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Homework Help: How much of a 5 gallon 40% salt solution should be replaced

  1. Jun 21, 2016 #1
    1. The problem statement, all variables and given/known data

    Here is the problem I can't solve, and will be grateful for your help on this - please, guid me to the understanding of how to solve such problems. The difficulty lies in replacing some of the solution with water. It would be much easier to solve if we add water.

    How much of a 5 gallon 40% salt solution should be replaced with pure water to obtain 5

    gallons of a 15% solution?

    I have to solve it by creating a system of linear equations.

    2. Relevant equations
    3. The attempt at a solution

    Here are my initial thoughts, but I couldn't go beyond:

    If we have 5 gallons of 40% salt solution, hence we have 2 gallons of salt in 5 gallons.
    The final solution should contain 0.75 gallons of salt (5 x 0.15); therefore, I have to replace X amount of 5 gallons of 40% solution with Y amount of water to get 5 gallons of 15% solution.

    X + Y = 5
    Thank you!
  2. jcsd
  3. Jun 21, 2016 #2

    Ray Vickson

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    Homework Helper

    You need to replace X gallons or 40% solution with X gallons of pure water, to keep the volume at 5 gallons.
  4. Jun 21, 2016 #3


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    You already figured out how much salt you want.
    What volume of 40% solution would give you that much salt?
    Keep that much in there, and fill the rest up with pure water.
  5. Jun 22, 2016 #4
    Oh ) Thank you very much! Had a "writer's block" yesterday )
    0.4x = 0.75
    x = 1.875 the amount of 40% solution in the final 15% solution
    5 - 1.875 = 3.125
    I don't see, though, why would I need to construct a system of linear equations, and then turn them into a matrix to solve with Cramer's rule
  6. Jun 22, 2016 #5


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    If you were going to build a system of equations, you would do something like:
    ## \frac{0.40x + 0y }{x+y} = 0.15 \\ x+y = 5, ##
    where x is the volume of 40% solution and y is the volume of pure water.
    To write this as a matrix equation, it would look like:
    ##\begin{bmatrix} 0.4& 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} .75\\ 5 \end{bmatrix}##
    However, this is a very simple system and you were able to solve it without using Cramer's rule.
    The simplified system works out to exactly what you did above:
    ##0.40 x = .75\\ x+y = 5.##
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