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Need Help to understand a simple way to solve this problem

  1. Feb 23, 2012 #1


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    If Joe produce 16 A everytime he produce 30 K , receives 7 additionnal A everytime he produce 40 K and receive 2 additionnal K everytime he produce 30 A , how many A will he have in total when he's gonna have 180 K?

    More clearly:

    16 A every 30 K
    +7 A every 40 K
    + 2 k every 30 A

    Finding the answer ain't that difficult , but I have a hard time putting it in one simple formula.Anyone can help to built such a formula?

    (sorry for my poor english , it is not my first language , and sorry if this is really easy , I might just be rusty , been a while )

    thanks !
  2. jcsd
  3. Feb 23, 2012 #2


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    also , I am not an academic but just a mathematic enthousiast , what academic level is this kind of problem on?

    I really don't know whether or not it's suppose to be easy or hard hehe
  4. Feb 23, 2012 #3


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    Hey reenmachine and welcome to the forums.

    For your problem its clear that we want to get a formula for A in terms of K.

    Now the thing that makes this a little 'tricky' is that we get some more 'K' for every 'A' we produce. This means that it's not completely 'explicit' in the way that you just get a simple A = f(K) because of this.

    So the first thing we have to do is take into account the 'A produces 2K' statement. Once this is done the rest is pretty straightforward.

    The function itself will be non-linear and use what is called floor functions.

    Basically for the '2K for every 30A' part you will add 2xFloor(A/30) 'K'. This accounts for the extra K but then since A depends on K you have to take this into account:

    Now without this factor we get the simple equation:

    F(K) = A = 16xFloor(K/30) + 7xFLOOR(K/40)

    Now adding the other factor we get

    A = 16xFloor([K+2xFloor(30/A)]/30) + 7xFloor([K+2xFloor(30/A)]/40)

    So as you can see this is an implicit equation since the 'A' term is embedded in the floor functions which makes this really really complex.
  5. Feb 23, 2012 #4


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    This is basically a really really hard implicit equation.

    Also I forgot to mention if you are only considering whole numbers for A then you need to add a Floor function around the entire thing which forces it to give 'whole number answers'.
  6. Feb 23, 2012 #5


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    I don't even think I know what floor functions are , but again I'm a french speaker first.I'll look into it and try to self-teach if it's possible.

    thanks a lot again really appreciated !!!!!
    Last edited: Feb 23, 2012
  7. Feb 23, 2012 #6


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    Think of the floor function as a 'step-case' function.

    Basically the floor function returns the 'integer' part of the number.

    So Floor(3.0) = 3, Floor(3.1) = 3, Floor(3.99999999999999999999) = 3 but Floor(4.00000001) = 4.

    You can visualize it as a step-case where the next 'step' occurs at every integer value.

    It's not an easy function by any means so keep that in mind.
  8. Feb 26, 2012 #7
    Where did you find this problem?
  9. Feb 26, 2012 #8


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    I created it myself randomly , I should've simply have said 8/15 instead of 16/30 but I just threw random numbers on the paper and forgot to simplify the ratio.Then I thought about a way to complicate things and I came up with these additionnal As and Ks , and then got angry because I failed to solve it within 2 hours. ( in fact I was capable of knowing how many As I would have at 180K , but couldn't do it without writing everything down instead of a simple equation )

    Keep in mind I'm not an academic at all , never entered university ( will enter soon at 29 years old , dropped out when I was younger but was always particularly good/interested at math , but I guess finding a simple equation for this problem was just over my head )

    pardon my poor english again , I'm a french speaker from canada.

    The problem also might be unclear the way I presented it , here's another presentation:

    Joe has a machine producing As and Ks at the same time , starting at zero_Once the machine will start , it will never stop producing until the end of time.The machine is more effective at producing Ks , since it produce 8A for 15K ( 8/15 ).

    7 additionnals As are falling from the sky as a gift everytime there's 40K produced/received as a gift.
    2 additionnals Ks are falling from the sky as a gift everytime there's 30A produced/received as a gift.

    How can you explain the effects of the gifts on each others?

    ( might be clearer , but then again I'm not a math teacher )
    ( btw a math ph.d told me chiro gave me the good answer , I was busy and didn't have time to try and understand the effect of the floor functions , but will try to understand it , so the fact I re-wrote the problem was not because I didn't like chiro's answers , I just like to be accurate :) )
    Last edited: Feb 26, 2012
  10. Feb 26, 2012 #9
    ^ I found your original post clear enough, just had no idea of how to tackle it! Don't really understand chiro's method yet...is this at a high school competition level?
  11. Feb 26, 2012 #10


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    I have no idea , but I'm pretty sure this problem is way too hard for highschool level.Might even be too hard for couple of first years in university , but since I never studied at university I can't really tell.Someone with a ph.d could answer this question better than me.
    Last edited: Feb 26, 2012
  12. Feb 27, 2012 #11
    I thought I'd have a go at answering this; please correct me if I'm wrong!


    A = 8xFloor([K+2xFloor(A/30)]/15)+7xFloor([K+2xFloor(A/30)]/40)

    K = 15xFloor([A+7xFloor(K/40)]/8)+2xFloor([A+7xFloor(K/40)]/30)
    Hope this helps
    Last edited: Feb 27, 2012
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