SOLUTION: How many gallons of a 5% solution should be mixed with a 10% solution to obtain 50 gallons of a mixture that is 8%?

Question 515503: How many gallons of a 5% solution should be mixed with a 10% solution to obtain 50 gallons of a mixture that is 8%?
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
You need to keep track of how much 'pure' stuff is involved.
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8% * 50 gallons = .08*50 = 4 gal of 'pure' stuff
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x = gallons of 5% solution
50-x = gallons of 10% solution
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5%*x + 10%*(50-x) = 4
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.05x + .10(50-x) = 4
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multiply by 100 to eliminate decimals.
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5x + 10(50-x) = 400
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5x +500 -10x = 400
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-5x = -100
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x = 20 gal of 5% solution
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50-x = 30 gal of 10% solution
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Check to be sure it is 50 gal of 8% solution.
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5% * 20 = 1 gal
10% * 30 = 3 gal
1+3 = 4 gal, which is what we need.
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Answer: Mix 20 gal. of 5% solution and 30 gal. of 10% solution to make 50 gal. of 8% solution.
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Done.
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