SOLUTION: A 90% acid solution is diluted with water to make a solution of 60% acid. When liters of water is added to dilute it again, the solution becomes 40% acid. How much water is added t

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Question 788263: A 90% acid solution is diluted with water to make a solution of 60% acid. When liters of water is added to dilute it again, the solution becomes 40% acid. How much water is added to the solution on the first instance? How much 90% acid was available at the start?
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Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
You need more information. The description is incomplete.

Let f=how much volume of 90% acid
Let y=how much water to add to get 60% acid.

90%2Af%2F%28y%2Bf%29=60
90f=60%28y%2Bf%29
90f=60y%2B60f
9f-6f=6y
3f=6y
highlight%28y=f%2F2%29 liters of water

The second dilution now starts with (f+f/2) liters of 60% acid.
You want now to dilute to 40% acid.
Let x=volume of water to add for diluting to 40%.

60%2A%28f%2Bf%2F2%29%2F%28x%2Bf%2Bf%2F2%29=40
Solve for x.