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Thread: Mixture problem

  1. #1
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    Mixture problem

    Q) A mixture of 125 galllons of wine and water contains 20% of water.How much water must be added to the mixture in order to increase the percentage of the water to 25% of the new mixture?

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  2. #2
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    Quote Originally Posted by a69356 View Post
    Q) A mixture of 125 galllons of wine and water contains 20% of water.How much water must be added to the mixture in order to increase the percentage of the water to 25% of the new mixture?

    Thanks
    1. The actual mixture contains $\displaystyle 0.2\cdot 125 = 25$ gallons of water.

    2. Let x denote the additional water then you know:

    $\displaystyle \dfrac{25+x}{125+x}=0.25$

    3. Solve for x.

    Spoiler:
    I've got $\displaystyle x = \dfrac{25}3\ gallons\ of\ water$
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  3. #3
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    Hello, a69356!

    A mixture of 125 galllons of wine and water contains 20% water.
    How much water must be added to increase the percentage of the water to 25%?
    When I first learned "Mixture Problems", I made a chart . . .


    The chart has the formula across the top:
    . . $\displaystyle \text{[Gallons of mixture]} \times \text{[Percent water]} \:=\:\text{[Gallons of water]}$

    . . $\displaystyle \begin{array}{c||ccccc}
    \qquad & \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline
    & &|& &|& \\ \hline
    & &|& &|& \\ \hline \hline
    & &|& &|& \end{array}$


    We start with 125 gallons of mixture which is 20% water.
    . . We have: .$\displaystyle 20\% \times 125 \:=\:25$ gallons of water.
    Write this in the first row:

    . . $\displaystyle \begin{array}{c||ccccc}
    & \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline
    \text{Start} & 125 &|& 20\% &|& 25 \\ \hline
    & &|& &|& \\ \hline \hline
    & &|& &|& \end{array}$


    We add $\displaystyle x$ gallons of water (which is 100% water).
    . . We add: .$\displaystyle 100\% \times x \:=\:x$ gallons of water.
    Write this in the second row.

    . . $\displaystyle \begin{array}{c||ccccc}
    & \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline
    \text{Start} & 125 &|& 20\% &|& 25 \\ \hline
    \text{Add} & x &|& 100\% &|& x \\ \hline \hline
    & &|& &|& \end{array}$


    The final mixture has $\displaystyle x+125$ gallons which is 25% water.
    . . It contains: .$\displaystyle 25\% \times (x+125) \:=\:0.25(x+125)$ gallons of water.
    Write this in the third row.

    . . $\displaystyle \begin{array}{c||ccccc}
    & \text{Gallons} & \times & \text{Percent} & = & \text{Water} \\ \hline\hline
    \text{Start} & 125 &|& 20\% &|& 25 \\ \hline
    \text{Add} & x &|& 100\% &|& x \\ \hline \hline
    \text{Mixture} & x+125 &|& 25\% &|& 0.25(x+125) \end{array}$


    Read down the last column.

    We start with 25 gallons of water, then add $\displaystyle x$ gallons of water.
    . . Hence, the mixture contains $\displaystyle x+25$ gallons of water. [1]


    Read across the last row.

    We have $\displaystyle x+125$ gallons which is 25% water.
    . . Hence, the mixture contains $\displaystyle 0.25(x+125)$ gallons of water. [2]


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

    There is our equation! $\displaystyle \hdots\quad x + 25 \:=\:0.25(x+125)$

    Got it?

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