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If Shannon has a project that will take 550 hours to complete and she only works on the project 3 days a week for only 3-4 hours a day, about how long will it take Shannon to complete her project?

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  • $\begingroup$ Is this a trick question? It will take 550 hours. $\endgroup$
    – John Douma
    Nov 13, 2015 at 21:44

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We could solve your problem with quite classical methods, but it is sometimes interesting to look at some problems differently to develop your intuition for other problems. Sometimes, it is useful to compare the quantities of the things involved in the wording of a problem.

Suppose she had a work requiring $h$ hours and she works $x$ hours a day in average, $y$ days a week. It takes $h/x$ days to complete the work and it takes $(h/x)/y$ weeks to complete it, i.e. $7\cdot(h/x)/y=\frac{7h}{xy}$ days in total.

Justifications: if she works $x$ hours per day of work, it would take her $h/x$ consecutive days to complete it. But she only works $y$ days per week, so it will take her $(h/x)/y$ weeks to complete it. There are $7$ days in a week, so it will take her a total amount of $7\cdot(h/x)/y$ days to complete the job.

The "most difficult" part of the reasoning may be the one from $(h/x)$ consecutive days to $(h/x)/y$ weeks. But look at the units: $h$ is in hours (unit: $\text{hours}$), $x$ in hours a day (unit: $\frac{\text{hours}}{\text{days}}=\text{hours}\cdot\text{days}^{-1}$) and $y$ in days per week (unit:$\frac{\text{days}}{\text{week}}$):

$$\frac{h}{x}\cdot\frac{1}{y}\Rightarrow\underbrace{\frac{\text{hours}}{\text{hours}\cdot \text{days}^{-1}}}_{(h/x)}\cdot\underbrace{\frac{\text{weeks}}{\text{days}}}_{1/y}=\text{weeks}$$

Then, you just have to know that a $1\,\text{week}$ is equivalent to $7\,\text{days}$ so that there are $7$ times more $\text{days}$ than weeks required to complete the job, i.e. $7\cdot\frac{h}{x}\cdot\frac{1}{y}=\frac{7h}{xy}$.

For your particular example, it takes her $7\cdot(550/3.5)/3\approx 336.67$ days in total.

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