Time and Work

Time and work questions ask how long a job takes when one or more people (or machines, or pipes) work on it. They look different on the surface, but they all rest on one idea: work is done at a steady rate. If you convert every statement into a rate, the problem becomes simple arithmetic.

The core idea

If someone can finish a whole job in n days, then in one day they finish 1/n of the job. That fraction is their rate. Work done = rate × time, and the whole job counts as 1.

So the time taken for a job at rate r is 1 / r.

Working together

When people work together, their rates add. If A takes a days and B takes b days, together they do 1/a + 1/b of the work per day, so the time together is

1 / (1/a + 1/b) = (a × b) / (a + b).

Example: A alone takes 12 days and B alone takes 36 days. Together: (12 × 36) / 48 = 9 days.

To find one person's time when the combined time and the other person's time are known, subtract rates. Example: A and B together finish in 15 days, and A alone takes 25 days. B's rate = 1/15 − 1/25 = 2/75, so B alone takes 75/2 = 37.5 days.

The LCM method

Fractions get untidy, so many people take the total work as the LCM of the given days. If A takes 10 days and B takes 15 days, let the total work be 30 units. Then A does 3 units a day and B does 2 units a day, together 5 a day, and the job takes 30 / 5 = 6 days. The units are imaginary, but they keep every number whole.

Alternate days and partial work

When people work in turns, find the work done in one cycle. A takes 10 days and B takes 15 days, and they work on alternate days starting with A. With 30 units of work, A does 3 and B does 2, so a two-day cycle completes 5 units. Six cycles finish 30 units, so the total is 12 days.

When a person joins later, work out how much has been done before they join, then use the combined rate for the rest.

Man-days

When the number of workers or hours changes, use man-days (or man-hours). The total work stays the same, so

M1 × D1 × H1 = M2 × D2 × H2

assuming all workers have the same efficiency. Example: 8 men finish a job in 15 days. How long would 12 men take? 8 × 15 = 12 × D, so D = 10 days. More workers mean fewer days.

If one person is, say, twice as efficient as another, their rates are in the ratio 2 : 1, and the time taken is in the ratio 1 : 2, so the faster one takes half the time.

Pipes and cisterns

This is the same idea with pipes. An inlet pipe fills a tank, so its rate is positive. An outlet pipe or a leak empties it, so its rate is negative. A pipe that fills a tank in 6 hours and a leak that empties it in 12 hours together fill at 1/6 − 1/12 = 1/12 per hour, so the tank fills in 12 hours.

Worked examples

Example 1. A can do a job in 14 days and B in 21 days. Working together, they take (14 × 21) / 35 = 8.4 days.

Example 2. A does a job in 24 days. He works for 8 days, then B finishes the rest in 10 days. A has done 8/24 = 1/3, leaving 2/3. So B does 2/3 in 10 days, and B alone would take 15 days.

Example 3. 10 men can do a job in 12 days working 6 hours a day. How many days do 15 men need if they work 8 hours a day? Man-hours = 10 × 12 × 6 = 720. Per day: 15 × 8 = 120. Days = 720 / 120 = 6.

Common mistakes

• Adding the days instead of the rates. Two people together are always faster than either alone.
• Forgetting that a leak or outlet has a negative rate.
• Mixing up days and hours when the working hours change.
• Leaving the total as 1 and carrying messy fractions when the LCM method would give whole numbers.

Exam tips

1. Write each person's rate on the first line of the solution.
2. Sense-check: the combined time must be smaller than the smaller of the two individual times.
3. For efficiency questions, work in the ratio of rates rather than in days.

Now test yourself with the quiz below, and read each explanation to compare your method with the rate method.