Algebra 2

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a chemistry lab assistant wants to prepare 10 liters of 48% acid solution. in stock he has 80% and 40% acid solutions. how many liters of each in stock solution should he mix to prepare a 10-liter mixture?

  • Algebra 2 -

    L = no. of liters of 80% solution
    .80L = value of 80% solution
    10 - L = no. of liters of 40% solution
    .40(10 - L) = value of 40% solution
    .48(10) = value of mixture

    .80L + .40(10 - L) = .48(10)
    Solve for L

    L = no. of liters of 80% solution
    10 - L = no. of liters of 40% solution

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