SOLUTION: John mixes a certain amount of 10% isopropyl mixture with 60 gallons of a 20% isopropyl solution, and the resulting mixture is 16% isopropyl. How much of the 10% isopropyl did he

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: John mixes a certain amount of 10% isopropyl mixture with 60 gallons of a 20% isopropyl solution, and the resulting mixture is 16% isopropyl. How much of the 10% isopropyl did he       Log On

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Question 134535: John mixes a certain amount of 10% isopropyl mixture with 60 gallons of a 20% isopropyl solution, and the resulting mixture is 16% isopropyl. How much of the 10% isopropyl did he add?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In words:
(gallons of isopropyl) / (gallons of mixture) = 16%
Let x = amount of 10% mixture to be added
60 gallons = amount of 20% mixture added
The total gallons of mixture = x+%2B+60
%28.1x+%2B+.2%2A60%29+%2F+%28x+%2B+60%29+=+.16
%28.1x+%2B+12%29+%2F+%28x+%2B+60%29+=+.16
multiply each sides by x+%2B+60
.1x+%2B+12+=+.16%28x+%2B+60%29
.1x+%2B+12+=+.16x+%2B+9.6
.06x+=+2.4
x+=+40
He added 40 gallons of the isopropyl mixture
check:
%28.1x+%2B+.2%2A60%29+%2F+%28x+%2B+60%29+=+.16
%28.1%2A40+%2B+.2%2A60%29+%2F+%2840+%2B+60%29+=+.16
%284+%2B+12%29+%2F+100+=+.16
16%2F100+=+.16
.16+=+.16
OK