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Math Help - Mixture math problem!

  1. #1
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    Mixture math problem!

    Here is the problem:

    Jane is asked to make a 37 milliliter solution of 12% HCL. However, only 8% and 15% HCL is available in the lab. How many milliliters of each solution should be use?

    I started out by setting up the variables like this, but then I don't know what to do afterwards. Here is how I set up the variables.

    Let x=number of ml of 8% HCl
    y=number of ml of 15% HCL


    Any help on how to solve this would be great appreciated!
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  2. #2
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    8%*x+15%*y=12%*(x+y) and x+y=37

    solve:x(ml)=111/7,y(ml)=148/7

    over

    Quote Originally Posted by AceComical View Post
    Here is the problem:

    Jane is asked to make a 37 milliliter solution of 12% HCL. However, only 8% and 15% HCL is available in the lab. How many milliliters of each solution should be use?

    I started out by setting up the variables like this, but then I don't know what to do afterwards. Here is how I set up the variables.

    Let x=number of ml of 8% HCl
    y=number of ml of 15% HCL


    Any help on how to solve this would be great appreciated!
    Follow Math Help Forum on Facebook and Google+

  3. #3
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    How did you get solve:x(ml)=111/7,y(ml)=148/7?
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  4. #4
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    Hello, AceComical!

    Jane is asked to make a 37 milliliter solution of 12% HCL.
    However, only 8% and 15% HCL is available in the lab.
    How many milliliters of each solution should be use?


    I started out by setting up the variables like this,
    but then I don't know what to do afterwards.
    Here is how I set up the variables.

    Let x = number of ml of 8% HCl
    Let y = number of ml of 15% HCL . . . . Good!

    We know that the total solution is 37 ml: . x + y \:=\:37\;\;{\color{blue}[1]}


    Next, we have to "talk" our way through the problem . . .

    We have x ml of 8% acid.
    . . It contains:. 0.08x ml of acid.

    We have y ml of 15% acid.
    . . It contains: . 0.15y ml of acid.

    Together, they contain: . {\color{blue}0.08x + 0.15y} ml of acid.


    But we are told that the final mixture is 37 ml which is 12% acid.
    . . It contains: . 0.12(37) \:=\:{\color{blue}4.4} ml of acid.


    We just described the final amount of acid in two ways.

    There is our equation: . 0.08x + 0.15y \:=\:4.4\;\;{\color{blue}[2]}


    We have: . \begin{array}{cccc}x + y &=& 37 & {\color{blue}[1]} \\ 0.08x + 0.15y &=& 4.4 & {\color{blue}[2]} \end{array}


    \begin{array}{ccccc}\text{Multiply {\color{blue}[1]} by 15:} & 15x + 15y &=& 555 \\ \text{Multiply {\color{blue}[2]} by 100:} & 8x + 15y &=& 440 \end{array}

    Subtract: . 7x \:=\:115 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{115}{7}}

    Substitute into [1]: . \frac{115}{7} + y \:=\:37 \quad\Rightarrow\quad\boxed{ y \:=\:\frac{144}{7}}

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  5. #5
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    Awesome, thanks for the help wood and Soroban!
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