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

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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) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = liters of 60% alcohol solution needed
Let +b+ = liters of 30% alcohol solution needed
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(1) +a+%2B+b+=+12+
(2) +%28+.6a+%2B+.3b+%29+%2F+12+=+.4+
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(2) +.6a+%2B+.3b+=+4.8+
(2) +6a+%2B+3b+=+48+
(2) +2a+%2B+b+=+16+
------------------
Subtract (1) from (2)
(2) +2a+%2B+b+=+16+
(1) +-a+-+b+=+-12+
+a+=+4+
and, since
(1) +4+%2B+b+=+12+
(1) +b+=+8+
4 liters of 60% alcohol solution are needed
8 liters of 30% alcohol solution are needed
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check:
(2) +%28+.6a+%2B+.3b+%29+%2F+12+=+.4+
(2) +%28+.6%2A4+%2B+.3%2A8+%29+%2F+12+=+.4+
(2) +%28+2.4+%2B+2.4+%29+%2F+12+=+.4+
(2) +4.8%2F12+=+.4+
(2) +4.8+=+4.8+
OK

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