Question 1199752: How many mL's of a 10% alcohol solution and how many mL's of a 20% solution must be mixed to obtain 30 mL's of a 16% solution? Found 2 solutions by ankor@dixie-net.com, greenestamps:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! How many mL's of a 10% alcohol solution and how many mL's of a 20% solution must be mixed to obtain 30 mL's of a 16% solution?
:
let x = amt of 20% solution
total is be 30 ml, therefore
(30-x) = amt of 10%
:
Solve this simple mixture equation
.10(30-x) + .20x = .16(30)
If a formal mathematical solution is required, then something like the setup shown by the other tutor is standard.
If formal math is not required, and if the speed of finding the answer is important, here is a quick and easy way to solve any 2-part mixture problem like this:
16% is 6/10 = 3/5 of the way from 10% to 20%;
therefore 3/5 of the mixture should be the ingredient with the higher percentage.
ANSWER: 3/5 of 30mL, or 18mL, of the 20% solution; the other 12mL of the 10% solution
