Time and Work Tricks: The LCM Method for SSC, RRB and Bank Exams

Many students get through school with time and work by writing "A's one day work = 1/12, B's one day work = 1/15, together = 1/12 + 1/15" and then adding fractions with a pen. It works. It is also slow, and it is where careless errors creep in. The exam does not need you to do it that way.

There is a cleaner method that sits behind almost every shortcut you will find in coaching notes: pick the total work as the LCM of the given days, and turn every person's speed into a whole number of "units per day". After that, time and work is integer arithmetic. This post builds the chapter around that one idea and then adds the handful of special cases (leaving and joining, wages, pipes, alternate days, men-days) that the exam uses over and over. Every worked answer here has been calculated and checked.

Time and work sits in the arithmetic block alongside percentage and profit and loss. If you found the fraction thinking from the percentage tricks post useful, you will find this chapter is the same habit with different clothing. Check your exam's latest notification for how many arithmetic questions to expect; the chapter is small, but the same idea also powers pipes, cisterns, and even some speed-distance questions.

The core idea: work is a number, not a fraction

Suppose A finishes a job in 12 days and B in 15 days. Instead of calling the job "1", call it the LCM of 12 and 15, which is 60 units. Now A does 60/12 = 5 units a day, and B does 60/15 = 4 units a day. Together they do 9 units a day, so the job takes 60/9 = 6 and 2/3 days.

That is all the method is. Choose total work = LCM of the days given. Divide to get each person's daily output. Add or subtract daily outputs. Divide total work by the combined rate.

The reason it works is that you can pick any number for the total work, and the LCM is just the one that keeps every daily rate a whole number. If you had picked 1 you would get 1/12 and 1/15. If you had picked 120 you would get 10 and 8, and the answer would still be 120/18 = 6 and 2/3. The LCM is the smallest number that avoids fractions.

Three people: A alone 10 days, B alone 15 days, C alone 30 days. LCM = 30. A does 3, B does 2, C does 1 units a day. Together 6 a day. So 30/6 = 5 days. You never wrote a fraction.

What "efficiency" means

The daily rate you just computed is called efficiency in most coaching material. A person with efficiency 5 does 5 units a day. So person A (5 units) is more efficient than B (4 units), and their efficiency ratio is 5 : 4. Time taken is inversely proportional to efficiency, so the ratio of days taken is 4 : 5, and check: 12 : 15 = 4 : 5. Correct.

This gives a second way to enter the same problem. If a question says "A is twice as efficient as B", assign A = 2, B = 1. If it says "A is 50% more efficient than B", assign A = 3, B = 2. If it says "A can do in 3/4 the time B takes", the efficiency ratio is the opposite, so A : B = 4 : 3.

Try this. "A is twice as efficient as B and together they finish the work in 18 days. How long would A alone take?" Efficiencies are 2 and 1, together 3 a day. Total work = 3 x 18 = 54 units. A alone: 54/2 = 27 days. B alone would be 54 days. Check: 1/27 + 1/54 = 3/54 = 1/18. Correct.

And one with a percentage: "A is 50% more efficient than B. A alone takes 12 days. How long do they take together?" A : B = 3 : 2. A does the job in 12 days, so total = 36 units and A does 3 a day (36/12 = 3). B does 2 a day, so B alone takes 18 days. Together 5 a day, and 36/5 = 7.2 days. Check: 1/12 + 1/18 = 5/36, so 36/5 = 7.2. Correct.

Finding one person's time from two pieces of information

The other classic: "A and B together finish a job in 6 days. A alone takes 10 days. How long does B alone take?" LCM of 6 and 10 is 30. Together: 5 units a day. A: 3 units a day. So B = 5 - 3 = 2 a day, and B alone takes 30/2 = 15 days. Check: 1/10 + 1/15 = 5/30 = 1/6. Correct.

This subtraction pattern is the most common way a time and work question hides. Remember: the total work is the LCM of whatever days are given, the rates are total/days, and any unknown rate is found by subtracting known rates from a combined rate.

Three pairs, one triple

"A and B together can do a job in 12 days, B and C in 15 days, A and C in 20 days. In how many days can they finish it working all three together, and how long does each take alone?"

LCM of 12, 15, 20 is 60. So A + B = 5, B + C = 4, A + C = 3 units per day. Adding the three: 2(A + B + C) = 12, so A + B + C = 6. Then A = 6 - 4 = 2, B = 6 - 3 = 3, C = 6 - 5 = 1. All three together take 60/6 = 10 days. Alone: A takes 60/2 = 30 days, B takes 60/3 = 20 days, C takes 60/1 = 60 days.

Check with pairs: A and B alone rates 2 + 3 = 5, and 60/5 = 12. B and C: 3 + 1 = 4, 60/4 = 15. A and C: 2 + 1 = 3, 60/3 = 20. All match.

The pattern "sum the pairs, halve, subtract each pair" solves every three-pair question in about twenty seconds. If you do it with fractions, it takes a page.

Leaving and joining

This is where the LCM method really pays. The problem changes who is working in the middle, but each stage is just units per day times days.

"A can do a job in 20 days and B in 30 days. They start together, but A leaves after 6 days. In how many days is the job completed?" LCM = 60. A = 3, B = 2, together 5 a day. In 6 days they complete 30 units. Remaining 30 units is done by B alone at 2 a day, which takes 15 days. Total = 6 + 15 = 21 days.

Now a variation: "A alone starts and works for 5 days, then B takes over and finishes the remaining in 12 days. A alone takes 20 days to do the whole job. How long would B alone take?" A does 5/20 = 1/4 of the job. B does 3/4 in 12 days, so the whole job takes B 12 / (3/4) = 16 days. Using units: total 20 units (assuming A = 1 unit/day), A completed 5, B completed 15 in 12 days, so B does 1.25 a day, and the total 20 takes B 16 days. Both routes agree.

And the "leaving at the end" version: "A takes 10 days and B takes 15 days. They work together, but A leaves 2 days before the work is finished. Find the total time." Let the total time be x days. B works all x days. A works x - 2 days. LCM = 30, so A = 3, B = 2. Total work: 3(x - 2) + 2x = 30. That gives 5x = 36, so x = 7.2 days. Check: A works 5.2 days (5.2 x 3 = 15.6), B works 7.2 days (7.2 x 2 = 14.4), and 15.6 + 14.4 = 30. Correct.

The trick with "leaves x days before completion" is to count the person who leaves as having worked (total - x), never as if they left after x days. Mixing up those two readings is a classic mistake.

Wages: pay in proportion to work done

If two people work on a job for the same number of days, wages are divided in the ratio of their efficiencies. If they worked different numbers of days, the ratio is units of work done, so (efficiency x days).

"A can do a job in 12 days and B in 18 days. They complete it together and are paid 1,500 in total. What is each person's share?" Efficiencies: LCM = 36, A = 3, B = 2. Ratio 3 : 2. A's share = 3/5 of 1,500 = 900, and B's = 600. Check: 900 + 600 = 1,500, and 900 : 600 = 3 : 2.

Note that the ratio of wages is the inverse of the ratio of days: 12 : 18 = 2 : 3 in days, so wages go 3 : 2. If you accidentally split the money 2 : 3, the person who works faster gets less, which is obviously wrong. Whenever you finish a wages question, ask whether the more efficient person got more.

If the work times differ, multiply first. Say A works 4 days at 3 units a day (12 units) and B works 6 days at 2 units a day (12 units). Their shares are equal, even though B worked longer. Wages follow the amount of work, which is efficiency times days.

Men-days, hours and the chain rule

The chapters are often split as "time and work" and "chain rule", but they are the same. Work is measured in man-days (or man-hours), and the formula is:

Total work = men x days x hours per day (x efficiency, if different).

"12 men can build a wall in 18 days. After 6 days, 6 men leave. How many more days will it take?" Total work = 12 x 18 = 216 man-days. Work done in 6 days = 12 x 6 = 72. Remaining = 144 man-days. With 6 men, it takes 144/6 = 24 more days. So the full job took 6 + 24 = 30 days.

Adding the hours. "15 men working 8 hours a day can do a piece of work in 20 days. How many days will 16 men working 5 hours a day take?" Total work = 15 x 8 x 20 = 2,400 man-hours. Now 16 x 5 = 80 man-hours a day. 2,400/80 = 30 days.

The habit: compute the total work once, in the largest unit that you have all the data for, then divide by the new rate. Do not set up proportions with arrows; they are error-prone under pressure.

Men and women

"5 men can do a job in 24 days, or 8 women can do it in 24 days. How long will 10 men and 8 women take?" Since 5 men and 8 women take the same time, 5 men = 8 women in efficiency. That means 1 man = 8/5 women = 1.6 women. Then 10 men = 16 women, and adding the 8 women gives 24 women. Now 8 women take 24 days, meaning the total work is 8 x 24 = 192 woman-days, and 24 women take 192/24 = 8 days. Check in man units: 10 men + 8 women = 10 + 5 = 15 man-equivalents (since 8 women = 5 men), and 5 men take 24 days so the total work is 120 man-days, and 120/15 = 8 days. Both routes give 8.

Alternate days

Alternating problems trip students because the pattern needs to be worked out cycle by cycle. The method: compute the work done in one full cycle (two days), find how many full cycles fit, and then handle the remainder.

"A can do a job in 10 days and B in 15 days. They work on alternate days, starting with A. In how many days will the job be finished?" LCM = 30, so A = 3, B = 2. One two-day cycle does 5 units. 30/5 = 6 cycles, so 12 days. (If B starts, it is still 12, because the work divides exactly into cycles.)

Now a case where it does not divide exactly. "A takes 12 days and B takes 16 days. They work on alternate days, A starting." LCM = 48. A = 4, B = 3. One cycle does 7 units. Six cycles = 12 days = 42 units. Remaining 6. On day 13, A works and completes 4, leaving 2. On day 14, B works and needs 2/3 of a day. So the job takes 13 and 2/3 days. Check: 48 - 42 = 6, 6 - 4 = 2, 2/3 = 0.667, so 13.67 days.

The trap: the answer is not "14 days". The work finishes partway through the fourteenth day. Read the options; if 13 2/3 is there, that is the answer. If the question says "on which day is the work finished" the answer is the 14th.

Pipes and cisterns

This chapter is time and work in a different costume. A pipe filling a tank is a worker; a pipe emptying it (an outlet or a leak) is a worker with negative efficiency. Everything above applies: total capacity = LCM, inlet rate positive, outlet rate negative.

"An inlet pipe fills a tank in 6 hours. An outlet empties it in 10 hours. If both are open, how long to fill the tank?" LCM = 30. Inlet = +5, outlet = -3, net = +2 a hour. 30/2 = 15 hours.

Three pipes: "Pipes A and B fill a tank in 12 and 15 hours, and pipe C empties it in 20 hours. All three are open. How long to fill?" LCM = 60. A = 5, B = 4, C = -3. Net = 6 a hour. 60/6 = 10 hours.

The leak version is very common: "A pipe fills a tank in 8 hours, but because of a leak it takes 10 hours. How long does the leak take to empty a full tank?" The leak's rate is the difference between the pipe's rate and the net rate. 1/8 - 1/10 = 1/40, so the leak empties the tank in 40 hours. In units: LCM of 8 and 10 = 40. Pipe = 5, net = 4, leak = 1. 40/1 = 40 hours. Check: 40/5 = 8 hours for the pipe alone, and 40/4 = 10 hours with the leak.

If a pipe is closed halfway, treat it like a person leaving. "Two pipes fill a tank in 20 and 30 hours. Both are opened together, but the first is closed after 6 hours" is identical to the A and B leaving problem earlier: 60 units, rates 3 and 2, 30 units in 6 hours, and the remaining 30 units at rate 2 takes 15 more hours, total 21 hours.

A quick sanity test for every answer

After every time and work question, check three things. The combined time should always be less than the smaller of the individual times. If A takes 12 days and B takes 15, together must be under 12. Second, when someone leaves, the time must go up relative to the case where they stay. Third, in wages questions, the faster worker should earn more. If any of those fail, look for a swapped ratio.

A short list of shortcut formulas (and where they come from)

You can derive all of these from the LCM method, but it helps to recognise them.

• Two people, days a and b: together = ab/(a + b). For 12 and 15: 180/27 = 6 and 2/3. Correct.
• If A + B take x days, and A alone takes a, then B alone takes ax/(a - x). For A = 10 and x = 6: 60/4 = 15. Matches the earlier example.
• Three people, days a, b, c: together = abc/(ab + bc + ca). For 10, 15, 30: 4,500/(150 + 450 + 300) = 4,500/900 = 5. Correct.
• If A is n times as efficient as B, A takes 1/n of B's time. If A + B together take t days, then A alone takes t(n + 1)/n and B alone takes t(n + 1). For n = 2 and t = 18: A = 18 x 3/2 = 27, B = 18 x 3 = 54. Matches the example above.

Use these when the numbers are ugly enough that the LCM is large. For 7, 11 and 13 days the LCM is 1,001, and the formula gives no advantage either, but that is unusual in a real paper. Exam setters choose days that give small LCMs (6, 10, 12, 15, 20, 24, 30, 36, 40, 45, 60).

Common problem families and how to recognise them

You will see roughly six families. Recognising which family the question belongs to is half the work.

Direct combined work. "A and B can do a job in x and y days. How long together?" LCM, add rates, divide.

Reverse: find one from combined. Given the combined time and one individual time, subtract rates.

Leaving or joining. Split into stages. Each stage: rate times duration. Keep a running total against the LCM.

Wages. Ratio of work done, not ratio of days.

Men, women, children. Convert everyone to one unit (a man-equivalent) and use man-days.

Pipes and cisterns. Outlets get negative rates.

Alternate days. Compute per cycle and handle the remainder.

If you can name the family in five seconds, you will not need a formula. If you cannot name it, reread. Nine times out of ten the family has a name.

Mistakes that keep repeating

The biggest one is adding times instead of rates. A takes 12 days and B takes 15 days, and students write "together 27 days" or "13.5 days" (the average). Neither is anything close. Together must be less than either alone.

Second is picking the wrong LCM. If the days are 12, 15 and 20, the LCM is 60, not 3,600 and not 180. A smaller LCM means smaller numbers. A quick way to find it: list multiples of the biggest number until the others divide in.

Third is mishandling the "leaves before completion" problems, described earlier. Draw a timeline: a line of x days, a bracket for who worked how long. It takes 5 seconds.

Fourth is forgetting efficiency changes, such as "A is 25% more efficient after the first 3 days". Split into two stages and compute each stage's rate.

Fifth is ignoring hours. If the question says "8 hours a day" and then "10 hours a day", you need man-hours, not man-days. Read the units in every sentence.

Sixth is doing every question with fractions even after you have learned the LCM method. For the first week the LCM method feels slower because you have to think about the setup. By the second week it is faster than fractions. Stay with it through the awkward stage.

How to build speed in this chapter

You will not need hundreds of questions for time and work; you need about 60 well-chosen ones and a careful review. A plan that works.

Days one to three: 10 questions a day using only the LCM method. Write the total work at the top of every question, and the rates below it. Do not use any other approach. This is to make the setup automatic.

Days four and five: leaving and joining, wages, and alternate days. Do 8 to 10 of each. In each, write the timeline.

Day six: pipes and cisterns. It should feel easy because it is the same thing. Do 15 questions in 15 minutes.

Day seven: a timed set of 20 mixed questions in 20 minutes. Mark anything that took more than a minute.

After that, revisit once a fortnight with a set of 10 questions. Time and work does not decay as fast as some chapters, but a fortnightly refresh keeps the setup automatic.

When you feel ready, test yourself inside a full paper. Pareeksha's mock tests give you the section-wise breakdown so you can see whether arithmetic is taking more than its share of time, and the free sectional practice is a good place to drill a set of 20 arithmetic questions under a timer. If time and work is where your paper is losing seconds, the section-wise time management post will help you decide how long to give it.

Where this chapter sits with the rest of quant

Time and work links closely to profit and loss, because both use the "assume a convenient number" idea, and to percentage, because efficiency comparisons are percentage comparisons (A is 50% more efficient than B is a percentage statement). It is also the neighbour of time, speed and distance, where the same LCM trick applies to two travellers or trains.

For RRB exams, the questions tend to be direct and shorter. For SSC, expect a little more layering, like a leave-and-join with wages. For banking, time and work often shows up inside a longer word problem or a data interpretation caption, so knowing the efficiency method helps you read fast. Check the latest notification for what the exam you are targeting includes; the syllabus can be tweaked between cycles.

If you are preparing for more than one exam at the same time, you can treat the arithmetic block as common. That is covered in the post on preparing two exams together.

Your next step

Take a sheet of paper and solve these five without fractions, only using the LCM method: A and B in 12 and 15 days; A, B, C in 10, 15, 30 days; A and B in 20 and 30 days with A leaving after 6; a tank with an inlet of 6 hours and an outlet of 10 hours; and alternate days for 10 and 15. You should arrive at 6 and 2/3 days, 5 days, 21 days, 15 hours and 12 days. If any answer differs, redo it and see where the setup went wrong.

FAQ

Do I have to use the LCM method, or can I use fractions? Either gets the right answer. The LCM method is faster once you are used to it and produces fewer arithmetic slips, especially with three people or leaving and joining.

What if the LCM is a big number? Take the LCM of only the numbers you are given. If it is large, you can also choose any common multiple that is convenient, like 100, and accept a decimal rate. Or use the formula ab/(a + b).

How do I handle "twice as fast" or "60% more efficient"? Convert to a ratio of efficiencies (2 : 1, or 8 : 5) and assign those as units per day. Time taken has the inverse ratio.

How are wages divided if people join on different days? By total work done by each, which is efficiency times days worked. Not by days alone and not by efficiency alone.

Are pipes and cisterns really the same chapter? Yes. Inlets are workers and outlets are workers with negative efficiency. The method is unchanged.

Should I attempt a long alternate-days question in the exam? If the numbers divide neatly into cycles it takes under a minute. If a remainder is involved and you are behind on time, mark it and come back.