Algebra

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Word problem:

Radiator in a certain make needs to contain 50 liters of 40% antifreeze. The radiator now contains 50 liters of 20% antifreeze. How many liters of this solution must be drained and replaced with 100% antifreeze to get the desired strength?

  • Algebra -

    Let X be the number of liters that must be drained and replaced with 100% antifreeze.

    [0.2(50-X)+X]/50 = 0.4

    Solve that for X.

  • Algebra -

    You want to double the concentration?

    If you remove X liters, you remove .2X antifreeze, and .8X water.

    You have left in the radiator after draining X, 10-.2X liters of pure antifreeze. You need a total of 20 liters to get 40percent antifreeze.

    So you add 10+.2X of antifreeze.

    You drain out 10-2X of solution.

    X=10+.2X
    .8X=10
    x=12.5 liters

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