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  1. #1
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    word problem

    A saline solution is 20% salt. How many gallons of water must be added to dilute the mixture to 8 gals of a 15% saline solution?

    Usually with word problems I would try to make 2 equations, solve one for a variable, then substitue that variable in the other equation and then solve for the remaining variable. However, I have a hard time coming up with an equation for this problem because I feel like there's not enough informtation. Don't I at least need to know how many gallons were there at the beginning?
    Any help is appreciated...
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  2. #2
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    Quote Originally Posted by tami-chan87 View Post
    A saline solution is 20% salt. How many gallons of water must be added to dilute the mixture to 8 gals of a 15% saline solution?

    Usually with word problems I would try to make 2 equations, solve one for a variable, then substitue that variable in the other equation and then solve for the remaining variable. However, I have a hard time coming up with an equation for this problem because I feel like there's not enough informtation. Don't I at least need to know how many gallons were there at the beginning?
    Any help is appreciated...
    Since you don't know how much water you have in the beginning, the saline concentration is 20%x, where x is the original gallons of water.

    Pure water is 0% saline solution.

    Resulting mixture of 8 gallons is 15%(8) saline solution.

    So,

    .20x + .0x = .15(6)
    .2x = 1.2
    x = 6

    So, you would have to add 8-6 or 2 gallons of water.
    Last edited by masters; August 1st 2008 at 10:09 AM.
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