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Simple linear algebra problem

  1. Oct 26, 2009 #1
    Hi All!

    I hope i posted this in the correct forum. Technically its not homework, just a problem from a puzzle-book which i enjoyed very much until i got to this problem.


    1. The problem statement, all variables and given/known data

    Calvin Carpenter needed screws of three different sizes. He bought the same number of screws of each size at $.03, $.04, and $.05 per screw respectively.
    Wendy Woodall spent the same amount of money as Calvin, but divided her money equally among screws of the three sizes. In all, Wendy purchased six more screws than Calvin did. How much money did they each spend?

    2. Relevant equations

    3. The attempt at a solution

    Calvin gets the same number of screws of each size:

    3x+4x+5x=k (cents, x=number of screws, k=total price in cents)

    Wendy gets screws for the same price each size:

    3a+4b+5c=k
    3a=4b=5c

    I could mess around some more, but i just dont get it.
    Im trying to understand this since more than a week and every day i feel dumber. If anyone could explain this to me i would be very gratefull.

    thanks in advance,
    schmolch
     
  2. jcsd
  3. Oct 26, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    i think its in your definition of how wendy spends, but were close

    first for Calvin as you wrote
    3x+4x+5x= 12x = k

    now wendy spends her money evenly but divides it equally, as there are three types of screws, each one gets k/3 spent on it

    so if she buy a,b,c of respectively

    then
    3a = 4b = 5c = k/3

    so now knowing k, give a,b,c & x

    however there is one unused fact, wendy got 6 more screws than Calvin...
     
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