SOLUTION: How many liters of a 30% alcohol solution and how many of a 60% alcohol solution must be mixed to produce 18 liters of a 50% alcohol solution?
Question 189311: How many liters of a 30% alcohol solution and how many of a 60% alcohol solution must be mixed to produce 18 liters of a 50% alcohol solution? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! How many liters of a 30% alcohol solution and how many of a 60% alcohol solution must be mixed to produce 18 liters of a 50% alcohol solution?
.
Let x = liters of 30% alcohol
and y = liters of 60% alcohol
.
Then, since we have two unknowns we need two equations:
x + y = 18 (equation 1)
.30x + .60y = .50(x+y) (equation 2)
.
Solve equation 1 for y:
x + y = 18
y = 18-x
.
Substitute the above into equation 2 and solve for x:
.30x + .60y = .50(x+y)
.30x + .60(18-x) = .50(x+18-x)
.30x + 10.8 - .60x = .50(18)
10.8 - .30x = 9
-.30x = -1.8
x = 6 liters (of 30% alcohol)
.
Substitute the above into equation 1 and solve for y:
x + y = 18
6 + y = 18
y = 12 liters (of 60% alcohol)
