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Math Help - Math Word Problem : Mixture. I don't get it. T-T

  1. #1
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    Math Word Problem : Mixture. I don't get it. T-T

    A radiator contains 25 quarts of a water and antifreeze solution of which 60% is antifreeze. How much of this solution should be drained and replaced with water for the new solution to be 40% antifreeze?

    Can somebody please explain how it's done?
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  2. #2
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    Quote Originally Posted by strigy View Post
    A radiator contains 25 quarts of a water and antifreeze solution of which 60% is antifreeze. How much of this solution should be drained and replaced with water for the new solution to be 40% antifreeze?

    Can somebody please explain how it's done?
    1. The actual mixture contains

    25\ quarts \cdot 0.6 = 15\ quarts of pure anti-freeze.

    2. The desired mixture should contain

    25\ quarts \cdot 0.4 = 10\ quarts of pure anti-freeze.

    3. That means you have to remove that amount of the actual mixture (call it x) which contains exactly 5 quarts of pure anti-freeze:

    x \cdot 0.6 = 5~\implies~x = \dfrac{25}3\ quarts
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  3. #3
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    I still don't get it. . Sorry. Thanks for the answer anyway.
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    Oh, wait scrap what I said. I think I get it now!!!!!!!! omg THANK YOU.

    Err..okay I have another one:

    A gardener has 26lb of a mixture of fertilizer and weed killer. If 1 lb of the mixture is replaced by weed killer the result is a mixture that is 5 % weed killer. What percent of the original mixture was weed killer?

    There doesn't need to be an answer, just steps on how to get there. Thanks!
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  5. #5
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    Hello, strigy!

    A gardener has 26lb of a mixture of fertilizer and weed killer.
    If 1 lb of the mixture is replaced by weed killer the result is a mixture that is 5% weed killer.
    What percent of the original mixture was weed killer?
    Let x = percent of weed killer (WK) in original mixture.

    He starts with 26 lbs of mix which is x\% weed killer.
    . . It contains: . {\color{blue}26x} lbs of WK.

    One pound of mix is removed.
    . . It contains: . {\color{blue}x} lbs of WK.

    One pound of pure (100%) WK is added.
    . . It contains: . {\color{blue}1} lb of WK.

    Hence, the final mix contains: . 26x - x + 1 \:=\:{\color{red}25x + 1} lbs of WK.


    But we know that the final mix will be 26 lbs which is 5% WK.
    . . So it contains: . (26)(0,05) \;=\;{\color{red}1.3} lbs of WK.


    We just described the amount of WK in the final mix in two ways.

    There is our equation! . . . \boxed{25x + 1 \;=\;1.3}


    Solve for x\!:\quad 25x \:=\:0.3 \quad\Rightarrow\quad x \:=\:\frac{0.3}{25} \:=\:0.012 \;=\;1.2\%

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