OK, so embarrassingly I've forgotten how to do this type of problem which I'll generalize:

$n$ workers of type $A$ can do a job in $x$ hours. $m$ workers of type $B$ can do the same job in $y$ hours. How long would it take $n_1$ workers of type $A$ and $m_1$ workers of type $B$ to do the job working together?

Logically I can see that $1$ type $A$ worker should spend $n$ times as much time as $n$ workers take. Same with $B$. So a type $A$ worker can do the job in $nx$ hours and a type $B$ worker could do the job in $my$ hours. I just don't seem to be able to figure out how long a mixture of type $A$ and type $B$ workers take.

Could someone please explain the process to me? Thanks.

  • $\begingroup$ Hint: How many jobs could a type $A$ worker complete in $1$ hour? How many jobs could $2$, or $3$, or $n_1$ workers complete in $1$ hour? $\endgroup$ – John Joy Sep 9 '15 at 13:31
  • $\begingroup$ Hint: $n_1$ workers of type A work at a rate of $n_1$ jobs per $nx$ hours, or $n_1/nx$ jobs per hour. Do the same thing for B; then consider how much work they'd do together per hour. $\endgroup$ – symplectomorphic Sep 9 '15 at 13:31
  • $\begingroup$ Also note that the problem tacitly assumes the workers work at a constant rate. $\endgroup$ – symplectomorphic Sep 9 '15 at 13:34

$nx$ man-hours of type A = $ym$ man-hours of type B

thus 1 man-hour of type B = $\dfrac{nx}{ym}$ man-hours of type A.

Convert to type A

so hours needed by given mix = $\dfrac{nx}{n_1 + \dfrac{nx}{ym}\cdot m_1}$


The rate of work of a worker from group $A$ is $\frac{1}{nx}$. The rate of a worker from group $B$ is $\frac{1}{my}$. Adding more workers from either group, you just add the rates. Now if you add up $m_1$ workers and $n_1$ workers from the groups you get a rate of $$n_1 \frac{1}{nx} + m_1\frac{1}{my}$$ And to get the time it takes, use $$\text{time} = \frac{1}{\text{rate}}$$

  • $\begingroup$ Your first two sentences are wrong. $\endgroup$ – symplectomorphic Sep 9 '15 at 13:35
  • $\begingroup$ @symplectomorphic Thank you! $\endgroup$ – Eric Auld Sep 9 '15 at 13:45
  • 1
    $\begingroup$ You might want to mention that the reason $$\text{time} = \dfrac 1{\text{rate}}$$ is because the workers are only doing one job in this problem. In general, rate is defined by $$\text{rate} = \dfrac{\#\text{ of jobs}}{\text{time}}$$ $\endgroup$ – user137731 Sep 9 '15 at 14:43

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