SOLUTION: Suppose that one solution contains 40% alcohol and another solution contains 80% alcohol. How many liters of each solution should be mixed to make 9.5 liters of a 70% alcohol solut
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: Suppose that one solution contains 40% alcohol and another solution contains 80% alcohol. How many liters of each solution should be mixed to make 9.5 liters of a 70% alcohol solut
Log On
Question 969304: Suppose that one solution contains 40% alcohol and another solution contains 80% alcohol. How many liters of each solution should be mixed to make 9.5 liters of a 70% alcohol solution? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! Let x= number of liters of 40% or .40 alcohol solution.
Let 9.5 -x = number of liters of 80% or .80 alcohol solution.
This is for me a lot easier than x and y. I have one variable, and x +9.5-x=9.5
Now set it up:
x(.4) + (9.5-x)(.8)=9.5(.7)
I am using decimals for alcohol percentage.
.4x +7.6-.8x =6.65 0.8 *(9.5-x) = 0.8 *9.5 - 0.8*x
-.4x =6.65-7.6= -0.95 notice that both sides of the equation are negative. I make both sides positive.
x=0.95/.4 or 10.5/4 or 2.375 liters 9.5-x= 7.125 liters
Check with alcohol
2.375 liters *.4 alcohol/liter =0.95 alcohol
7.125 liters *.8 alcohol/liter=5.7 alcohol
Total is 6.65 alcohol
Total requested was 9.5 liters *.7 alcohol/liter = 6.65 alcohol
