SOLUTION: How many liters of 60% alcohol solution and 30% alcohol solution must be mixed to obtain 12 liters of 40% alcohol solution?

Question 802380: How many liters of 60% alcohol solution and 30% alcohol solution must be mixed to obtain 12 liters of 40% alcohol solution?

Found 2 solutions by josmiceli, richwmiller:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = liters of 60% alcohol solution needed
Let = liters of 30% alcohol solution needed
--------------
(1)
(2)
--------------------------
(2)
(2)
(2)
------------------
Subtract (1) from (2)
(2)
(1)

and, since
(1)
(1)
4 liters of 60% alcohol solution are needed
8 liters of 30% alcohol solution are needed
------------
check:
(2)
(2)
(2)
(2)
(2)
OK

Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
x+y=12,
.6x+.3y=.4*12
x=4., y=8.
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