Question 1018556: Three solutions contain a certain acid. The first contains 10% acid, the second 30%, and the third 50%. A chemist wishes to use all three solutions to obtain a 100-liter mixture containing 20% acid. If the chemist wants to use twice as much of the 50% solution as of the 30% solution, how many liters of each solution should be used?
L of the 10% solution
L of the 30% solution
L of the 50% solution
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x=10%, y=30%, z=50%
x+y+z=100
.10x+.30y+.50z=20, which is 100*0.20 or the amount of "pure acid".
But we know that z=2y
Therefore x+3y=100
and .10x +.30y+.50(2y)=20
multiply second by 10
x+3y+10y=200
x=100-3y
substitute back
100-3y+3y+10y=200
10y=100
y=10 liters
z=20 liters
x=70 liters
10%*70 liters is 7 liters of pure acid
30% of 10 liters is 3 liters
50% of 20 liters is 10 liters
They add to 20 liters
10%:70 liters
30%: 10 liters
50%: 20 liters
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