$\textrm{The problem is as follows:}$

In an office there are two printers, a brand new model and an older one. The older printer needs 8 hours to complete a job. But both models working together complete the same job in 3 hours. How long will it take for the newer printer to do the same task alone?.

I am not sure how to solve this problem. I thought about using functions to tackle this situation like:


therefore if the job is given as a function of time $\textrm{w}$ thus I wrote it this way:


For the older printer $w_{1}$:


For the newer printer $w_{2}$:


Since both complete the job in three hours then:





However $-5$ is not the correct answer. Moreover it does not make sense to have a negative sign for time, although in this case it would indicate that the function is linear but the slope is negative.

I'm stuck at here, although in my attempt to solve this problem I tried to use functions (linear), maybe there is another way to solve this kind of situations? Can somebody help me to compare between a method using functions and different one?.

Since $-5$ is not right. What would be the incorrect interpretation of the problem?


Velocities ("Completed jobs per hour") add together. The old printer has velocity $\frac18$, as the sum of the two velocities is $\frac13$. Thus the velocity of the new printer alone is $\frac13-\frac18=\frac{5}{24}$.

The new printer completes $\frac5{24}$ jobs every hour, so it takes $\frac{24}{5}$ hours${}=4$ hours, $48$ minutes to complete a single job.

  • $\begingroup$ your method is very straightforward. But I wonder, if the same answer can be reached using a linear function?. $\endgroup$ – Chris Steinbeck Bell Dec 11 '17 at 12:40
  • $\begingroup$ @ChrisSteinbeckBell To be quite honest, I'm having some difficulty following exactly what the argument in your question post says. What exactly is this function $w(t)$ meant to represent? What is the input, what is the output? You can't use functions to solve something if you aren't clear on what exactly the functions are meant to do. $\endgroup$ – Arthur Dec 11 '17 at 12:54
  • $\begingroup$ I defined $w(t)$ as the work each printer has to do. Being $t$ the time required (input).and $m$ would be the "speed" and $b$ the y-intercept. Since the speed is constant or I assumed this to be the case I thought that a function of time could be used to solve the problem. $\endgroup$ – Chris Steinbeck Bell Dec 11 '17 at 13:10
  • $\begingroup$ @ChrisSteinbeckBell I see. So you want to do basically the same thing that I have done, only phrase it in terms of a function that takes as input the amount of time spent, and as output the number of works completed, is that right? In that case, the $y$-intercepts are clearly $0$, since no time spent means no work done. As for the slopes, $w_1(t) = \frac18t$, since we know that $w_1(8) = 1$. And we have $w_2(t) = at$ for some $a$. Finally, we know that $w_1(3) + w_2(3) = 1$, and you may use this to find $a$. $\endgroup$ – Arthur Dec 11 '17 at 13:17
  • $\begingroup$ Why $w_{1}(3)+w_{2}(3)=1$? I have a problem trying to understand where does this relationship comes from?. If I follow the rest which you mentioned I reach to $\frac{5}{24}$ which becomes into the answer. $\endgroup$ – Chris Steinbeck Bell Dec 11 '17 at 13:33

I would tackle the problem as follows:

Define $W$ to be the workload the printers have to do and $p_1,p_2$ be the effect of the printers respectively. Now we can say the following, since $$ W=P\cdot t $$ and we have $$ W=p_1\cdot 8,\qquad W=(p_1+p_2)\cdot 3 $$ and from this data it is not hard to get $$ t=\frac{W}{p_1}. $$ I found it to be


Hope this helps :)

  • $\begingroup$ Is it just me or your final answer appears blinking?. The way how you get to that answer is not clear. I think it would be a good idea to add some steps. $\endgroup$ – Chris Steinbeck Bell Dec 11 '17 at 12:38
  • $\begingroup$ @ChrisSteinbeckBell I only wanted to give some ideas and hints how to think here not solving the exercise for the OP. I included the final answer in a "hidden" enviroment so the OP can think himself and check whether the anser is correct $\endgroup$ – Vinyl_cape_jawa Dec 11 '17 at 12:42
  • $\begingroup$ Got it. How did you made that hidden effect?. $\endgroup$ – Chris Steinbeck Bell Dec 11 '17 at 13:04
  • $\begingroup$ @ChrisSteinbeckBell I ususally try to give only hints and not the answer, it is better if one finds the answers himself :) you type ´>!´ in the beginning of the line $\endgroup$ – Vinyl_cape_jawa Dec 11 '17 at 13:07
  • $\begingroup$ I also agree with your view at some extend, although there are times when after several attempts the OP (original poster?) becomes frustrated and is unable to get to the answer and gives up. Hopefully this is not the case. :) $\endgroup$ – Chris Steinbeck Bell Dec 11 '17 at 13:14

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