SOLUTION: I have problems on knowing where to start on story problems and what to do with them. Here is an example. How much pure acid must be mixed with 80 liters of 10% acid to get 30%

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: I have problems on knowing where to start on story problems and what to do with them. Here is an example. How much pure acid must be mixed with 80 liters of 10% acid to get 30%       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 460415: I have problems on knowing where to start on story problems and what to do with them. Here is an example.
How much pure acid must be mixed with 80 liters of 10% acid to get 30% acid?
Found 2 solutions by robertb, richwmiller:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
let x = # liters of pure acid to be added. Pure acid is 100% acid (This sounds stupid, but it really is important.)
Then
1.00x + 0.10*80 = 0.30(x+80)
==> x + 8 = 0.30x + 24
==> 0.70x = 16
==> x+=+22%266%2F7 liters.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
from the view of acid
x+ .1*80=.30(x+80)
x=22.8571
from the view of water
.9*80=.7*(x+80)
x=22.8571