2
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

$nx$ 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 = $\dfrac{nx}{n_1 + \dfrac{nx}{ym}\cdot m_1}$

|cite|improve this answer
$\endgroup$
0
$\begingroup$

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 $$\text{time} = \frac{1}{\text{rate}}$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.