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How to set up a rate problem with two rates

  1. Sep 14, 2009 #1
    Ok. I am helping my cousin do word problems but I am having trouble explaining the method or principles behind why I set up the equations the way I do.

    1. The problem statement, all variables and given/known data
    Terence and Mao are mowing a lawn together. Terence can mow 3 of these lawns in 5 hours. Matt can mow 2 of these lawns in 3 hours. Assuming they work independently, how long does i take them to mow the lawn together?

    I let T = the time it takes in hours to mow the lawn with both people, what we are solving for.

    I let A = Area of the lawn.

    R_T = 3 lawns / 5 hrs = 3A / 5 hrs
    R_M = 2A / 3 hrs

    2. Relevant equations
    I set up the equation:

    A = (R_T + R_M)*T and solve for T.

    So A = (3/5A + 2/3A)*T = (19/15 * T)A

    --> T = 15/19 hrs

    3. The attempt at a solution

    I am having trouble explaining why we set up the rates as lawns per hour instead of hours per lawn.

    I know it will yield the wrong answer if we try to solve T = (R_T + R_M)*A where our rates are in hours per lawn because it doesn't really make sense to add the rates that way. Of course I can show 3/4 + 1/2 != 4/3 + 2/1 and that when we add speeds we add them time on the bottom distance on the top, but is there a reason behind this?

    Is there a theoretical or generalized reason why we do this?

    In many other word problems it is OK to invert the ratios or rates.. I suspect we run into trouble here and have to do it the specific, right way (time on the denominator) because the area being mowed by Terence depends on the are being mowed by Mao.

    Can someone give me any insight?

    If this is more suited for the math general forum please move it there.
     
  2. jcsd
  3. Sep 14, 2009 #2

    symbolipoint

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    You can set up the rate any way you like as long as you use the dimensioning consistantly.

    When both mowers work together, the rates are additive (uniform rates exercise).
    If you choose "Lawns per Hour", then your fundamental rule is, for R = the rate, T= time (in this case in hours), J= number of jobs, in this case mowing 1 lawn, then R*T=J.

    What is most convenient to help track you information is to arrange a table: The rows are "Terence", "Matt", and "Together". The columns are Rate, Time, Jobs. Fill-in each piece of information that you know from the problem description, assign variables for what you do not know or what you want to find; trick is, some information you fill-in should be expressions based on the given information. Set up appropriate equation or equations. This varies depending on what is asked, what is given, what is not given. You will usually develop a fractional equation. Most of the rest is fairly simple algebra.

    Note that in your example, Area is not important. The count of jobs is important.
     
    Last edited: Sep 14, 2009
  4. Sep 15, 2009 #3
    I know how to solve it and explain it when we are using uniform rates and adding them together.

    I just don't quite know how to explain it if we choose to use the inverted rates, or hours it takes per a lawn mowed.

    I know now the method to combine these inverted rates is to take one rate, find out how long it takes to do a certain task (like mow 1 lawn and say it takes that person 1 hour to do so), and then find out how many of those tasks are completed by the other rate in that time (say the other person is slower and mows 0.5 lawns in the 1 hour).

    Then we see that in 1 hour 1.5 lawns were mowed and that is our combined inverted rate, 1 hour / 1.5 lawns.

    For the regular rates it would have been 1 lawn / 1 hr + 0.5 lawns / 1 hr = 1.5 lawns / 1 hour which is just the inverse of our combined inverted rate.

    I am trying to explain the how and why and the intuition behind the method of combining inverted rates.

    When do we do it by jobs and not just adding uniform rates?

    What if we don't have our rates or inverted rates involving time?

    Is there any rule of thumb?

    When are rates additive and when are they not and what is the logic behind this?
     
  5. Sep 15, 2009 #4

    symbolipoint

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    As simply as might be said, rates are additive when the agents performing the task are working on the same job at the same time.

    An exercise must have enough information given to permit filling in all other important or relevant expressions to continue analyzing and then solve the problem. A rate, ANY RATE, is a comparison between two numbers. Further complication is often not necessary. Some exercises will directly give a rate, and other exercises will give two numbers which are combinable to constitute a rate. Still other exercises might give information which relate time or jobs to other time or jobs, and expressions for time, jobs, or rates may sensibly be found. The only frequent "rule of thumb" is the simple rate equation, R*T=J (Rate times time equals the number of jobs, appropriate units expected).
     
  6. Sep 15, 2009 #5
    R*T = J works even if R = 1 hr / 5 lawns?

    So J is in some sort of squared time over lawns or some other unit?

    I am talking about inverted rates, not rates where where some sort of unit is over time. That is easy to explain how to combine.

    An example:

    Let T = the time it takes for both people to mow the lawn, A = Area of the lawn.
    R_T = 3 lawns / 5 hours or 3A/5hr = (3/5)(A/hr). R_M = 2 lawns / 3
    hours or 2A/3hr = (2/3)(A/hr).

    Our combined rate R* = (3/5 + 2/3) = 19 lawns / 15 hours

    Let T, A be the same.
    S_T = 5hrs/3A = (5/3)(hr/A). S_M = (3/2)(hr/A).

    Terence mows 1 lawn in 5/3 hours: (5/3)(hr/A) * A = 5/3hr.

    In 5/3hr how much does Mao mow? we want the answer in terms of A. So
    we do (5/3)hr / (3/2)(hr/A) = 10/9 * A.

    So Mao mows 10/9A in 5/3 hour while Terence mows A = 9/9A in 5/3 hour.
    So our S* = 5/3 hr / 19/9*A = 5/3hr * 9/19(1/A) = 5/1hr * 3/19(1/A) =
    15 hr / 19 A which is the inverse of R*.

    Are you just saying we should always write rates the standard way (something over time) and create a table and write relevant equations and solve the standard way?

    My question is what happens if I choose HOURS PER LAWN. Rates are then not additive. Why not?

    Can you please read this post and my examples and respond to it. I don't really disagree with what you have said it just seems like you did not even read my problem or just told me to choose LAWNS per HOUR and not bother with HOURS per LAWN.
     
    Last edited: Sep 15, 2009
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