The problem states the following:

  • To complete an order at a milk factory $8$ workers had processed in $9$ days working $6$ hours per day, $30$ gallons of milk.
  • To complete the job the engineer at the plant hired $2$ more workers.
  • How many days those $10$ men working $8$ hours a day would need to complete the remaining $50$ gallons ?

In my attempt to solve this problem I tried to use these conversion factors: For the gallon part I assumed that the job the workers do is per gallon.

$$\textrm{8 workers}\times \textrm{9 days}\times \frac{\textrm{6 hour}}{\textrm{day}}\times\frac{1}{\textrm{30 gallon}}=\frac{8\times 9}{5}\,\frac{\textrm{worker hour}}{\textrm{gallon}}$$

Then the second condition would become into something like this, I used $\textrm{80 gallons}$ as for the second part would be a computation for the whole processing job of the milk:

$$\frac{1}{\textrm{10 workers}}\times\frac{\textrm{day}}{\textrm{8 hours}}\times \textrm{80 gallons} \times \frac{8\times 9}{5}\,\frac{\textrm{worker hour}}{\textrm{gallon}}=\textrm{14.4 days}$$

Therefore by subtracting this value with what was already elapsed $\textrm{9 days}$ would become into $\textrm{14.4 - 9.0 = 5.4 days}$.

However by comparing on the existing alternatives on my book being:

  • 10 days, 9 days, 12 days, 11 days, 13 days

none of these seem to check with what I've found, therefore I'm not very convinced on my result.

Although that in the denominator there is the number of workers hours for the new condition yet this approach does not seem to solve correctly the problem. Could it be that my solution is wrong? or the alternatives given are not good?

Can somebody please instruct me which thing could I did it wrong?

Since this problem involves speed I understand that I can also use linear functions such as the likes of $f(x)=mx+b$ But in this case, would the $y$-intercept be zero? Therefore in the proposed answer I'd appreciate someone can also include this method to compare both.


Continuing by trial-and-error I got to one of the alternatives by replacing $\textrm{50 gallons}$ instead of $\textrm{80 gallons}$.

$$\frac{1}{\textrm{10 workers}}\times\frac{\textrm{day}}{\textrm{8 hours}}\times \textrm{50 gallons} \times \frac{8\times 9}{5}\,\frac{\textrm{worker hour}}{\textrm{gallon}}=\textrm{9 days}$$

But again, I am not very convinced from the result. How does $\textrm{9 days}$ equate the work of 10 men processing the milk. It feels as the same result from the original 8 workers.

Should the number of gallons be summed as I did in the first part or left as it is in the second computation?. Can somebody help me to clear these doubts and also a solution by going on to linear functions route?.

  • $\begingroup$ You did the math correctly, the fact that the workers are in the denominator is not wrong. You are using days per worker hour. $\endgroup$ – TSF Mar 2 '18 at 20:12
  • $\begingroup$ @TonyS.F. I'm glad that it seems my initial thoughts were not wrong. I missed that part regarding days per worker hour. But yet the second part of my "request" is still unanswered. How can I solve this problem using linear function method? $\endgroup$ – Chris Steinbeck Bell Mar 2 '18 at 21:23
  • $\begingroup$ @Leucippus Can you please inspect this problem and use your superpowers on it?. :) Thanks in advance. $\endgroup$ – Chris Steinbeck Bell Mar 3 '18 at 6:43

Before I begin. I find it funny how the question forgot to add 'minimum days' because the fact they didn't means 34738974143232 is a viable answer:


The hours worked to gallons of milk ratio is:


10 workers working 8 hours a day is 80 hours.

Since you need 50 gallons, to find the minimum hours needed you would have to boost your ratio so 5 becomes 50. Therefore 72 becomes 720.

720/8 is 9.

9 Days

Does my working help?


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