SOLUTION: How much water should be mixed to a 500 mL of 30% solution of alcohol to have a resulting concentration of 20%?

Algebra ->  Pythagorean-theorem -> SOLUTION: How much water should be mixed to a 500 mL of 30% solution of alcohol to have a resulting concentration of 20%?      Log On


   

Question 1062652: How much water should be mixed to a 500 mL of 30% solution of alcohol to have a resulting concentration of 20%?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
you have 500 milliliters of 30% alcohol solution.

you want to add water to this solution so that it becomes a 20% solution.

the amount of alcohol in the 500 milliliters is .3 * 500 = 150 milliliters.

you want the same amount of alcohol divided by (500 + x) milliliters to be equal to 20%.

your equation would be 150 / (500 + x) = .2

x represents the number of milliliters of water to be added.

multiply both sides of this equation by (500 + x) and divide both sides of this equation by .2 to get:

150 / .2 = 500 + x

simplify to get 750 = 500 + x

solve for x to get x = 250.

you need to add 250 milliliters of water to get a solution that is 20% alcohol.

150 / 500 = .3 = 30%

150 / 750 = .2 = 20%