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Thread: Word problem method needed

  1. #1
    Jun 2012

    Word problem method needed

    Hi, I am looking for a method similar to age problems or rate, dist problems for getting a proportion of a mixture.

    In this example I have to mix coffee1 at $6.50/unit with cofee2 at $9.00/unit to give me $7.50/unit.

    I got the answer by doing it this way;

    (6.50 + 9.00 )/7.5 = x

    6.5/x + 9/x = 7.5

    So 6.5/x is my coffee1 component.

    Is this a accepted method or how would a diagram be used here? Anything that would make life easier will be good and will help remember what to do next time.

    Thanks for the help.
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  2. #2
    Super Member
    Jun 2012

    Re: Word problem method needed

    Quote Originally Posted by SVDM View Post

    (6.50 + 9.00 )/7.5 = x
    Ehh, I don't think that works. That equation yields x = 2.066..., but we don't want x (also, what does x represent?).

    A better way to solve it is let a and b be the number of units of coffee1 and coffee2, respectively. We take a weighted average:

    $\displaystyle \frac{6.5a + 9b}{a+b} = 7.5$ For the LHS, the numerator represents the total cost of the coffee, the denominator represents the total # of units.

    $\displaystyle 6.5a + 9b = 7.5a + 7.5b \Rightarrow 1.5b = a$. That means, for every 1 unit of b (coffee2), we need 1.5 units of a (coffee1) to stay at $7.50 a unit. We like integer ratios so I'll multiply by 2, for every 3 units of coffee1 we need 2 units of coffee2.
    Thanks from SVDM
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