Time and work problems

Two workers, two different speeds, one shared job — how long do they take together? Time and work problems live inside the ICSE Class 8 chapter on Direct and Inverse Variations, and they reward a single trick: stop thinking in days and start thinking in fraction of the job done per day. Days cannot be added across different workers, because "12 days plus 18 days" makes no sense once two people are sharing the load. Rates, however, are just fractions of the same fixed job, and fractions add normally. Once the whole job is treated as the number 1, every worker becomes a fraction of that 1, and the rest of the topic is ordinary fraction arithmetic wearing a word-problem disguise.

The core insight

Treat the whole job as 1 unit of work, convert each worker's time into a rate of work per day (1n\frac1n), and add rates — never days — to find how long people take working together.

The core idea

The job is 1 unit. If a worker finishes an entire task alone in nn days, working at a constant steady pace, then in one day that worker completes 1n\frac1n of the job:

Rate of work=1n of the job per day\text{Rate of work} = \frac{1}{n}\ \text{of the job per day}

where: nn is the number of days the worker takes to finish the job alone.

Rates add; days don't. If A alone takes AA days and B alone takes BB days for the same job, their combined daily output is the sum of the two rates:

Combined rate=1A+1B\text{Combined rate} = \frac{1}{A} + \frac{1}{B}

This works because a rate always refers to the same fixed job, so two rates measured on that job can be stacked directly — unlike the two "day" numbers, which belong to different paces and cannot be added.

Time is the reciprocal of the combined rate. If the combined rate simplifies to 1T\frac{1}{T} of the job per day, then working together the job is finished in TT days:

T=A×BA+BT = \frac{A \times B}{A + B}

where: AA and BB are the number of days each worker takes alone, and TT is the number of days they take together — always smaller than either AA or BB, since two people working together can never be slower than either one alone.

Work done in a fixed time. If a rate of 1n\frac1n per day is kept up for tt days, the fraction of the job finished is:

Work done=Rate×Time=tn\text{Work done} = \text{Rate} \times \text{Time} = \frac{t}{n}

This is the same inverse-variation idea seen elsewhere in the chapter: for a fixed total job, the number of workers and the number of days needed vary inversely — doubling the workers roughly halves the days, because the total work (workers times days, in the simplest case) stays constant.

Figure
A bar showing one whole job split into two shares one from worker A one from worker B that together fill the bar in less time than either worker alone
One job, two rates: A does 1/12 per day, B does 1/18 per day A alone, 12 days: fills 1/12 of the bar each day B alone, 18 days: fills 1/18 of the bar each day Together, one day: A's share plus B's share, 5/36 of the bar Combined rate 5/36 per day means the whole bar fills in 36/5 = 7.2 days, faster than either alone Bar length represents one full job (1 unit of work)
Derivation ATime taken together from two individual times

The goal is to show, starting only from the definition of a work rate as a fraction of the job done per day, exactly why the combined time two workers take together equals T=A×BA+BT = \dfrac{A\times B}{A+B} rather than some other combination of AA and BB, and why this must always come out smaller than both AA and BB.

Step 1. Write each worker's daily rate as a fraction of the whole job. If A alone finishes in AA days, A's rate is 1A\frac1A of the job per day; likewise B's rate is 1B\frac1B per day. These are shares of the same job, so they are directly comparable fractions.

Step 2. Add the rates, since working together simply means both fractions of the job get completed within the same single day. The combined rate is 1A+1B\frac1A + \frac1B. Putting this over a common denominator, 1A+1B=B+AAB\frac1A + \frac1B = \frac{B+A}{AB}.

Step 3. A combined rate of A+BAB\frac{A+B}{AB} per day means that in one day, A+BAB\frac{A+B}{AB} of the job gets done. To find the number of days TT needed to complete exactly 1 whole job at this steady combined rate, use Rate×T=1\text{Rate} \times T = 1, so TT is the reciprocal of the rate: T=1A+BAB=ABA+BT = \dfrac{1}{\frac{A+B}{AB}} = \dfrac{AB}{A+B}.

Step 4. Check the result behaves sensibly. Since AA and BB are both positive, ABA+B\dfrac{AB}{A+B} is always less than the smaller of AA and BB — algebraically, ABA+B<A\dfrac{AB}{A+B} < A because B<A+BB < A + B makes the fraction BA+B\frac{B}{A+B} less than 1, so T=A⋅BA+B<AT = A \cdot \frac{B}{A+B} < A. This matches common sense: two people working together can never take longer than the faster person alone. Verified: with A=12, B=18A=12,\ B=18, T=12×1812+18=21630=7.2T = \dfrac{12\times18}{12+18} = \dfrac{216}{30} = 7.2 days, matching 112+118=536\frac1{12}+\frac1{18}=\frac5{36} inverted to 365=7.2\frac{36}{5}=7.2.

Worked example

A can complete a piece of work in 12 days, and B can complete the same work in 18 days. They work together for 3 days, and then A leaves. In how many more days will B alone finish the remaining work?

Step 1. Find the combined rate of A and B, using the LCM of 12 and 18, which is 36.

112+118=336+236=536 of the job per day\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}\ \text{of the job per day}

Step 2. Find the work done together in the first 3 days: rate times time.

Work done=3×536=1536=512 of the job\text{Work done} = 3 \times \frac{5}{36} = \frac{15}{36} = \frac{5}{12}\ \text{of the job}

Step 3. Find the work still remaining by subtracting from the whole job (1 unit).

Remaining work=1−512=712 of the job\text{Remaining work} = 1 - \frac{5}{12} = \frac{7}{12}\ \text{of the job}

Step 4. B now finishes this remaining fraction alone, at B's own rate of 118\frac{1}{18} per day. Time needed is remaining work divided by B's rate.

Time=7/121/18=712×18=12612=212=1012 days\text{Time} = \frac{7/12}{1/18} = \frac{7}{12} \times 18 = \frac{126}{12} = \frac{21}{2} = 10\tfrac12\ \text{days}

B alone needs 101210\tfrac12 more days to finish the remaining 712\frac{7}{12} of the work after A leaves.

Common traps

  • Adding days instead of rates. "A takes 12 days, B takes 18 days, so together 30 days" is always wrong — working together is always faster than either worker alone, so rates (fractions per day) must be added, then inverted.
  • Forgetting the reciprocal step. The sum 1A+1B\frac1A + \frac1B is a rate (work per day), not the time itself; the time taken is the reciprocal of that sum.
  • Subtracting work instead of rates when one person leaves partway. Track fraction of work completed first, subtract from 1 to get what remains, and only then convert the remaining fraction into days using the relevant rate — don't just subtract days.
  • Mixing up a rate with a time. A rate of 112\frac{1}{12} means one-twelfth of the job per day; it is not "12" of anything, and it must never be added directly to a number of days.

Ask the tutor: if you're stuck on whether to add rates or add days in a specific problem, or want another partial-work example worked through step by step, ask the AI tutor with your exact numbers.

Why can't I just average the two days, like (12+18)/2, to get the time together?

Averaging days treats "days" as if they were quantities that combine directly, but they don't — a day for A represents a different fraction of the job than a day for B. The quantities that genuinely add are the rates, each expressed as a fraction of the same fixed job per day. Once you add 112+118=536\frac{1}{12}+\frac{1}{18}=\frac{5}{36} and invert, you get 7.27.2 days, not the average of 1212 and 1818, which would be 1515 — a very different, and wrong, number.

How do I handle a problem where one person starts a few days before the other joins?

Split the timeline into stages and track work done, not days elapsed. In the first stage, only one person works, so multiply that person's rate by the number of days they work alone to get the fraction of the job finished. Subtract that from 1 to see what remains. In the next stage, both rates apply together, so use the combined rate on the remaining fraction to find how many more days are needed. The key habit is always converting time into work-done using rate times time, then working with fractions of the job, never with raw days across different people.

Is there a difference between someone undoing another person's work, like a pipe draining a tank while another fills it?

That is a related idea using the same rate machinery, but it goes beyond the direct-and-inverse-variation reasoning in this chapter, where every worker is assumed to be helping toward the same job. In the simplest version you would still add rates, but a draining or undoing effect is treated as a negative rate, subtracted rather than added. If your syllabus and textbook do not cover tank-and-pipe problems in this chapter, you can safely set that variant aside for now and focus on people or machines working together toward the same finished job.

Before this

Where this leads