SOLUTION: What quantity of a 55% acid solution must be mixed with a 25% solution to produce 300 mL of a 45% solution?

Question 905670: What quantity of a 55% acid solution must be mixed with a 25% solution to produce 300 mL of a 45% solution?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Do it like this one. Only the numbers are different.
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Amanda wants to make 8 gal. of a 20% saline solution by mixing together a 56% saline solution and a 8% saline solution. How much of each solution must she use?
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e = amount of 8%
f = amount of 56%
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e+ f = 8 (total solution)
8e + 56f = 20*8 (total saline)
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e+ f = 8
e + 7f = 20
------------ Subtract
-6f = -12
f = 2
e = 6
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