SOLUTION: A lab assistant needs 15L of a 30% alcohol solution. She must mix a 15% solution and a 40% solution. How may liters of each solution will be required to create the final mixture?

Question 807654: A lab assistant needs 15L of a 30% alcohol solution. She must mix a 15% solution and a 40% solution. How may liters of each solution will be required to create the final mixture?
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%
---
e+ f = 8 (total solution)
8e + 56f = 20*8 (total saline)
---
e+ f = 8
e + 7f = 20
------------ Subtract
-6f = -12
f = 2
e = 6
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