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?

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.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) (Show Source): You can put this solution on YOUR website!
Let = liters of pure acid to be added
Let = liters of 10% solution to be added
given:
(1)

(2)
Subtract (1) from (2)
(2)
(1)

From (1),

15 liters of pure acid and 30 liters of 10% acid solution must be mixed
RELATED QUESTIONS
Pure acid is to be added to a 10% acid solution to obtain 45L of a 40% acid... (answered by josgarithmetic,math_tutor2020,greenestamps,ikleyn)

pure acid is to be added to a 10% acid solution to obtain 42 L of a 40% acid solution.... (answered by oberobic)

pure acid is to be added to a 20% acid solution to obtain 40L of a 40% acid solution.... (answered by josgarithmetic)

Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution.... (answered by longjonsilver)

Pure Acid is to be added to a 10% acid solution to obtain 27 litters of 20% acid... (answered by robertb)

Pure acid is to be added to a 10% acid solution to obtain 90 L of a 20% acid solution.... (answered by josgarithmetic)

Pure acid is to be added to a 10% acid solution to obtain 90 L of a 20% acid solution.... (answered by Alan3354)

Pure acid is to be added to a 10% acid solution to obtain 81 L of a 20% acid solutions.... (answered by ankor@dixie-net.com)

pure acid is to be added to a 10%acid solution to obtain 90l of 89% solution . what... (answered by vleith)