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[Algebra] Workload (rate) problems

  1. Feb 2, 2012 #1
    [Algebra] "Workload" (rate) problems

    Intuitively, I don't even understand how the basic Q = rt works.

    For a simpler problem, such as: Sarah takes 10 minutes to mow a single lawn and Jessica takes 15 minutes to mow that same lawn, apparently the algebra is:
    (1/10) + (1/15) = 1/t
    For the solution to how long it will take both to mow the same lawn if they do it together.

    I don't understand why it's adding the two efforts; though I understand that the '1' is that they can mow 1 lawn every number of minutes which is why it's a fraction. Isn't adding them just increasing the total time it takes to mow the lawn, by adding both Sarah's time to mow one lawn and Jessica's time to mow one lawn while doing it at the same time?

    So back to the original question, I'll attempt to just plug in numbers without even understanding why and how the formula even works:

    A (Andy) = 75/20
    B (Ben) = 100/30
    C = 75/25
    D = 70/15

    So I guess the first step is to subtract the square feet that Andy already mowed at his pace for 5 minutes:

    500/t - [(75/20)*5]
    500/t - 75/4

    Actually, I'm not even sure. My head hurts right now so I'll probably attempt it tomorrow after I sleep, but I'll just leave this here for any hints...
  2. jcsd
  3. Feb 2, 2012 #2

    Ray Vickson

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    Re: [Algebra] "Workload" (rate) problems

    In your simpler example: Sarah mows 15 lawns in 150 minutes and Jessica mows 10 lawns in 150 minutes, so in 150 minutes they together mow 25 lawns. Their combined rate is 25/150 lawns/min, and 25/150 = 1/10 + 1/15. Of course, this assumes that if they both work on the same lawn they do not get in each other's way, etc.

  4. Feb 3, 2012 #3


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    Re: [Algebra] "Workload" (rate) problems

    Sarah mows 1/10 of a lawn in 1 minute.
    Jessica mows 1/15 of a lawn in 1 minute.

    Combined, they mow 1/10 + 1/15 of a lawn in 1 minute.

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