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Math Help - Mixture problem - end up with a negative answer???

  1. #1
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    Mixture problem - end up with a negative answer???

    One alloy is one part copper to three parts tin. A second alloy is one part copper to four parts tin. How much of a second alloy shold be combined with 24 pounds of the first allow to obtain a new alloy that is two parts copper to seven parts tin?
    I made this chart:

    Code:
                Amount (lbs)  % Copper  Total Copper
    Alloy 1   24                .25           .25(24)
    Alloy 2   x                  .20           .20x
    Mix        24+x             2/7           2/7(24+x)
    And then got:

    .25(24) + .20x = .29(24+x)
    6 + .20x = 6.96 + .29x
    -.96 = .09x
    -10.67 = x


    Where did I go wrong?
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  2. #2
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    Quote Originally Posted by cdbmath View Post
    I made this chart:

    Code:
                Amount (lbs)  % Copper  Total Copper
    Alloy 1   24                .25           .25(24)
    Alloy 2   x                  .20           .20x
    Mix        24+x             2/7           2/7(24+x)
    And then got:

    .25(24) + .20x = .29(24+x) <<<<<< here
    6 + .20x = 6.96 + .29x
    -.96 = .09x
    -10.67 = x


    Where did I go wrong?
    The first line must read like this:

    .25(24) + .20x = \dfrac29 \cdot (24+x)

    And \dfrac29 \neq 0.29

    You should come out with x = 30
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