Aptitude · Time and Work

Stop adding days. Add rates.

Q: Give the two-person shortcut.

Together time = ab/(a+b) , where a and b are the individual times. For 10 and 15 it is 150/25 = 6 days. It is fast, but it only works for exactly two people doing the whole job together throughout — it fails for three workers or for anyone joining or leaving.

Q: If a person finishes a job in N days, what is their one-day work?

One N-th of the job, 1/N . And the reverse holds: if someone does 1/N of a job per day, they finish in N days. Days and one-day work are reciprocals, which is the same as saying they are inversely related.

A takes 10 days and B takes 15. Together they do not take 25 days, or 12.5. The mistake is adding the wrong quantity — and the fix is one choice made at the start of every question in this chapter.

Set each person’s days and watch the LCM remove the fractions →

Times never add. Rates add. Take the total work as the LCM of the given times and every rate becomes a whole number.

01 The idea

Work, rate, time — and choosing the size of the job

A can paint a wall in 10 days and B can paint the same wall in 15 days. Working together, how long? Adding the days gives 25, which is absurd — help cannot make a job slower. Averaging gives 12.5, which is also wrong, because that is slower than A working alone.

The quantity that adds is the rate. A does a tenth of the wall a day, B does a fifteenth, so together they do a tenth plus a fifteenth. That is the whole method, and every question in this chapter is a variation on it.

Working in fractions like 1/10 and 1/15 is correct and slow. So use the trick that defines this chapter: the answer cannot depend on how big the job is, so choose a convenient size. Take the total work as the LCM of the times — here LCM(10, 15) = 30 units.

Now A does 30/10 = 3 units a day and B does 30/15 = 2 units a day. Together 5 units a day, and 30 units at 5 a day is 6 days. No fractions appeared anywhere. That single choice is worth more in this chapter than any formula.

Choose the total work to be the LCM of the given times. Every rate becomes a whole number, and the fractions vanish from the entire question.

WorkThe size of the job, written W. Arbitrary, so you may choose it — and choosing the LCM of the times is what makes everything else easy.

EfficiencyHow much someone does in one day, written E. Also called the one-day work. If a person finishes in N days, their efficiency is W/N, or 1/N of the job.

The core relationW = E × T, so T = W/E. Time is work divided by rate — the only formula the chapter really has.

02 Worked example

A in 10 days, B in 15 — together in 6

This pair runs the whole module. A can do a piece of work in 10 days and B can do the same work in 15 days. How long do they take working together?

Choose the size of the jobThe answer cannot depend on how big the wall is, so pick a size that divides by both times. The LCM of 10 and 15 is 30.total work = LCM(10, 15) = 30 units

Find A’s rateA clears the whole 30 units in 10 days, so three units a day.A = 30 / 10 = 3 units/day

Find B’s rateB clears 30 units in 15 days, so two a day. Note both rates came out whole — that is what the LCM bought.B = 30 / 15 = 2 units/day

Add the ratesThey work at the same time, so their outputs accumulate together.A + B = 3 + 2 = 5 units/day

Divide the job by the combined rateThirty units at five a day.30 / 5 = 6 days

Check the answer against two bounds before moving on. Six days is less than A’s 10, as it must be — adding a helper cannot slow the job. And it is more than 10/2 = 5, because B is slower than A so two of them are worth less than two A’s. Any together-time outside that window is an arithmetic error, and these two checks catch almost every one.

03 The method

The formula, and the shortcut for exactly two people

The LCM route works for any number of workers and any complication. The product formula below is faster but only covers the simplest case.

W = E × T, so a person finishing in N days has efficiency 1/N. Working together, E_total = E₁ + E₂ + … and T = W / E_total.

For exactly two people, together time = ab/(a + b). Here 10 × 15 / 25 = 6 days. It is quick and it is a trap to rely on: it does not extend to three people, and it breaks the moment anyone joins late, leaves early or works alternate days. Learn the LCM method as your default and keep this for two-person questions only.

Days to finish alone	One-day work	Units/day if total = 30
10	`1/10`	`3`
15	`1/15`	`2`
6	`1/6`	`5`
30	`1/30`	`1`
N	`1/N`	`30/N`
Adding the times	`never valid`	10 + 15 = 25 is not an answer
Averaging the times	`never valid`	12.5 is slower than A alone

05 Cheat sheet

The core on one page

Four rows of method and three of the errors that this chapter punishes hardest.

Case	Route	On A = 10, B = 15
Total work	`LCM of the times`	LCM(10,15) = 30 units
Individual rate	`W / days`	A = 3, B = 2 units/day
Working together	`add the rates`	5 units/day
Time from rate	`W / rate`	30/5 = 6 days
Two people, shortcut	`ab/(a+b)`	150/25 = 6 days
Adding the times	`wrong`	25 days is not an answer
Averaging the times	`wrong`	12.5 is slower than A alone

The answer beats the fastest workerTogether time is always less than the quickest person’s solo time. If it is not, you added times instead of rates.

But not by more than the headcountWith two workers the together time cannot be below half the faster one’s time. Those two bounds together catch nearly every slip.

The job size is yours to chooseNothing in the answer depends on it, so pick the LCM. Fractions in a time-and-work solution usually mean this choice was not made.

06 Where & why

Where this shows up

Time and work is one of the highest-yield chapters in the syllabus, and everything in it rests on this one lesson.

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Two or three workers together

The direct question, set as a speed item. With the LCM method it is fifteen seconds and no fractions.

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Everything harder in this module

Alternate days, people leaving, wages and efficiency all reduce to rates in units. The LCM choice is what keeps them arithmetic rather than algebra.

Pipes and cisterns

The same chapter with a negative rate

An emptying pipe is a worker with a negative efficiency. If this lesson is solid, that whole module is already half learned.

Interviews

“Why don’t the times add?”

A clean answer — because rate is work per unit time and it is rates that combine — reads as understanding rather than recall.

If you take one habit from this module, make it writing “total work = LCM” as the first line of every time-and-work question. It converts the entire chapter from fraction arithmetic into whole-number counting.

07 Interview questions

What gets asked

Ten, starting from the misconception and ending at the bounds worth checking every answer against.

A takes 10 days and B takes 15. Why isn’t the answer 25 days together?

Because times do not add — rates do. Adding times would mean help makes a job slower, which is nonsense. A does a tenth of the job a day and B a fifteenth, so together they do a tenth plus a fifteenth, which is a sixth. Six days.

What is the LCM method and why use it?

Take the total work as the LCM of the given times. For 10 and 15 that is 30 units, making A three units a day and B two. Every rate becomes a whole number, so no fractions appear anywhere. The answer is unchanged because the size of the job is arbitrary.

Are you allowed to just choose the size of the job?

Yes, and it is not a trick. The question asks for a time, and that time is the same whether the wall is one wall or thirty units of wall — the job size cancels out of the calculation. Choosing the LCM is picking a convenient unit, not making an assumption.

Give the two-person shortcut.

Together time = ab/(a+b), where a and b are the individual times. For 10 and 15 it is 150/25 = 6 days. It is fast, but it only works for exactly two people doing the whole job together throughout — it fails for three workers or for anyone joining or leaving.

A in 10, B in 15, C in 30. How long together?

Five days. On 30 units A does 3 a day, B does 2 and C does 1, so together 6 units a day and 30/6 = 5 days. Note the product shortcut is no help here, which is why the LCM method should be your default.

How do you check a together-time answer quickly?

Two bounds. It must be less than the fastest person’s solo time, because help cannot slow the work. And with n workers it cannot be less than the fastest time divided by n. For A = 10 and B = 15 the answer must lie between 5 and 10 days, and 6 does.

If a person finishes a job in N days, what is their one-day work?

One N-th of the job, 1/N. And the reverse holds: if someone does 1/N of a job per day, they finish in N days. Days and one-day work are reciprocals, which is the same as saying they are inversely related.

A and B together finish in 6 days and A alone in 10. Find B alone.

Fifteen days. On 30 units the pair does 5 a day and A does 3, so B does 2, giving 30/2 = 15 days. Subtracting rates is how every “find the other person” question works — and you subtract rates, never times.

What happens to the time if the number of workers doubles?

Hard

(a) Starting from rates, derive the two-person formula ab/(a+b) for the together time. (b) Write down and simplify the equivalent formula for three people with times a, b and c. (c) A student uses ab/(a+b) on a question where A works for the whole job but B joins only halfway through, and gets a wrong answer. Explain precisely which assumption in your derivation fails.

Follow-up

Part (c) is the reason not to lean on the shortcut. The derivation assumes both workers act for the entire duration, so the combined rate is constant — and that assumption is exactly what a late joiner breaks. Every harder model in this module breaks it in some way.

Show the hint

For (a), add 1/a and 1/b, then take the reciprocal. For (c), look at which step in that derivation requires both rates to apply for the same length of time.