SOLUTION: how many ounces of a 40% alcohol solution must be mixed with 5 ounces of a 45% alcohol solution to make a 41% alcohol solution?

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Question 285941: how many ounces of a 40% alcohol solution must be mixed with 5 ounces of a 45% alcohol solution to make a 41% alcohol solution?
Found 2 solutions by josmiceli, richwmiller:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In words:
(ounces of alcohol in final solution)/(total ounces of final solution) = 41%
Let x = ounces of 40% alcohol solution needed
.45%2A5+=+2.25ounces of alcohol in 45% solution
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%28.4x+%2B+2.25%29%2F%28x+%2B+5%29+=+.41
.4x+%2B+2.25+=+.41%2A%28x+%2B+5%29
.4x+%2B+2.25+=+.41x+%2B+2.05
.01x+=+2.25+-+2.05
.01x+=+.2
x+=+20
20 ounces of a 40% alcohol solution are needed
check:
%28.4x+%2B+2.25%29%2F%28x+%2B+5%29+=+.41
%28.4%2A20+%2B+2.25%29%2F%2820+%2B+5%29+=+.41
%288+%2B+2.25%29%2F25+=+.41
10.25+=+25%2A.41
10.25+=+10.25
OK

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
just have a shot of either and you won't care and it won't matter!