Question 626508: How many milliliters of a 31% alcohol solution must be added to 6 milliliters of a 11% alcohol solution to produce a 21% alcohol solution? Found 2 solutions by Alan3354, Maths68:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! How many milliliters of a 31% alcohol solution must be added to 6 milliliters of a 11% alcohol solution to produce a 21% alcohol solution?
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21% is the average, so it's equal amounts
6 ml
Solution B
Amount = 6 ml
Concentration = 11% = 0.11
Mixture
Amount = 6+x ml
Concentration = 21% = 0.21
(Amount of Solution A)*(Concentration of Solution A)+ (Amount of Solution B)*(Concentration of Solution B)=(Amount of Mixture)*(Concentration of Mixture)
(x)*(0.31)+ (6)*(0.11)=(6+x)*(0.21)
0.31x+0.66=1.26+0.21x
0.31x-0.21x=1.26-0.66
0.1x=0.6
0.1x/0.1=0.6/0.1
x=6
6ml of a 31% alcohol solution must be added to 6 ml of a 11% alcohol solution to produce 12ml of a 21% alcohol solution.
