Question 277315
With 'solution' problems, you have to keep track of how much pure stuff you need.
This question is tricky because the question presents the complementary percentages.
Solution 1 is 70% water, so it is 30% pure acetic acid.  We normally call that a 30% solution.
Solution 2 is 30% water, so it is 70% pure acetic acid.  We normally call that a 70% solution.
The goal is 20 liters of a 60% acetic acid solution. 
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x = volume of 30% solution
20 - x = volume of 70% solution
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.3x = amount of pure acetic acid 
.7(20-x) = the other amount of pure acetic acid.
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20 liters of 60% acetic acid will have .6*20 = 12 liters of pure acetic acid.
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.3x + .7(20-x) = 12
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.3x + 14 - .7x = 12
-.4x = -2
x = -2/-.4 = 5 liters of 30% solution
which means there is
15 liters of 70% solution.
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checking to see if we have 12 L of pure acetic acid
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.3*5  + .7*15 = 1.5 + 10.5 = 12
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So, 5 liters of 30% acetic acid + 15 liters of 70% acetic acid results in 20 liters of 60% acetic acid solution.
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Reviewing the question, we need to give the complementary answer in terms of percentage of water.
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Answer:
5 liters of 70% water + 15 liters of 30% water = 20 liters of 40% water