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Math Help - Solving for an unknown variable

  1. #1
    Junior Member
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    Solving for an unknown variable

    Hi

    How best can I approach this question?

    A solution containing 45% water is mixed with another solution containing 10% bromine. If the chemist needs 30 liters of the mixture, how many liters of each solution are needed to make a 34.5% water/bromine solution?

    Let x= water and y= the salt solution

    "If the chemist needs 30 liters of the mixture"
    x + y = 30 ...is that right?
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  2. #2
    MHF Contributor
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    Re: Solving for an unknown variable

    Quote Originally Posted by KayPee View Post
    A solution containing 45% water is mixed with another solution containing 10% bromine. If the chemist needs 30 liters of the mixture, how many liters of each solution are needed to make a 34.5% water/bromine solution?
    Let x= water and y= the salt solution
    "If the chemist needs 30 liters of the mixture"
    x + y = 30 ...is that right?
    Easier this way: let w = water ; then bromine = 30 - w
    Code:
       w   @  .45
    30-w   @  .10
    ====      ====
      30   @  .345
    Can you turn that into an equation, then solve for w?
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