SOLUTION: The mad scientist has one solution that is 30% acid and another solution that is 18% acid. How much of each should he use to get 300 L of a solution that is 21% acid?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: The mad scientist has one solution that is 30% acid and another solution that is 18% acid. How much of each should he use to get 300 L of a solution that is 21% acid?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1163785: The mad scientist has one solution that is 30% acid and another solution that is 18% acid. How much of each should he use to get 300 L of a solution that is 21% acid?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
30% solution and 18% solution;
want 300 Liters of 21% solution.

v of 30% solution
300-v of 18% solution

30v%2B18%28300-v%29=21%2A300
-
30v-18v%2B18%2A300=21%2A300
%2830-18%29v=21%2A300-18%2A300
%2830-18%29v=300%2821-18%29
highlight_green%28v=300%28%2821-18%29%2F%2830-18%29%29%29-----------shown in this form intentionally; you compute this, and you should be able to finish.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

The mad scientist has one solution that is 30% acid and another solution that is 18% acid. How much of each should he use to get 300 L of a solution that is 21% acid?
Let amount of 30% to mix, be T

Then amount of 18% to mix = 300 - T

We then get: .3T + .18(300 - T) = .21(300)

.3T + .18(300) - .18T = .21(300)

.3T - .18T = .21(300) - .18(300)

.12T = .03(300)

Amount of 30% to mix, or 

You should be able to calculate the amount of the 18% that's needed.