Question

How many gallons of pure water must be added to 500 gallons of a 40% saline solution to reduce it to 25% saline solution

Answers

  1. Answer:
    300 gallons
    Step-by-step explanation:
    A saline solution is a solution of salt in water.
    The percent is the percent of salt in the total amount of solution.
    A 40% saline solution has 40% of salt out of the total volume.
    Let the amount of pure water needed = x.
    The amount of existing 40% solution is 500 gal.
    Let the total amount of 25% saline solution made = y.
    Equation of volumes of solutions:
    500 + x = y
    y = x + 500          Eq. 1
    Equation of volume of salt:
    500 gal of 40% saline solution has 0.4 × 500 gal = 200 gal of salt
    Pure water has 0% salt.
    The final product is a solution that is 25% saline. Its volume is y.
    The volume of salt in the y gallons of 25% saline solution is 25% of y = 0.25y
    The equation of salt content is:
    200 + 0 = 0.25y
    y = 800       Eq. 2
    Eq. 1 and Eq. 2 form a system of equations.
    y = x + 500
    y = 800
    Substitute 800 for y in Eq. 1.
    y = x + 500
    800 = x + 500
    x = 300
    Answer: 300 gallons of pure water

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