SOLUTION: How many liters of water must be mixed with 20 liters of a 90% alcohol mixture so that the alcohol content of the mixture is reduced to 50% Is my equation correct? 20(0.90) - x

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: How many liters of water must be mixed with 20 liters of a 90% alcohol mixture so that the alcohol content of the mixture is reduced to 50% Is my equation correct? 20(0.90) - x      Log On


   

Question 787646: How many liters of water must be mixed with 20 liters of a 90% alcohol mixture so that the alcohol content of the mixture is reduced to 50%
Is my equation correct?
20(0.90) - x (0.50) = x
thank you for helping me
Answer by mananth(16949) About Me  (Show Source):
You can put this solution on YOUR website!
--- percent ---------------- Amount
Alcohhol 90 ---------------- 20 liters
Water 0 ---------------- x liters
Mixture 50 ---------------- 20 + x liters

90 * 20 + 0 x = 50 ( 20 + x )
1800 + 0 x = 1000 + 50 x
0 x -50 x = -1800 + 1000
-50 x = -800
/ -50
x = 16 liters Water

m.ananth@hotmail.ca