Algebra Mixture problems?

A chemist has one solution which is 30% pure acid and another solution which is 60% pure acid. how many liters of each solution must be used to produce 60 liters of a solution which is 50% pure acid?

when I did the equation (.30x + .60(60)=.5(x+60)) I got that x equals 30 but I didn't know how to find the amount of the other pure acid

Comments

  • Your equation is set up wrong.

    Suppose that you have x liters of 30% pure acid. In order for the resulting solution to have 60 liters, the rest of the solution must consist of 60% acid, so there are 60 - x liters of 60% acid.

    The total solution has 60 liters of solution, so your equation should be:

    0.30x + 0.60(60 - x) = 0.5(60),

    which solves to yield x = 20. Therefore, x = 20 liters of 30% pure acid and 60 - x = 40 liters of 60% issue should be used.

    I hope this helps!

  • 30x + 60y = 50 * 60

    x + y = 60

    30 * (x + 2y) = 50 * 30 * 2

    x + 2y = 100

    x + 2y - x - y = 100 - 60

    y = 40

    x + y = 60

    x + 40 = 60

    x = 20

    20 liters of 30%

    40 liters of 60%

  • Use a table.

    type ----- % as dec ----- amount ----- pure

    30% ------- .30 ------------- x ------------- .30x

    60% ------- .60 ----------- 60 - x --------- .60(60 - x)

    50% ------- .50 ------------ 60 ------------ .50 * 60

    .30x + .60(60 - x) = .50 * 60

    3x + 360 - 6x = 300

    3x = 60

    x = 20

    20L of 30% solution plus 40L of 60% solution gives 60L of 50% solution

    Note-

    Your equation was using 60L of the 60% solution, not 60L of the 50% solution. The total in your problem was 90L of the 50% solution, with 30L of the 30% solution.

    Using a table can help to avoid mistakes like that in the future. It helps to keep things straight.

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