SOLUTION: pure acid is to be added to a 10% solution to obtain 45L of a 40% acid solution. What amounts of each should be used?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: pure acid is to be added to a 10% solution to obtain 45L of a 40% acid solution. What amounts of each should be used?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 349274: pure acid is to be added to a 10% solution to obtain 45L of a 40% acid solution. What amounts of each should be used?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = liters of pure acid to be added
Let b = liters of 10% solution to be added
given:
(1) a+%2B+b+=+45
%28a+%2B+.1b%29%2F45+=+.4
a+%2B+.1b+=+18
(2) 10a+%2B+b+=+180
Subtract (1) from (2)
(2) 10a+%2B+b+=+180
(1) -a+-+b+=+-45
9a+=+135
a+=+15
From (1),
b+=+45+-+15
b+=+30
15 liters of pure acid and 30 liters of 10% acid solution must be mixed