# Work problems: A Deeper Look

We all know that work problems are some applications of algebra in reality. But the equation that corresponds to work problems is not as clearly stated as investment, mixture and uniform-motion problems.

Now, suppose we have the set of workers $A$ and each worker $a_i \in A$ where $n$ is the number of workers. Each worker $a_i$ can do a certain job $J$ in $h_i$ hours. Now if all the workers do the job $J$ altogether and started at the same time, we come into an equation which is $$x\sum _{i=1} ^n \frac{1}{h_i} = 1$$

Now, we solve for $x$ which is the number of hours the whole workers can do the job altogether. Considering the fact that they started at the same time. What if there exists a worker who started earlier by an hour from the rest or what if the situation is more complicated.

The question now is, what could be possible explanations of the formula that could give as clarity of the formula?

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## 1 Answer

Looked at properly, this kind of work problem is just a kind of rate problem: since $h_i$ is the number of hours that it takes $a_i$ to do the job by himself, his rate of work is $$r_i=\frac1{h_i}\text{ jobs/hour}\;,$$ and $$r=\sum_{i=1}^n\frac1{h_i}=\sum_{i=1}^nr_i\text{ jobs/hour}$$ is the work rate of the entire crew of $n$ workers. Then, just as in uniform motion problems, if $j$ similar jobs are to be done, and $x$ is the amount of time required to do them, we must have $$rx=j\tag{1}$$ and therefore $$x=\frac{j}{r}\;.\tag{2}$$ This is just $$\begin{cases}\text{rate}\cdot\text{time}=\text{distance}\\\\\text{time}=\frac{\text{distance}}{\text{rate}}\end{cases}$$ in a slightly different guise.

People are used to thinking of rates in miles per hour or the like, but the reciprocal notion isn’t completely unfamiliar. Runners, for example, are very familiar with it: one normally speaks of running six-minute miles, for instance, not of running at a speed of ten miles per hour.

Variations in which one or more workers start early are easily handled in this framework. If worker $a_i$ starts $k_i$ hours early, he does $r_ik_i$ jobs before the official starting time, and the team of $n$ workers does $$\sum_{i=1}^nr_ik_i$$ jobs before the official starting time. Thus, only $$j-\sum_{i=1}^nr_ik_i$$ jobs remain to be done when work officially starts, so if $x$ now represents the number of hours needed after the official starting time, $(1)$ becomes $$rx=j-\sum_{i=1}^nr_ik_i\;,$$ and $(2)$ becomes $$x=\frac{j}r-\sum_{i=1}^n\frac{r_i}rk_i\;.$$

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