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 March 4th, 2018, 02:15 PM #1 Member     Joined: Jun 2017 From: Lima, Peru Posts: 97 Thanks: 1 Math Focus: Calculus How to solve this word problem using conversion factors and linear function? The situation is as follows 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?.
 March 4th, 2018, 02:28 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,458 Thanks: 1340 I would do this as $8w \cdot 9day \cdot 6\dfrac{hr}{day} = 30gal$ $232w\cdot hr = 30gal$ $1gal = \dfrac{232}{30}w\cdot hr$ in other words it takes $\dfrac{232}{30}$ worker hours to process a gallon of milk Given this we now have to determine how many days to process 50 gallons with 10 workers working 8 hours a day. $50 gal = 50\cdot \dfrac{232}{30}w\cdot hr = \dfrac{1160}{3} w\cdot hr$ we now have $10w \cdot 8 \dfrac {hr}{day} = 80 \dfrac{w \cdot hr}{day}$ so the total days will be $T = \dfrac{\dfrac{1160}{3}w\cdot hr}{80 \dfrac{w\cdot hr}{day}} = \dfrac{29}{6}day = {4.8\bar{3}}day$
March 4th, 2018, 05:32 PM   #3
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Quote:
 Originally Posted by Chemist116 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 will these 10 men, working 8 hours per day, need to complete the remaining 50 gallons?
Total hours for the 30 gallons:
8*9*6 = 432
So man-hours per gallon:
432/30 = 14.4 [1]

x = days for the 50 gallons.
Total hours for the 50 gallons:
10*x*8 = 80x
So man-hours per gallon:
80x/50

80x/50 = 14.4 [1]
x = 9 days

March 4th, 2018, 06:31 PM   #4
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 Originally Posted by romsek I would do this as $8w \cdot 9day \cdot 6\dfrac{hr}{day} = 30gal$ $232w\cdot hr = 30gal$
no.. it would be 432... i'm getting old

March 4th, 2018, 06:36 PM   #5
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Quote:
 Originally Posted by romsek no.. it would be 432... i'm getting old
I think that cat pic is putting you off

March 5th, 2018, 10:50 AM   #6
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Quote:
 Originally Posted by Denis Total hours for the 30 gallons: 8*9*6 = 432 So man-hours per gallon: 432/30 = 14.4 [1] x = days for the 50 gallons. Total hours for the 50 gallons: 10*x*8 = 80x So man-hours per gallon: 80x/50 80x/50 = 14.4 [1] x = 9 days
I this this looks like the method I used in the second part but explains that the key into solving these things is to find the number of man-hours per gallon or in general man-hours per job or anything that it is being built or done.

March 5th, 2018, 10:55 AM   #7
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 Originally Posted by romsek no.. it would be 432... i'm getting old
At first I didn't know what were you referring to but by doing the computations just figured out that it was about the result of man-hours per gallon. Don't feel bad about age. I'm 32 and should I feel embarrassed for asking these questions?. Just kidding! have a nice day! .

March 5th, 2018, 11:50 AM   #8
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 Originally Posted by Chemist116 I this this looks like the method I used in the second part but explains that the key into solving these things is to find the number of man-hours per gallon or in general man-hours per job or anything that it is being built or done.
Huh?

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