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11 December, 05:54

To obtain 890 liters of a production line solution that contains 2 percent of chemical a a sample of a solution containing 12 percent of chemical a and another containing 1 percent of chemical a are mixed how many liters of the 12 percent sample are required

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  1. 11 December, 09:06
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    the question is of a mixture of solutions containing a certain chemical

    finally the mixture should contain 2 % of chemical A with a volume of 890 L

    solution A has a percentage of 12 %

    solution B has a percentage of 1 %

    x volume is taken from solution A

    volume taken from A and B should be equal to 890 L

    therefore if x volume is taken from A then (890 - x) volume should be taken from solution B

    percentage multiplied by volume added gives the amount of chemical A

    therefore we can write the following equation

    12 % * x + 1 % * (890 - x) = 2 % x 890

    0.12x + 8.9 - 0.01x = 17.8

    0.11x = 8.9

    x = 80.9 L

    therefore 80.9 L of 12 % solution is required
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