Time & Work

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Work Rates & the LCM Method

⏱ 5 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Treat one whole job as the unit of work. A worker who finishes it in T days has a rate of $\dfrac{1}{T}$ per day, and rates add when people work together.

The LCM method keeps everything in whole numbers: set the job to the LCM of the given days, so each rate becomes small units per day.

Two workers who take $a$ and $b$ days alone finish together in $\dfrac{ab}{a+b}$ days.

What rate means

A job is one whole thing — a wall, a tank, a batch of shirts. A worker's rate is the share of that job he finishes in one day.

If A finishes the whole job in T days, his rate is $\dfrac{1}{T}$ per day. Two facts run the whole topic:

\text{work} = \text{rate} \times \text{days}, \qquad \text{rates add when people work together}

Each worker keeps his own rate inside a team. So the combined time is always less than the fastest single time.

Rule: If your combined answer came out larger than the fastest time, you added days instead of rates.

The LCM method

Fractions invite slips. Give the job a size in units instead — the LCM of the given days.

A finishes in 15 days, B in 10. Job = 30 units. A does $30 \div 15 = 2$ units a day, B does 3. Together 5 units a day → $30 \div 5 = 6$ days. Every number stays a small whole number to the end.

Tip: Convert every question here to units first. Rates, sums and remainders then stay integers.

Two workers

For exactly two workers with times a and b:

T = \frac{ab}{a+b}

6 and 12 days → $\dfrac{72}{18} = 4$ days. This is the LCM method compressed into one line — use it when the numbers divide cleanly.

The reverse is just as common. A and B together take 6 days; A alone takes 15. Subtract rates, never days: $\dfrac{1}{6} - \dfrac{1}{15} = \dfrac{1}{10}$ → B alone takes 10 days.

Three or more workers

Add all the rates in units. A, B and C take 9, 12 and 18 days: job = 36 units, rates $4 + 3 + 2 = 9$ a day → 4 days. As one formula:

T = \frac{abc}{ab + bc + ca}

Watch: Do not fold three workers two at a time with $\dfrac{ab}{a+b}$ . Add all three rates in a single line.

Pairwise data

Sometimes the question gives A+B, B+C and A+C instead of single times. Add the three pair-rates: each person appears exactly twice, so halve the sum for the trio rate.

Pairs of 10, 12 and 15 days → $\dfrac{1}{10} + \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{1}{4}$ → trio rate $\dfrac{1}{8}$ → 8 days together. For one person alone, subtract the pair that excludes him: A $= \dfrac{1}{8} - \dfrac{1}{12} = \dfrac{1}{24}$ → 24 days.

Partial work statements

'A can do $\dfrac{5}{6}$ of the work in 10 days.' Scale to the whole job first: full time $= 10 \times \dfrac{6}{5} = 12$ days. Then combine as usual.

In units it is equally direct: a 24-unit job with 20 units done in 10 days gives a rate of $\dfrac{20}{10} = 2$ units a day. Small numbers, same answer.

Careful: Multiply the days by the reciprocal of the fraction, not by the fraction itself.

Fraction of work left

A team that takes T days together finishes $\dfrac{t}{T}$ of the job in t days, leaving $1 - \dfrac{t}{T}$ .

A and B together take 12 days. After 5 days together, $\dfrac{5}{12}$ is done and $\dfrac{7}{12}$ is left. That remainder then belongs to whoever keeps working — usually the first step of a longer question.

Note: The leftover fraction is itself a common exam answer. Compute it exactly; do not round it.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Two workers together (or the reverse)

How to spot it:

Two individual times are given and the together time is asked — or the together time and one person's time are given and the other person is asked.

T = \frac{ab}{a + b}, \qquad \frac{1}{T_B} = \frac{1}{T_{A+B}} - \frac{1}{T_A}

Method

Set the job to the LCM of the two times, in units.
Add the two per-day rates; divide the job by the sum.
Reverse question: subtract one rate from the together rate, then invert.

Why it works:

Each worker carries his own share every day, so rates add or subtract — days never do.

Try this

A finishes a job in 15 days and B in 10 days. Working together, in how many days do they finish it?

Show solution

Job = LCM(15, 10) = $30$ units.
A: $30 \div 15 = 2$ /day; B: $30 \div 10 = 3$ /day.
Together $5$ /day → $30 \div 5 = 6$ days.

Answer

6 days

Type 2very common2 practice Q

Three or more workers together

How to spot it:

Three individual times are given (or two times plus the trio time) and the combined time — or one missing individual time — is asked.

\frac{1}{T} = \frac{1}{T_A} + \frac{1}{T_B} + \frac{1}{T_C}, \qquad T = \frac{abc}{ab + bc + ca}

Method

Take the job as the LCM of the given times.
Write each rate in units per day and add them all in one line.
Divide the job by the total rate; subtract rates from the trio rate if one is missing.

Why it works:

The whole job is split among the workers, so the daily shares add up to the trio's share.

Try this

A, B and C can do a piece of work in 9, 12 and 18 days respectively. In how many days will they finish it together?

Show solution

Job = LCM(9, 12, 18) = $36$ units.
Rates: $4 + 3 + 2 = 9$ a day.
$36 \div 9 = 4$ days.

Answer

4 days

Type 3common2 practice Q

Pairwise combination times

How to spot it:

'A and B together take …, B and C together take …, A and C together take …' — the trio time or one person's time is asked.

\frac{1}{T_{ABC}} = \frac{1}{2}\left(\frac{1}{T_{AB}} + \frac{1}{T_{BC}} + \frac{1}{T_{AC}}\right)

Method

Convert the three pair-times to pair-rates.
Add all three: each worker is counted twice, so halve the sum for the trio rate.
Any individual's rate = trio rate − the pair-rate that excludes him.

Why it works:

In the sum of the three pair-rates, A appears in two pairs — and so do B and C.

Try this

A and B can do a work in 10 days, B and C in 12 days, and A and C in 15 days. In how many days will all three finish it together?

Show solution

Pair rates: $\dfrac{1}{10} + \dfrac{1}{12} + \dfrac{1}{15} = \dfrac{1}{4}$ .
Each person counted twice → trio rate $= \dfrac{1}{8}$ .
$8$ days. A alone $= \dfrac{1}{8} - \dfrac{1}{12} = \dfrac{1}{24}$ → 24 days.

Answer

8 days

Type 4common2 practice Q

Fractional / partial work statements

How to spot it:

'A can do 3/5 of the work in 12 days', 'B does 2/7 of the work in 8 days' — the full-work time, or the together time, is asked.

\text{full time} = \text{days} \times \frac{\text{denominator}}{\text{numerator}}

Method

Scale each statement to the whole job: days × reciprocal of the fraction.
Combine the full-work times like any other question.
In LCM units the rate is simply (units done) ÷ (days taken).

Why it works:

A worker's rate is fixed, so a partial statement just describes a smaller stretch of the same job.

Try this

A can do $\dfrac{5}{6}$ of a work in 10 days and B can do $\dfrac{3}{8}$ of the same work in 9 days. Working together, in how many days will they complete the work?

Show solution

A's full time $= 10 \times \dfrac{6}{5} = 12$ days.
B's full time $= 9 \times \dfrac{8}{3} = 24$ days.
$\dfrac{1}{12} + \dfrac{1}{24} = \dfrac{1}{8}$ → $8$ days.

Answer

8 days

Type 5very common

Fraction of work left after t days together

How to spot it:

A team's together time and the days they worked are given; the fraction of work left — or who finishes the rest — is asked.

\text{left} = 1 - \frac{t}{T_{together}}

Method

Read off the together time T of the team as it stood.
Work done in t days $= \dfrac{t}{T}$ .
Left $= 1 - \dfrac{t}{T}$ ; if a new worker finishes the rest, multiply the leftover by his full time.

Why it works:

A team working at its combined rate covers the whole job in T days, so t days deliver exactly t/T of it.

Try this

A and B together can finish a work in 12 days. They work together for 5 days, after which B leaves. In how many days will C alone finish the remaining work if C alone can do the whole work in 24 days?

Show solution

Done in 5 days $= \dfrac{5}{12}$ ; left $= \dfrac{7}{12}$ .
C needs $\dfrac{7}{12} \times 24 = 14$ days.

Answer

14 days

Formula sheet

One-day work

\frac{1}{T}

Combined time (two workers)

T = \frac{ab}{a + b}

Combined time (three workers)

T = \frac{abc}{ab + bc + ca}

Work done and left

\text{done} = \frac{t}{T}, \quad \text{left} = 1 - \frac{t}{T}

Shortcuts that save time

⚡ LCM units, not fractions

Set the job to the LCM of the individual times; every rate becomes a small whole number.

Example

A finishes a job in 10 days and B in 20 days. Working together, they finish in:

Show solution

Work = LCM(10, 20) = $20$ units.
A: $2$ /day, B: $1$ /day → together $3$ /day.
$20 \div 3 = 6\dfrac{2}{3}$ days.

Answer

6 2/3 days

⚡ The ab over a+b reflex

For exactly two workers, one formula — no fraction addition at all.

Example

A does a job in 6 days, B in 12 days. Together they need:

Show solution

$T = \dfrac{6 \times 12}{6 + 12} = \dfrac{72}{18}$ .
$= 4$ days — less than either alone.

Answer

4 days

⚡ Three workers through the LCM

Same method with three rates added in one line.

Example

A, B and C take 6, 12 and 18 days respectively. Working together, they complete the work in:

Show solution

Work = LCM = $36$ units.
Rates: $6 + 3 + 2 = 11$ a day.
$36 \div 11 = 3\dfrac{3}{11}$ days.

Answer

3 3/11 days

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding the days (12 + 24 = 36) instead of adding the rates.

Rates add, days do not. The together time is ab/(a+b), and it must be less than the smaller time.

Mistake 02

For three workers, applying ab/(a+b) to a pair and merging again.

Add all three rates in one line, or use abc/(ab+bc+ca).

Mistake 03

Accepting a combined time larger than the fastest worker's time.

The team must beat the best single time; if it does not, rates were not added.

Mistake 04

In reverse questions, subtracting the times instead of the rates.

B's rate = together rate − A's rate; then invert for B's days.

Mistake 05

Mixing units — one rate per day, another per hour.

Convert every rate to the same time unit before adding.

Quick revision

Read this the night before the exam.

Rate = 1/time; rates add; work = rate × days.
Two workers: T = ab/(a+b). Three: T = abc/(ab+bc+ca).
Together minus one: subtract rates, never days.
Pairwise times: add the three pair-rates, halve for the trio.
Partial work: full time = days × reciprocal of the fraction.
Work left after t days together: 1 − t/T.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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