Time and Work — AAI ATC GK Lesson

Before you start

Learning objectives and outcomes

Objective 1Apply the one-day-work method to combine any number of workers.

Objective 2Use the men-days-work constant to scale problems across different worker counts.

Objective 3Solve pipes and cisterns problems by treating emptying pipes as negative work.

Outcome ARecognise any time-and-work variant as the same underlying method in disguise.

Subtopic 1

The one-day-work approach

Nearly every time-and-work problem is solved the same way: convert each worker's total time into a fraction of the work they complete in one day, then combine those fractions.

The core method

If A finishes a work in n days, A's one-day work = 1/n
Combined one-day work of A and B = 1/a + 1/b · Time together = 1/(combined one-day work)

This single technique — convert to one-day fractions, add, then invert — solves the overwhelming majority of time and work questions on this exam.

Memory trick — "Fractions add, days don't"

A common mistake is trying to average or add the number of days directly. Days never add directly — only the one-day work fractions do.

Easy

A can finish a piece of work in 10 days. In one day, A completes:

If A finishes the work in 10 days, A's one-day work is 1/10.

Moderate

A can do a piece of work in 10 days and B in 15 days. Working together, they will finish it in:

Combined one-day work = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6, so together they take 6 days.

Subtopic 2

Efficiency and men-days problems

A second family of problems scales work across a different number of workers, using the principle that total work stays constant.

The men-days-work constant

Men × Days = Constant (for a fixed amount of work)
M₁ × D₁ = M₂ × D₂

This is the same idea as inverse proportion: more workers means fewer days needed, in exactly inverse ratio, provided every worker has the same efficiency.

Figure 2.1 — Men and days balance like a seesaw A simple balance/seesaw diagram with "Men" on one side and "Days" on the other, showing that as one side increases, the other decreases to keep the product (total work) constant. Upload image here

✈ AAI ATC relevance — men-days scaling questions, especially where efficiency differs (like men vs women vs boys), are heavily tested and easy to solve once the constant-product idea clicks

Memory trick — "More hands, fewer days, same total work"

Whenever the number of workers changes, the total amount of work stays fixed — only how it's divided among the workers changes, which is exactly what the men×days=constant relationship captures.

Moderate

15 men can build a wall in 8 days. In how many days will 10 men build the same wall?

Using M₁D₁=M₂D₂: 15×8 = 10×D₂, so D₂ = 120/10 = 12 days.

Subtopic 3 · highest yield

Pipes and cisterns

Pipes and cisterns problems use exactly the same one-day-work method as ordinary time and work problems — the only twist is that an emptying (outlet) pipe contributes negative work.

Filling and emptying together

Filling pipe's one-hour work = +1/n · Emptying pipe's one-hour work = −1/n
Net one-hour work = sum of all filling rates minus sum of all emptying rates

If the net one-hour work comes out negative, the tank is actually being drained overall, even with a filling pipe open — this is a frequent trap in exam questions.

✈ AAI ATC relevance — pipes and cisterns questions are just time and work questions wearing a different costume — the same one-day-work method applies directly

Memory trick — "An outlet pipe is just negative work"

Treat an emptying pipe exactly like a worker doing negative work — add its rate as a negative fraction, and the whole problem collapses back into an ordinary time-and-work combination.

Hard

A tap fills a tank in 6 hours and another tap empties it in 8 hours. If both taps are opened together, the tank will be filled in:

Net one-hour work = 1/6 − 1/8 = 4/24 − 3/24 = 1/24, so the tank fills in 24 hours.

Exam strategy

Facts worth having automatic

One-day workIf A finishes in n days, A's one-day work is 1/n.

Combining workersAdd one-day-work fractions, then invert for the combined time.

Men×Days=constantFor a fixed job, more workers always means proportionally fewer days.

Emptying pipesTreat as negative work — subtract, don't add.

Never add days directlyOnly one-day-work fractions add; days never do.

Quick revision

Chapter summary sheet

One-day work1/n, if the job takes n days alone.

Combined time1 ÷ (sum of one-day-work fractions).

Men-days constantM₁D₁ = M₂D₂, for the same job.

Filling pipePositive one-hour work, +1/n.

Emptying pipeNegative one-hour work, −1/n.

Net rate negativeTank drains overall, even with a filling pipe open.

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End of chapter

Comprehensive assessment

Full solutions are in the downloadable PDF above.

Multiple choice questions

1. If A finishes a job in 10 days, what fraction of the work does A do in one day?

2. A does a work in 10 days, B in 15 days. Find their combined time working together.

3. 15 men build a wall in 8 days. Find the time for 10 men to build the same wall.

4. State the men-days-work relationship for a fixed job.

5. A tap fills a tank in 6 hours, another empties it in 8 hours. Find the time to fill with both open.

6. Explain why an emptying pipe is treated as negative work in these problems.

7. A is twice as efficient as B. Together they finish a job in 12 days. Find how long B alone would take.

8. 6 men and 8 boys finish work in 10 days; 26 men and 48 boys finish it in 2 days. Find the ratio of a man's to a boy's work rate.

9. A works alone for 5 days on a 20-day job, then leaves. Find how much work remains.

10. 15 men or 24 women complete a job in 20 days. Find how many days 8 men and 10 women would take.

Time and Work — one more topic locked in.

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