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Math Help - Concentration

  1. #1
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    Concentration

    A 55 gallon barrel contains a mixture with a concentration of 30%. How much of this mixture must be withdrawn and replaced by a 100% concentrate to bring the mixture up to 50% concentration?

    I know that in the 55 gallon barrel with 30% concentration there is 16.5 gallons of solute and 38.5 gallons of solvent. Is that info useful in this problem? Iím not sure what to do from there...
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  2. #2
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    Let x=the number of gallons to be drained.

    Since there are 55 gallons in the original 30%, we can set it up like so:

    Original 30% less amount drained: .3(55-x)

    Pure amount added back: x

    New 50% mixture: .5(55)

    So, we have this to solve for x:

    .3(55-x)+x=.5(55)
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  3. #3
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    Hmm.. How did you know to multiply the .3 to the 55?
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  4. #4
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    Hello, Kitty!

    I'll show you my favorite approach to "Mixture Problems."
    For reference, assume the solution is water and acid.


    A 55-gallon barrel contains a mixture which is 30% acid.
    How much of this mixture must be withdrawn and replaced by 100% acid
    to bring the mixture up to 50% concentration?
    We start with 55 gallons which is 30% acid.
    . . The barrel contains: . 0.30 \times 55 \:=\: 16.5 gallons of acid.
    Write that in the first row of our chart . . .

    \begin{array}{c|c|c|c|}<br />
& \text{sol'n} & \% & \text{acid} \\ \hline<br />
\text{Start} & 55 & 30\% & 16.5  \end{array}


    We remove x gallons of mixture.
    . . This contains: . 30\%\times x \:=\:0.3x gallons of acid.
    Write this in the second row of our chart . . .

    \begin{array}{c|c|c|c|}<br />
& \text{sol'n} & \% & \text{acid} \\ \hline<br />
\text{Start} & 55 & 30\% & 16.5 \\<br />
\text{Remove} & \text{-}x & 30\% & \text{-}0.3x \\\end{array}


    We add x gallons of pure (100%) acid.
    . . This contains: . 100\% \times x \:=\:x gallons of acid.
    Write this in the third row of our chart . . .

    \begin{array}{c|c|c|c|}<br />
& \text{sol'n} & \% & \text{acid} \\ \hline<br />
\text{Start} & 55 & 30\% & 16.5 \\<br />
\text{Remove} & \text{-}x & 30\% & \text{-}0.3x \\<br />
\text{Add} & +x & 100\% & +x \end{array}


    We know that the final mixture is 55 gallons which is 50% acid.
    . . This contains: . 50\% \times 55 \:=\:27.5 gallons of acid.
    Write this in the last row of our chart . . .

    \begin{array}{c|c|c|c|}<br />
& \text{sol'n} & \% & \text{acid} \\ \hline<br />
\text{Start} & 55 & 30\% & 16.5 \\<br />
\text{Remove} & \text{-}x & 30\% & \text{-}0.3x \\<br />
\text{Add} & +x & 100\% & +x \\ \hline<br />
\text{Mixture} & 55 & 50\% & 27.5  \end{array}


    Look at the last (rightmost) column.
    It describes the amount of acid in the various stages.

    We start with 16.5 gallons of acid,
    . . we remove 0.3x gallons of acid,
    . . we add x gallons of acid,
    and we end up with 27.5 gallons of acid.


    And there is our equation! . . . . . {\color{blue}16.5 - 0.3x + x \:=\:27.5}

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  5. #5
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    WOW! That makes sense! Thanks for explaining it so well!!
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