A container has 10 liters of a 20% acid solution. How many liters of a 50% solution must be added to make a 30% solution?
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Let x be the number of liters of the 50% solution that must be added.
The total amount of acid in the 10 liters of 20% solution is 10 liters * 20% = <<10*20*.01=2>>2 liters.
The total amount of acid in the x liters of the 50% solution is x * 50% = (x/2) liters.
The total amount of acid in the final 30% solution is (10 + x) liters * 30% = (10 + x)/10 * 3 liters.
The equation to solve is (10 + x)/10 * 3 = 2 + (x/2).
Multiplying both sides of the equation by 10 gives 3(10 + x) = 20 + 5x.
Expanding the equation gives 30 +3x = 20 + 5x.
Subtracting 3x from both sides gives 30 = 20 + 2x.
Subtracting 20 from both sides gives 10 = 2x.
Dividing both sides by 2 gives x = 5.
Therefore, 5 liters of the 50% solution must be added to make a 30% solution. Answer: \boxed{5}.
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