SOLUTION: How many liters of a mixture containing 80% pure acid should be added to 5 liters of 20% solution to give a 30% solution?
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Question 1188904: How many liters of a mixture containing 80% pure acid should be added to 5 liters of 20% solution to give a 30% solution?
Found 2 solutions by VFBundy, greenestamps:
Answer by VFBundy(438) (Show Source): You can put this solution on YOUR website!
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Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
The formal algebraic solution from the other tutor is a standard way of solving mixture problems like this; you should understand it, because if is applicable to a wide variety of problems.
Here is an informal method for solving any "mixture" problem like this, if a formal algebraic solution is not required. While the numbers in this particular problem make a formal algebraic solution relatively easy, this informal method will often lead to a much faster and easier solution.
(1) You are starting with a 20% acid solution and adding some 80% acid solution, stopping when the mixture is 30% acid.
(2) From 20% to 80% is a difference of 60; from 20% to 30% is a difference of 10; so 30% is 10/60 = 1/6 of the distance from 20% to 80%.
(3) That means 1/6 of the final mixture is the 80% acid you are adding.
(4) So the 5 liters of the original 20% acid is 5/6 of the final mixture.
(5) Since 5 liters is 5/6 of the final mixture, the 1/6 of the final mixture you are adding is 1 liter.
ANSWER: 1 liter
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