Time & Work

Work, time and rate

In these questions "work" means one whole job — building a wall, filling a tank, stitching shirts. The rate of a worker is the part of the job he finishes in one day (or one hour). If A finishes the whole job in T days, his rate is $\frac{1}{T}$ of the job per day. Two rules run the whole topic:

$\text{Work done} = \text{rate} \times \text{time}, \qquad \text{rates add when people work together}$

When two people work together, their rates simply add — each does his own bit every day. So the combined time is always less than the fastest individual time. If your answer is bigger, you added days instead of rates.

The LCM method — whole numbers, no fractions

Fractions slow you down and cause sign slips. Instead, give the job a size in units: take the LCM of all the individual times. A finishes in 15 days, B in 10 days. Let the job = 30 units (LCM of 15 and 10). A does $30 \div 15 = 2$ units a day, B does 3 units a day. Together $5$ units a day → $30 \div 5 = 6$ days. Every number here is a small whole number — that is the whole point of the method.

Combining two workers

For exactly two workers with times a and b: $T = \frac{ab}{a + b}$ Example: 6 and 12 days → $\frac{6 \times 12}{18} = 4$ days. This is the same as the LCM method — try it once to convince yourself, then use whichever is faster. The reverse question is just as common: "A and B together take 6 days, A alone takes 10 — find B alone." Subtract the rates: $\frac{1}{6} - \frac{1}{10} = \frac{1}{15}$ → 15 days.

Three or more workers

Add all three rates. A, B, C take 9, 12, 18 days: job = 36 units, rates $4 + 3 + 2 = 9$ a day → 4 days. As a formula, three workers take $T = \frac{abc}{ab + bc + ca}$

Pairwise data (A+B, B+C, A+C)

When the three pair-times are given, add the three pair-rates — each person's rate now appears twice — and halve: $\frac{1}{T_{AB}} + \frac{1}{T_{BC}} + \frac{1}{T_{AC}} = 2\left(\frac{1}{T_A} + \frac{1}{T_B} + \frac{1}{T_C}\right)$ Pairs take 10, 12 and 15 days: sum of pair rates $= \frac{15}{60} = \frac14$ → trio rate $\frac18$ → all three together take 8 days. To find one person alone, subtract a pair from the trio: A = trio − (B+C).

Partial work statements

"A can do $\frac{3}{5}$ of a work in 12 days." Scale to the whole job: full time $= 12 \times \frac{5}{3} = 20$ days. Convert every worker to a full-work time first, then combine as usual. With the LCM method this is equally quick: if the job is 35 units and A does 21 units in 12 days, his rate is $\frac{21}{12} = \frac74$ units a day.

Common traps

• Adding the individual days (15 + 10 = 25) instead of the rates.
• In "together minus one" questions, subtracting the times instead of the rates.
• Partial work: scaling the days by the given fraction instead of its reciprocal.
• Forgetting the sanity check — a combined time must beat the best single time.

Quick revision

• Rate = 1/time; rates add; Work = rate × time.
• Two workers: $T = \frac{ab}{a+b}$. Three: $T = \frac{abc}{ab+bc+ca}$.
• Together − A = B: subtract rates, never days.
• Pairwise times: add the three pair-rates, halve → trio rate.
• Partial work: full time = days × (reciprocal of fraction).