Time and Work

This is the starting type. Only one person (or one machine) is working, and the question changes the time or the part of the work. Everything rests on a constant rate: in equal time, equal work.

So the method has two steps. First find the work done in one unit of time (per day, per minute). Then multiply by the new time, or divide the required work by the rate to get the time.

Radhika types 45 pages in 3 hours, so her rate is 15 pages per hour. In 7 hours she will type 105 pages, and 60 pages will take her 4 hours.

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Parts and percentages follow the same thinking. If Radhika completes 30% of a project in 9 days, she does 10% in 3 days, so the whole project takes 30 days and 70% takes 21 days. You don't even need the whole separately; go from the given part to the asked part in one small step.

The second kind of question gives the work day by day: "on the first day he did 12 of the work, on the second day 13 of the remaining work…". Here the word remaining decides everything: apply each fraction to what is left at that moment. Keep a running balance after each day.

For example, take a work of 60 units. Radhika does 12 of it on day 1 (30 units, 30 left). On day 2 she does 13 of the remaining, that is 10 units, so 20 units are left.

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When two or more people work together, their efficiencies add up. The whole type is that one line.

Radhika can clean a hall in 6 hours and Meera in 9 hours. Total work = LCM(6, 9) = 18 units. Radhika = 3 units/hour, Meera = 2 units/hour, together 5 units/hour. So together they take 185 = 3.6 hours, which is 3 hours 36 minutes.

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Look at what we did. We never added the times; 6 + 9 = 15 hours would be absurd, because two people working together must take less time than the faster one alone. We added speeds. So the together time is always less than the smallest individual time, and you can use this to strike out wrong options at once.

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For three people the steps don't change: LCM of three numbers, three efficiencies, add, divide. With 12, 18 and 36 days, the total is 36 units and the efficiencies are 3, 2 and 1, so together they do 6 units a day and finish in 6 days.

The second form asks about a part of the work: "what part is done in 3 days?", "how much work is left after 4 days?", "in how many days will they finish 25% of the work?". Same units, one more line: work done = efficiency × days, and work left = total − work done. Write the answer as a fraction or percentage of the total.

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The third form gives the rates directly, such as pages per hour or items per minute. Then you don't need an LCM at all, because rates add in the same way. If Radhika types 12 pages an hour and Meera 8, together they type 20 pages an hour.

Work given as a quantity. Not every question speaks of a "whole job". Many give the output: trees cut, acres ploughed, pages typed, bricks laid. Then you do not need an LCM at all. Turn each worker into a rate, that is, quantity per hour (or per day), add the rates, and divide the total quantity by the combined rate.

Radhika cuts 24 trees in 6 hours, so she cuts 4 trees an hour. Meera cuts 30 trees in 5 hours, so she cuts 6 an hour. Together they cut 10 trees an hour, and 85 trees take 8510 = 8.5 hours. Keep the unit the same for everybody: one worker's "per hour" must never be added to another worker's "per day".

This is Type 2 run backwards. The question tells you how long the group takes and how long some members take, and you have to find the missing member. The rule is just as simple: efficiency of the missing person = efficiency of the group − efficiencies of the known persons.

Radhika and Meera together finish a project in 8 days, and Radhika alone takes 12 days. Total work = LCM(8, 12) = 24 units. Together = 3 units/day, Radhika = 2 units/day, so Meera = 1 unit/day and Meera alone takes 24 days.

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The same subtraction works for three or four people. Suppose A, B and C together take 6 days, and A and B alone take 9 and 18 days. With 18 units: group = 3, A = 2, B = 1, so C = 0, which means C does nothing at all. If you ever get such a result, read the question again; a valid question leaves a positive efficiency for the missing person.

Sometimes the known person is given through a part of the work: "George does 35% of the work in 6 days". Convert it into whole-work time first (as in the foundation section on parts of the work), then subtract. And when the question finally asks for a part of the work by the missing person ("C alone will do 49 of the work in how many days?"), find C's full time first and then take that part.

A quick sense check: the missing person's time must be more than the group's time. If your answer is smaller than the group's time, you have subtracted the wrong way round.

One standard result is worth remembering. If A alone takes x days more than A and B together, and B alone takes y days more, then together they take √xy days. Say Radhika alone needs 4 days more than Radhika and Meera together, and Meera alone needs 9 days more. Together time = √4 × 9 = 6 days, so Radhika alone takes 10 days and Meera 15 days. Check: 10 × 1525 = 6.

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In this type nobody's individual time is given. Pairs are given instead: "A and B together take 12 days, B and C take 15 days, C and A take 20 days." Don't sit down to solve three equations. See what happens when you add the three pairs:

(A + B) + (B + C) + (C + A) = 2(A + B + C)

Every person appears exactly twice. So the sum of the three pair efficiencies is double the efficiency of all three together. Halve it and you have A + B + C. After that each person comes out by one subtraction: A = (A + B + C) − (B + C), and so on.

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Say Radhika and Meera together need 9 days, Meera and Sita 12 days, and Sita and Radhika 18 days. Take total work = 72 units (double the LCM 36, so that halving later still gives whole numbers). Pair efficiencies: 8, 6 and 4. Their sum is 18 = 2(R + M + S), so all three together do 9 units/day and take 8 days. Radhika = 9 − (Meera + Sita) = 9 − 6 = 3, so Radhika alone takes 24 days.

Sometimes only two pairs and the whole group are given. Then there is nothing to add: if A + B + C and A + B are known, C is a direct subtraction.

The second family in this type gives a relation between people: "A can do the work in the same time as B and C together". This means efficiency of A = efficiency of B + efficiency of C. Write it in units and the rest is one line. A wording like "A takes 2.4 times as long as B and C together" means A's efficiency is 12.4 of the efficiency of (B + C).

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When a question asks whether the given statements are enough (data sufficiency), use the same thinking. Two pair sums such as (Q + R) and (P + R) can never give P + Q + R, because you still don't know any one person.

Now we compare two workers. Only one rule is needed: for the same work, time ∝ 1efficiency. The time ratio is the efficiency ratio turned upside down.

If the efficiencies of A and B are 3 : 2, their times are 2 : 3. If A takes 10 days, B takes 10 × 32 = 15 days.

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Percentages here are ratios, so use the fraction values you learnt in Percentage:

• "A is 25% more efficient than B" ⇒ A : B = 125 : 100 = 5 : 4 ⇒ time ratio 4 : 5.
• "A is 20% less efficient than B" ⇒ A : B = 80 : 100 = 4 : 5 ⇒ time ratio 5 : 4.
• "A is 33.33% more efficient" ⇒ 33.33% = 13, so A : B = 4 : 3 ⇒ time ratio 3 : 4.

Meera finishes a model in 21 days, and Radhika is 75% more efficient. Efficiency ratio R : M = 175 : 100 = 7 : 4, so the time ratio is 4 : 7. Radhika takes 21 × 47 = 12 days.

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"n times as efficient", "n times as fast" and "n times as good a workman" all mean the same thing: efficiency ratio n : 1 and time ratio 1 : n.

When the two workers are described through parts ("A does 13 of the work in 10 days"), first find each one's time for the whole work, then take the reciprocal ratio. "Who is more efficient, and by what percent?" is then a plain percentage question: divide the difference by the smaller efficiency.

This type joins Type 5 with working together. Take the efficiency ratio itself as units per day; the total work then comes from whatever time is given.

Radhika is 4 times as efficient as Meera, and together they finish a mural in 10 days. Take Radhika = 4 units/day and Meera = 1 unit/day. Together they do 5 units a day, so the total work = 5 × 10 = 50 units. Radhika alone takes 504 = 12.5 days and Meera alone takes 50 days.

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The second form gives a difference of times: "A is 3 times as efficient as B, and A finishes 40 days earlier." Efficiency 3 : 1 means time 1 : 3. The difference of 3 − 1 = 2 parts is 40 days, so 1 part = 20 days. A takes 20 days, B takes 60 days, and together they take 20 × 6080 = 15 days.

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The third form asks for a share of the work: "A is 50% more efficient than B; what percent of the work does A do?" When they work together, each does work in the ratio of efficiencies. A : B = 3 : 2, so A does 35 of the work, that is 60%.

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Sometimes the efficiencies hide inside group statements, like "Q and R together are 2 times as efficient as P". Read it as (Q + R) = 2P. If Q and R together take 50 days, P works at half their speed and takes 2 × 50 = 100 days. Turn the sentence into an equation first; the arithmetic after that is always light.

Here the team changes in the middle. A and B start together and then A leaves, or A starts alone and B joins later. The method stays the same: break the work into phases and add the units of each phase.

Phase 1: who is working, and for how many days? Work done = (their efficiency) × (days).
Phase 2: remaining work = total − phase 1 work. Time = remaining workefficiency of the new team.

Radhika can stitch a quilt in 10 days and Meera in 40 days. They start together, and Meera leaves after 6 days. Total = LCM(10, 40) = 40 units; Radhika = 4, Meera = 1 unit/day. In 6 days they make 6 × 5 = 30 units. The remaining 10 units are done by Radhika alone in 104 = 2.5 more days, so the quilt takes 8.5 days in all.

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Joining works the same way. If Radhika works alone for 5 days and then Meera joins, phase 1 = 5 × 4 = 20 units, and the remaining 20 units at 5 units/day take 4 days. Total time = 5 + 4 = 9 days.

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With three people and two changes, make three phases. When a person "joins after 8 days from the beginning", count the days from day 1, not from the previous joining. A small timeline on rough paper helps.

And read the last line carefully: "how many more days" wants only the last phase, while "total days" or "in how many days was the work completed" wants all the phases added.

Type 7 gave you everyone's speed and asked for the time. This type gives you the time of every phase and hides one person's speed. Same phases, opposite direction: find the work done in the known phases; whatever is left belongs to the unknown person.

Radhika can bind an album in 24 days. She works for 8 days, then Meera joins, and together they finish the rest in 8 more days. Take 24 units, so Radhika = 1 unit/day. Her 8 days give 8 units, which leaves 16 units for the two of them in 8 days, that is 2 units/day together. Meera = 2 − 1 = 1 unit/day, so Meera alone also takes 24 days.

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A very common form: "A and B together can do it in 30 days. A works for some days, then B finishes the rest alone in some days." The way in is to split the given days into "together days + extra days". If A works 10 days and B works 25 days, read it as (A + B) for 10 days plus B alone for 15 days. The (A + B) part is known, so B comes out directly.

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Another form brings in a new worker at the end: "after 5 days, C joins and they finish the rest in 4 days". Find the remaining work, divide by 4 to get the team's speed, and subtract the known members. The question may then ask for C's time for only a part of the work, so take that part at the end.

When a percentage of the work is given ("A does 60% in 18 days"), turn it into whole-work time first. After that it is the same subtraction again.

"A left 5 days before the completion of the work." Now you don't know when A left, because you don't know when the work ended. Two simple ways handle every such question.

Way 1: start from the end. In the last 5 days only the others worked. Their work in those days is known, so subtract it from the total. The rest was done by the full team, and its time follows. Add the two parts.

Radhika can paint a room in 12 days and Meera in 24 days. They start together, but Radhika leaves 3 days before the work is finished. Total = 24 units; Radhika = 2, Meera = 1. In the last 3 days Meera alone does 3 units. The other 21 units were done together at 3 units/day, which takes 7 days. Total time = 7 + 3 = 10 days.

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Way 2: let the total time be T. Write each person's working days in terms of T: the one who worked till the end worked T days, the one who left 5 days early worked (T − 5) days, and so on. Add everybody's units and equate the sum to the total work. When two or three people leave at different times, this is the safest way.

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The reverse question also comes: "they started together, after some days A left, and B finished the rest in 17 days. After how many days did A leave?" Let A work x days; then B worked x + 17 days. Again in units: A·x + B·(x + 17) = total.

In this type people do not work together every day; they take turns. Think in terms of a cycle: find the work done in one complete cycle, fit in as many full cycles as possible, and then finish the last bit day by day.

Radhika can finish a project in 10 days and Meera in 15 days. They work on alternate days, Radhika first. Total = 30 units; Radhika = 3, Meera = 2. One cycle (2 days) = 5 units, so 5 cycles give 25 units in 10 days. Day 11 is Radhika's: 3 units, total 28. Day 12 is Meera's: 2 units, total 30. The work finishes in 12 days.

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Look at the end very carefully. Do not simply divide the total by the work of one cycle. Here 305 = 6 cycles = 12 days happened to be right, but if Radhika could have finished the remaining bit on her own day, the answer would have been shorter. So stop the full cycles a little early and finish by hand. If the last worker needs only part of his day, take that fraction: needing 1 unit out of his 3 means 13 of a day.

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The same idea covers many wordings:

• Three people in rotation (A on day 1, B on day 2, C on day 3, and repeat): the cycle is 3 days.
• "A works every day, and B and C help him on every third day": the cycle is 3 days with A, A, A + B + C.
• "The man starts on the first day" versus "the woman starts": compare the two cycles to find each person's speed.
• Building and breaking: one adds units, the other removes them. The job is done the first time the wall reaches full, and that happens on a building day, so stop there and not at the end of a cycle.
• Holidays: "A takes a day off after every four days" is a 5-day pattern for A.

A worker who undoes the work. Sometimes one person builds and another breaks down: a mason raises a wall and a labourer demolishes it, or one worker keeps spoiling part of what the team has done. Treat the breaker as a worker with negative efficiency, exactly like a leak in the Pipe and Cistern chapter. If they work together every day, the net efficiency is builder − breaker. If they work on alternate days, one cycle adds (builder − breaker) units.

Radhika builds a wall in 3 days and Meera breaks the same wall in 5 days. Take the wall as 15 units: Radhika adds 5 units a day and Meera removes 3. Working together every day, the net gain is 2 units a day, so the wall needs 7.5 days. On alternate days, starting with Radhika, the wall stands at 5, 2, 7, 4, 9, 6, 11, 8, 13, 10 after days 1 to 10, and Radhika's 11th day takes it to 15. So the wall is complete in 11 days.

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The wall is complete the first time the total touches the full height, and that can happen only on a building day. So do not divide the full height by the gain of one cycle. Stop a little before the end and finish day by day.

Now the workers come in groups: "12 men can build a wall in 20 days working 8 hours a day". The idea is the same as before, only counted in a new unit. If all the men are equally efficient, the total work is measured in man-days (or man-hours).

12 men for 20 days = 240 man-days. Whether 24 men work for 10 days or 6 men for 40 days, it is the same 240 man-days and the same wall. If the hours per day also change, count man-hours: 12 × 20 × 8 = 1920 man-hours.

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This gives the well-known formula:

M₁ × D₁ × H₁W₁ = M₂ × D₂ × H₂W₂

M = number of men, D = days, H = hours per day, W = amount of work (walls, metres of canal, pages, bottles…). If the work is the same in both situations, W₁ = W₂ and it cancels. If the men have different efficiencies, multiply by the efficiency E as well.

For example, 12 men dig a 60 m trench in 10 days working 6 hours a day. How many days will 15 men need for a 90 m trench working 8 hours a day? 12 × 10 × 660 = 15 × D × 890 ⇒ 12 = 120D90 ⇒ D = 9 days.

Food and ration questions use the same formula with "eaters × days". Food for 500 people for 40 days is 20,000 person-days of food. If 100 people leave after 10 days, the food left is 500 × 30 = 15,000 person-days, enough for 400 people for 37.5 days.

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When two different persons come with their own hours ("R works 6 hours a day for 18 days"), first convert each into the total hours for the whole work (here 108 hours), then use Type 2 on those hours.

Work measured as volume. In digging questions the amount of work is the volume of earth removed, so W in the MDH formula is a volume: length × breadth × depth for a trench or a tank, and 227 r² h for a round well. If the breadth and depth do not change, W is simply proportional to the length. Everything else is as before: men, days and hours stay on one side and the volume goes on the other.

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Radhika's team of 8 men digs a trench 20 m long, 3 m wide and 2 m deep in 10 days. The earth removed is 20 × 3 × 2 = 120 m³ for 8 × 10 = 80 man-days, so every m³ costs 23 man-day. A trench of 24 m × 5 m × 2 m is 240 m³, so it needs 160 man-days, and 10 men finish it in 16 days. For a round well, remember that the volume depends on r², so doubling the diameter makes the volume four times.

This type is Type 7 for groups. Keep a simple account of man-days: total work (in man-days) = sum of the man-days of every phase.

20 workers can finish a road in 30 days, so the road is 600 man-days of work. After 12 days (240 man-days done), 4 workers leave. The remaining 360 man-days are done by 16 workers in 36016 = 22.5 more days.

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The contractor version asks: "how many extra workers are needed to finish on time?" Find the work left and the days left, divide to get the men needed, and subtract the men already there. Suppose 50 men have done only half the work after 40 of the 60 days. The other half also needs 50 × 40 man-days, but in 20 days, which means 100 men, so 50 extra.

Two more patterns are worth knowing:

• Fewer men, more days. "If there were 10 persons less, it would take 10 days more." Same work means n × D = (n − 10)(D + 10). One linear equation gives n.
• Men dropping out every day. "A group planned to finish in 5 days, but 4 men left every day, so it took 6 days." The daily strength is n, n − 4, n − 8, … for 6 days, which is an arithmetic progression. Its sum 6n − 4(0 + 1 + … + 5) = 6n − 60 must equal the planned 5n man-days, so n = 60.

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If the hours per day also change in the middle, count man-hours instead of man-days. The bookkeeping stays the same.

In Type 11 all the workers were equal. Here a man, a woman and a boy work at different speeds, so counting heads is not enough. The method: convert everyone into one kind of worker (or find each one's units per day) and then use the MDH rule.

Form 1: "or" statements. "4 men or 6 women can do the work in 15 days" means 4 men = 6 women in work. So 1 woman = 46 = 23 man. Now 6 men + 3 women = 6 + 2 = 8 men, and the time is 4 × 158 = 7.5 days.

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Form 2: two groups, same work. "2 men and 5 women take 6 days; 1 man and 4 women take 9 days." The work is the same, so 6(2M + 5W) = 9(M + 4W). This gives the ratio M : W. Put it back to find the total work in "woman-days" or "man-days".

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Form 3: each kind given separately. "6 men do it in 18 days, 8 women in 12 days, 5 boys in 20 days." Then one man needs 108 days, one woman 96 days and one boy 100 days. Use the LCM method on these individual times, multiply each efficiency by the number of workers, and add.

Radhika's case shows Form 1 once more: 3 men or 5 women can paint a hall in 10 days, and Radhika hires 6 men and 5 women. 5 women = 3 men, so the team is 9 men, and the time is 3 × 109 = 313 days.

The same thinking gives an efficiency ratio: "20 women do it in 16 days and 10 men in 12 days" ⇒ 320 woman-days = 120 man-days ⇒ one man does as much as 320120 = 83 women.

The last type is about money, and it has one firm rule: wages are shared in the ratio of work done. Not in the ratio of days, not in the ratio of times, but of work.

When everybody works for the same number of days, work done ∝ efficiency, so wages are in the ratio of efficiencies. If A takes 20 days and B 30 days, their efficiencies are 3 : 2, and so are their wages.

Radhika (14 days) and Meera (21 days) take a contract for ₹5,000 and work together till the end. Efficiencies = 3 : 2 (total 42 units), so Radhika gets 35 × 5000 = ₹3,000 and Meera ₹2,000. The faster worker gets more money, even though she needs fewer days alone.

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When people work for different numbers of days, work done = efficiency × days. If A works 5 days at 4 units/day and B works 6 days at 3 units/day, the work done is 20 : 18, and the wages follow 20 : 18 = 10 : 9.

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A very common question: "A and B take a job for some amount; with C's help they finish it in a few days. How much does C get?" Find what fraction of the job A and B did in those days. Whatever is left was C's work, and C's money is that fraction of the total. You never need C's own time.

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Daily wage questions ("₹650 is paid every day to A, B and C together") work the same way: share the daily amount in the ratio of their efficiencies. And when wages are given for groups of men and women, set up two equations in "one man's daily wage" and "one woman's daily wage", just as in Type 13.

Overtime and different daily wages. So far the money was a fixed contract, shared in the ratio of work done. Often the money is given as a rate instead: so much per day or per hour, and sometimes more for overtime. Then each person's pay is simply rate × time worked. The work only decides how many days each person needs.

Radhika is paid ₹60 an hour for the first 8 hours of a day and 1.5 times that, ₹90, for every extra hour. For an 11-hour day she gets 8 × 60 + 3 × 90 = ₹750.

The cost of a job done alone = daily wage × days taken alone. A is paid ₹600 a day and takes 12 days alone, so the job costs ₹7,200 with him. B is paid ₹800 a day and takes 8 days, so it costs ₹6,400 with B. B earns more per day, yet the job is cheaper with him, because he finishes it sooner. When the two work together they take 12 × 812 + 8 = 4.8 days, and the total paid is 4.8 × (600 + 800) = ₹6,720.