SOLUTION: A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solution contains 85% pure antifreez

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Question 1189920: A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture. The premium antifreeze solution contains 85% pure antifreeze. The company wants to obtain 255 gallons of a mixture that contains 30% pure antifreeze. How many gallons of water and how many gallons of the premium antifreeze solution must be mixed?
Found 3 solutions by Boreal, Theo, josgarithmetic:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website!
x*0.85=0.85x pure (100%)
end is 255*0.30=76.5 gallons pure
so 0.85x=76
x=89.41 gallons of the 85% antifreeze and 165.59 gallons of water.

Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
x = number of gallons of premium antifreeze solution.
y = number of gallons of water.

x + y = 255
.85 * x + 0 * y = .3 * 255
multiply both sides of the first equation by .85 and simplify the second equation to get:
.85 * x + .85 * y = 216.75
.85 * x = 0 * y = 76.5
subtract the second equation from the first to get:
.85 * y = 140.25
solve for y to get:
y = 140.25 / .85 = 165
solve for x to get:
x = 255 - 165 = 90
x + y = 90 + 165 = 255.
this confirms total gallons at 255.
.85 * x + 0 * y = .85 * 90 = 0 * 165 = 76.5 / 255 = .3 = 30% antifreeze.
this confirms total antifreeze solution in the inexpensive mixture = 30%.
your solution is:
90 gallons of premium antifreeze and 165 gallons of water must be mixed to get a solution that is 30% antifreeze.

Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
Simple Dilution ----

x of the 85%

pure = pure

----------the 85% to use
----------the water to use