / W2. The product of men, days and hours is total man-hours, and man-hours per unit of work stays constant for a fixed difficulty. Drop H if hours are not given and drop W if the same job is done in both scenarios.
How are pipes-and-cisterns problems different from worker problems?
They are mathematically identical, except that an emptying pipe (outlet or leak) does negative work. The net filling rate is the sum of inlet rates minus the sum of outlet rates, and the fill time is the tank size divided by that net rate. If the net rate is zero or negative, the tank never fills.
How should wages be divided among workers?
Wages are divided in the ratio of work actually done. When everyone works the same number of days this reduces to the ratio of efficiencies; when the days worked differ, multiply each worker's rate by the days worked to get units done and split the pay in that ratio. Splitting by efficiency alone when days differ is the most common mistake.