Problema Solution
how many liters of a 10% alcohol solution must be mixed with 40 L of a 50% solution to get a 40% solution?
Answer provided by our tutors
Bottle A
10% solution
X liters
Bottle B
50% solution
40 liters
bottle C (mixture)
40% solution
Xliters + 40liters
So the equation is:
(10%)X + (50%)40 = (40%) (X+40)
.1x + (.5)40 = .4(x+40)
.1x + 20 = .4x + .4(40)
.1x + 20 = .4x + 16
.4x - .1x = 20 - 16
.3x = 4
X = 4/.3
X = 13.33 liters