SOLUTION: A chemist has two solutions. One is 40% alcohol, and the other is pure alcohol. How much of each solution should be mixed together to make 1.6 liters of a solution that is 52% alco

Question 956211: A chemist has two solutions. One is 40% alcohol, and the other is pure alcohol. How much of each solution should be mixed together to make 1.6 liters of a solution that is 52% alcohol?

I have tried x + y= 1.6 x=1.6-y
Also using grids.
Came up with
0.40(1.6-y) + 1.00y= .832
0.64-0.40y + 1.00y =.832
0.64 +0.6y= .832
0.6y= 1.3
y=2.16

I feel I am doing something very wrong. Help is greatly appreciated.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
A chemist has two solutions. One is 40% alcohol, and the other is pure alcohol. How much of each solution should be mixed together to make 1.6 liters of a solution that is 52% alcohol?
***
let x=amt of 40% alcohol to mix
1.6-x=amt of pure(100%) alcohol to mix
..
40%x+100%(1.6-x)=52%*1.6
.40x+1.6-x=0.832
.60x=0.768
x=1.28
1.6-x=0.32
amt of 40% alcohol to mix=1.28 liters
amt of pure(100%) alcohol to mix=0.32 liters
note: your setup is ok, you just added instead of subtracting in the 4th step.
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