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 Algebra Pre-Algebra and Basic Algebra Math Forum

 April 13th, 2018, 12:26 AM #1 Member   Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0 Working rates with algebra One day, it takes p hours for q factory workers to construct r items. The following day three more worked join the team and four more items are required to be constructed. 1)Find how many hours it will take, in terms of p,q and r. 2)If in fact it takes the same length of time, find an equation and show that it reduces to 4q=3r. Instead of answering the questions, unless you would like to, can someone please clarify how to express rates of work by algebra, like in this question? Thanks April 13th, 2018, 05:05 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,112 Thanks: 1002 Make up a simple example; as example: 5 workers working for 8 hours construct 400 items. So q=5, p=8 and r=400. Which means each worker constructs 400/(5*8 ) = 10 items per hour. Now what happens if (as example) 2 more workers are hired? 7 workers @ 10 items per hour = 70 items per hour. Play with that !! Thanks from topsquark and Student2018 April 22nd, 2018, 12:30 AM   #3
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Quote:
 Originally Posted by Student2018 One day, it takes p hours for q factory workers to construct r items. The following day three more worked join the team and four more items are required to be constructed.
If it takes "p hours for q factory workers to construct r items" then they, altogether, are working at rate r/p "items per hour". If there are q workers, they each have (average) rate (r/p)/q= r/(pq) "items per hour per worker". Adding three more workers, assuming they work at the same rate, they work, together, at rate (r/(pq))(q+ 3)= r(q+3)/(pq) "items per hour".

Quote:
 1)Find how many hours it will take, in terms of p,q and r.
Working at r(q+3)/(pq) "items per hour", to construct "four more items", that is, r+ 4 items, requires (r+ 4)/(r(q+3)/(pq)) (items divided by items per hour)= (pqr)/(r(q+3)) hours.

Quote:
 2)If in fact it takes the same length of time, find an equation and show that it reduces to 4q=3r.
The length of time, initially, was p hours. After adding 3 workers, but requiring 4 more items it took (pqr)/(r(q+3)) hours. If those are the same, p= (pqr)/(r(q+3)) . What does that reduce to?

Quote:
 Instead of answering the questions, unless you would like to, can someone please clarify how to express rates of work by algebra, like in this question? Thanks
If a person (or machine) constructs "r items" in "q hours" then they work at a rate of r/q "items per hour. If people work together then their rates add. Here, adding p works at the same rates was the same as multiplying by p,

Last edited by Country Boy; April 22nd, 2018 at 12:34 AM. April 22nd, 2018, 06:59 AM   #4
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 Originally Posted by Country Boy If it takes "p hours for q factory workers to construct r items" then they, altogether, are working at rate r/p "items per hour". If there are q workers, they each have (average) rate (r/p)/q= r/(pq) "items per hour per worker". Adding three more workers, assuming they work at the same rate, they work, together, at rate (r/(pq))(q+ 3)= r(q+3)/(pq) "items per hour". Working at r(q+3)/(pq) "items per hour", to construct "four more items", that is, r+ 4 items, requires (r+ 4)/(r(q+3)/(pq)) (items divided by items per hour)= (pqr)/(r(q+3)) hours. The length of time, initially, was p hours. After adding 3 workers, but requiring 4 more items it took (pqr)/(r(q+3)) hours. If those are the same, p= (pqr)/(r(q+3)) . What does that reduce to? If a person (or machine) constructs "r items" in "q hours" then they work at a rate of r/q "items per hour. If people work together then their rates add. Here, adding p works at the same rates was the same as multiplying by p,
I looked at this for a few minutes wondering when my mind had finally snapped because

$p = \dfrac{ p q r}{r(q + 3)} \implies 1 = \dfrac{qr}{r(q + 3)} \implies 1 = \dfrac{q}{q + 3} \implies q + 3 = q \implies 3 = 0.$

It shows how greatly I respect Country Boy that I literally checked my work on that twice. What was going on? I had been following right along, hadn't I? Not carefully enough it turned out.

$\dfrac{r + 4}{\dfrac{r(q + 3)}{pq}} = \dfrac{r + 4}{1} * \dfrac{pq}{r(q + 3)} = \dfrac{pq(r + 4)}{r(q + 3)} \ne \dfrac{pqr}{r(q + 3)}.$

$\therefore p = \dfrac{pq(r + 4)}{r(q + 3)} \implies 1 = \dfrac{qr + 4q}{qr + 3r} \implies qr + 3r = qr + 4q \implies 3r = 4q.$ April 22nd, 2018, 08:26 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,112 Thanks: 1002 Jeffroi, dans le coin pour 12 minutes.... April 22nd, 2018, 08:47 AM   #6
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 Originally Posted by Denis Jeffroi, dans le coin pour 12 minutes....
That will make miss my bus. How about 8 minutes? April 22nd, 2018, 08:53 AM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,112 Thanks: 1002 Oh hokay; but finish off the 4 minutes when you come back.... Tags algebra, rates, working Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post srecko Calculus 5 December 13th, 2016 08:00 PM JamesBrady Algebra 6 May 18th, 2015 08:52 PM Lambin Algebra 2 January 3rd, 2013 05:38 PM

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