SOLUTION: a chemist has 20% and 60% solutions of acid available. how many liters of each solution should be mixed to obtain 10 liters of a 30% acid solution
Question 887111: a chemist has 20% and 60% solutions of acid available. how many liters of each solution should be mixed to obtain 10 liters of a 30% acid solution Answer by Alan3354(69443) (Show Source):
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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
