Problema Solution
how many liters of a 90% acid solution must be mixed with a 15% acid solution to get 225 L of a 80% acid solution?
Answer provided by our tutors
x = liters of 90% solution
y = liters of 15% solution
x + y = 225
(.90x + .15y)/225 = .80
y = 225 - x
(.90x + .15(225 - x))/225 = .80
(.90x + 33.75 - .15x)/225 = .80
(.75x + 33.75) = 180
.75x = 146.25
x = 195
(195) + y = 225
y = 30
195L of 90% solution and 30L of 15% solution are needed.