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22 September, 23:54

A 10 L mixture is made up of 40% antifreeze. How much of the mixture needs to be removed and filled with pure antifreeze to reach a concentration of 60% antifreeze

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  1. 23 September, 03:32
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    3.3 liters of the mixture is needed to be removed and filled with 3.33 liters pure antifreeze to reach a concentration of 60% antifreeze.

    Explanation:

    A 10 L mixture is made up of 40% antifreeze.

    Initial Volume of antifreeze in 40% mixture = 40% of 10 L = 4L

    Let volume of pure antifreeze added = x

    Volume of antifreeze removed = 40% of x = 0.4 x

    Volume of antifreeze in 60% mixture = 60% of 10 L = 6 L

    Volume of antifreeze left after removal of 0.4 x L of antifreeze and addition of x L of pure antifreeze will be equal to 6 L of antifreeze in the final solution.

    4L - (0.4 x) + x = 6L

    x = 3.33 L

    3.3 liters of the mixture is needed to be removed and filled with 3.3 liter pure antifreeze to reach a concentration of 60% antifreeze.
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