Problema Solution
A 90% acid solution is diluted with water to make a solution of 60% acid. When 3 gallons of water is added to dilute it again, the solution becomes 40% acid. How much water was added to the solution on the first dilution? How much 90% acid was available from the start?
Answer provided by our tutors
let
x = gal of the 90% acid solution available from the start
y = gal of water added to the solution on the first dilution
A 90% acid solution is diluted with water to make a solution of 60% acid:
0.90x + 0*y = 0.60(x + y)
When 3 gallons of water is added to dilute it again, the solution becomes 40% acid:
0.60(x + y) + 0*3 = 0.40(x + y + 3)
by solving the system of equations:
0.90x = 0.60(x + y)
0.60(x + y) = 0.40(x + y + 3)
we find
x = 4 gal
y = 2 gal
click here to see the step by step solution of the system of equations:
there were 2 gal of water added to the solution on the first dilution.
there was 4 gal of 90% acid available from the start.