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  1. #1
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    MIxture word problem

    How would I go about setting this up?


    How many liters of a 30% alcohol solution must be mixed with 80 liters 0f a 90% solution to get a 40% solution ?

    Thanks
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  2. #2
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    One more problem I'm having trouble with:


    Walt made an extra $7000 last year from a part time job. He invested part of the money at 6% and the rest at 8%. He made a total of $500 in interest. How much was invested at 8%?
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  3. #3
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    Quote Originally Posted by Jerry99 View Post
    How would I go about setting this up?


    How many liters of a 30% alcohol solution must be mixed with 80 liters 0f a 90% solution to get a 40% solution ?

    Thanks
    30x + 90(80) = 40(x+80)

    solve for x
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  4. #4
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    Quote Originally Posted by Jerry99 View Post
    One more problem I'm having trouble with:


    Walt made an extra $7000 last year from a part time job. He invested part of the money at 6% and the rest at 8%. He made a total of $500 in interest. How much was invested at 8%?
    assuming simple interest ...

    0.06(7000-x) + 0.08(x) = 500

    solve for x
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  5. #5
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    Thanks man...
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  6. #6
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    Hello, Jerry99!

    \text{How many liters of a 30\% alcohol solution must be mixed}
    \text{with 80 liters 0f a 90\% solution to get a 40\% solution?}

    Here is one way to reason it out . . .


    We will use \,x liters of the 30% solution.
    . . It contains: . 0.30x liters of alcohol.

    We have 80 liters of the 90% solution.
    . . It contains: . 0.90 \times 80 \:=\:72 liters of alcohol.

    The mixture will have: . 0.30x + 72 liters of alcohol. .[1]


    But we know that the mixture will be x+80 liters which is 40% alcohol.
    . . The mixture will have: . 0.40(x+80) liters of alcohol. .[2]


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

    There is our equation! . . . 0.30x + 72 \;=\;0.40(x + 80)

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