SOLUTION: How many ounces each of a 12% alcohol solution and a 30% alcohol solution must be combined to obtain 60 ounces of an 18% solution?
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Question 35037: How many ounces each of a 12% alcohol solution and a 30% alcohol solution must be combined to obtain 60 ounces of an 18% solution?
Found 2 solutions by Paul, Prithwis:
Answer by Paul(988) (Show Source): You can put this solution on YOUR website!
Let the solution be x
12x+30(60-x)=18(60)
-18x=-720
x=40
60-40=20
Hence, for 12% about 40 ounces is needed and for 30% about 20 ounces is needed.
PAul.
Answer by Prithwis(166) (Show Source): You can put this solution on YOUR website!
Let the answer be x
Then .12x + .30(60-x)= .18(60)
=> 12x + 30(60-x) = 18(60)
=> -18x = -720
=> x=40
...
40 ounces of 12% alcohol is needed to mix with 20 ounces of 30% alocohol to prepare 60 ounces of 18% alocohol.
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