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%?

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Question 901260: 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%?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
let x be the number of liters of water
--- percent ---------------- Amount
Alcohol 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