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April 4th, 2012, 03:39 PM  #1 
Newbie Joined: Apr 2012 Posts: 14 Thanks: 0  Mixture Problem
What would my equation(s) be to solve this mixture story problem? How much 12% Na2SO4 and pure water must be mixed to produce 22.8 gallons of a mixture that is 7% sodium sulfate? I started with: Let x = the # of gallons of 12% Na2SO4 And Let y = the # of gallons of 100% water or should it be Let 22.8x = the # of gallons of 100% water needed? 
April 4th, 2012, 03:49 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Mixture Problem
I would start by finding the amount in gallons of pure sodium sulfate we need: Now we want to find how many gallons x of the 12% mixture contains that amount: So, we need 13.3 gallons of the 12% sodium sulfate and 22.8  13.3 = 9.5 gallons of water. We could also simply the process somewhat (combine the two steps into one) and state: and then the amount of water is 22.8  13.3 = 9.5. 
April 4th, 2012, 07:38 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Mixture Problem Hello, Kimmysmiles0! Quote: Since we are adding pure water, we will consider the amount of water at each stage.  

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