Time and work

Work done = Rate of work × Time taken. This is the anchor equation behind every time-and-work problem, and it is just the unitary method applied to "how much of a job gets done."

where: Work done is measured as a fraction of the whole job (the completed job = $1$ unit), Rate of work is the fraction of the job finished per day, and Time taken is in days. If someone can finish an entire job alone in $A$ days, their rate is $\dfrac{1}{A}$ of the job per day — this is simply the unitary method run in reverse: instead of asking "how much for one item," it asks "how much in one day."

Combining two workers. When two people work together, their daily fractions simply add, because on any given day each person independently contributes their own share toward finishing the same job, and those shares stack up rather than interfere with each other:

and since Rate $\times$ Time $= 1$ (one whole job), the time together is the reciprocal of that sum.

The inverse-variation view. For a fixed job, the number of workers and the days needed vary inversely — double the workers, half the time — so

where: $m_1, d_1$ are one group's worker-count and days, and $m_2, d_2$ are another group's worker-count and days for the same total job, all workers assumed to work at the same steady rate. This is exactly the "total worker-days needed" staying constant.

Rahul can paint a fence alone in 12 days, and Priya can paint the same fence alone in 15 days. Working together, how many days will they take?

Step 1. Write each rate: Rahul's rate is $\dfrac{1}{12}$ of the fence per day, Priya's rate is $\dfrac{1}{15}$ of the fence per day.

Step 2. Add the rates on a common denominator of 60: $\dfrac{1}{12} + \dfrac{1}{15} = \dfrac{5}{60} + \dfrac{4}{60} = \dfrac{9}{60} = \dfrac{3}{20}$ of the fence per day together.

Step 3. Take the reciprocal to get the time: $T = \dfrac{1}{3/20} = \dfrac{20}{3} = 6\dfrac{2}{3}$ days.

Step 4. Cross-check with the shortcut formula: $T = \dfrac{AB}{A+B} = \dfrac{12 \times 15}{12+15} = \dfrac{180}{27} = \dfrac{20}{3} = 6\dfrac{2}{3}$ days. Both routes agree.

Rahul and Priya together take $6\dfrac{2}{3}$ days, i.e. 6 days and $\dfrac{2}{3} \times 24 = 16$ hours.

Competitive stretch

15 labourers can dig a trench in 8 days. To finish it in only 6 days, how many labourers (at the same rate) are needed? Using $m_1 d_1 = m_2 d_2$: the total labourer-days needed is $k = 15 \times 8 = 120$. For 6 days, labourers $= \dfrac{120}{6} = 20$. Since fewer days demands more hands, $20 > 15$ makes sense, and $20 \times 6 = 120 = 15 \times 8$ confirms it. 20 labourers are needed — 5 more than before.

• Adding days instead of rates. Writing "Rahul + Priya take $12 + 15 = 27$ days" is wrong — you must add the rates $\frac1A + \frac1B$, never the days directly.
• Forgetting the final reciprocal. After finding the combined rate (work per day), students sometimes leave it as the answer instead of inverting it to get days.
• Using $m_1 d_1 = m_2 d_2$ without the same-rate assumption. It only holds if every worker keeps the same pace and hours per day; if hours per day change too, that factor must be included.
• Mixing up who is faster. Fewer days taken means a higher rate (faster worker), not slower — reversing this gives a combined time larger than either person's individual time, which is never possible.
• Forgetting the combined time must be shorter than both. Whatever $T = \dfrac{AB}{A+B}$ comes out to, it must be smaller than the faster worker's own time alone; if a computed "combined" answer is larger than either individual time, a rate has been subtracted or misplaced somewhere in the working.

Ask the tutor: Stuck on setting up a combined-rate problem, or want to check whether your workers-and-days answer makes sense — ask the AI tutor to walk through it with you step by step.