SOLUTION: A hospital needs to dilute a 60% boric acid solution to a 10% solution. If it needs 40 liters of the 10% solution, how much of the 40% solution and how much water should it use? F

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A hospital needs to dilute a 60% boric acid solution to a 10% solution. If it needs 40 liters of the 10% solution, how much of the 40% solution and how much water should it use? F      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 155010: A hospital needs to dilute a 60% boric acid solution to a 10% solution. If it needs 40 liters of the 10% solution, how much of the 40% solution and how much water should it use? For some reason i always come up with a negative 24 and that isn't right.
Answer by ptaylor(1894) About Me  (Show Source):
You can put this solution on YOUR website!
I THINK YOU NEED TO RECHECK YOUR PROBLEM. I THINK THE "40%" SHOULD BE "60%". ANYWAY, I WILL MAKE THAT ASSUMPTION.
Let x=amount of 60% solution needed
Then 40-x=amount of water needed
Now we know that the amount of pure boric acid in the water(0) plus the amount of pure boric acid in the 60% solution (0.60x) has to equal the amount of pure boric acid in the final mixture(0.10*40). So our equatuion to solve is:
0.60x=0.10*40
0.60x=4 divide each side by 0.60
x=6.67 liters --------------of the 60% solution
40-x=40-6.67=33.33 liters-------------of water
Ck
6.67*0.60=40*0.10
4=4

Hope this helps---ptaylor