SOLUTION: How many gallons of a 60% antifreeze solution must be mixed with 90 gallons of 25% antifreeze to get a mixture that is 50% antifreeze?

Question 1145618: How many gallons of a 60% antifreeze solution must be mixed with 90 gallons of 25% antifreeze to get a mixture that is 50% antifreeze?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x = the number of gallons of the 60% solution.
you get .6 * x + .25 * 90 = .5 * (x + 90).
simplify to get .6 * x + 22.5 = .5 * x + 45
subtract .5 * x from both sides of the equation and subtract 22.5 from both sides of the equation to get:
.1 * x = 22.5
solve for x to get x = 225.
you have 225 gallons of 60% antifreeze solution added to 90 gallons of 25% antifreeze solution.
total gallons is 315.
total antifreeze is .6 * 225 + .25 * 90 = 157.5 gallons of antifreeze.
157.5/315 = .5 = 50% antifreeze in the combined solution of 315 gallons.
your solution is that 225 gallons of 60% antifreeze solution are required.
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