SOLUTION: Karina needs 20 liters of a 51% alcohol solution but she has 10% and 90% solutions available. How many liters of the 10% and how many liters of the 90% solutions should she mix to

Question 1192669: Karina needs 20 liters of a 51% alcohol solution but she has 10% and 90% solutions available. How many liters of the 10% and how many liters of the 90% solutions should she mix to make the 51% solution?
Found 2 solutions by josgarithmetic, Boreal:
Answer by josgarithmetic(39616) (Show Source): You can put this solution on YOUR website!
Lesson Mixing two concentrations for known amount of mixture, https://www.algebra.com/my/Mixtures%3A-All-in-Symbols.lesson?content_action=show_dev

v of the 90%
20-v of the 10%

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Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
x liters of 10%=0.10x pure alcohol
20-x liters of 90%=.9(20-x)=18-.9x pure alcohol.
they add to 20*0.51 or 10.2 l pure alcohol.
so -0.8x+18=10.2
-0.8x=-7.8
x=9.75 l of 10% or 0.975 l. pure
20-x=10.25 l of 90% or 9.225 l. pure
They are very similar volumes and the midpoint between the percentages is 50%, so they should be and slightly more of the more concentrated since 51% is more than 50%.
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