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  1. #1
    Member Jonboy's Avatar
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    Solution

    Man these are so tough to find the equation for me.

    Here's the problem, I just want a push in the right direction to find the equation:

    How many grams of sugar must be added to 60 grams of a solution that is 32% sugar to obtain a solution that is 60% sugar?
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by Jonboy View Post
    Man these are so tough to find the equation for me.

    Here's the problem, I just want a push in the right direction to find the equation:

    How many grams of sugar must be added to 60 grams of a solution that is 32% sugar to obtain a solution that is 60% sugar?
    Hello,

    1. calculate the amount of sugar in the solution
    2. add x grams of sugar

    3. solve for x:

    \frac{0.32 \cdot 60 +x}{60+x}= 0.6

    For confirmation only: x = 42 g
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  3. #3
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    Hello, Jonboy!

    I usually place these "Mixture Problems" in a chart.


    How many grams of sugar must be added to 60 grams
    of a solution that is 32% sugar to obtain a solution that is 60% sugar?

    Make a chart.
    Make one row for the Original solution, one for the Added solution, one for the Mixture.

    \begin{array}{cccccccc} & |& \text{Solution} & \times & \text{Percent} & = & \text{Sugar} &| \\ \hline \text{Original} &|&  &|&  &|&  &| \\ \text{Add} &|&  &|&  &|&  & | \\ \hline \text{Mixture} &|& &|& &|& &| <br />
\end{array}



    We have 60 grams of solution which is 32% sugar.
    . . It has: . 32\% \times 60 \:=\:0.32\times60 \:=\:19.2 grams of sugar.
    Write that in the first row.

    \begin{array}{cccccccc} & |& \text{Solution} & \times & \text{Percent} & = & \text{Sugar} &| \\ \hline \text{Original} &|& 60 &|& 32\% &|& 19,2 &| \\ \text{Add} &|& &|& &|&  & | \\ \hline \text{Mixture} &|& &|& &|& &| \end{array}



    We will add x grams of sugar (which is 100% sugar, of course).
    . . It has: . 100\% \times x \:=\:x grams of sugar.
    Write that in the second row.

    \begin{array}{cccccccc} & |& \text{Solution} & \times & \text{Percent} & = & \text{Sugar} &| \\ \hline \text{Original} &|& 60 &|& 32\% &|& 19,2 &| \\ \text{Add} &|& x &|& 100\% &|& x & | \\ \hline \text{Mixture} &|& &|& &|& &| \end{array}


    The Mixture will have x + 60 grams of solution which is 60% sugar.
    . . It has: . 0.60(x+60) grams of sugar.
    Write that in the third row.

    \begin{array}{cccccccc} & |& \text{Solution} & \times & \text{Percent} & = & \text{Sugar} &| \\ \hline \text{Original} &|& 60 &|& 32\% &|& 19,2 &| \\ \text{Add} &|& x &|& 100\% &|& x & | \\ \hline \text {Mixture} &|& x+60 &|& 60\% &|& 0.60(x+60) &| \end{array}


    The equation is waiting for us in the last column: . 19.2 + x \;=\;0.60(x + 60)

    . . . Got it?

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  4. #4
    Member Jonboy's Avatar
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    Thank you both for your input!
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