SOLUTION: How many gallons of a 60% antifreeze solution must be mixed with 90 gallons of 10% antifreeze to get a mixture that is 50% difference?
Algebra ->
Percentage-and-ratio-word-problems
-> SOLUTION: How many gallons of a 60% antifreeze solution must be mixed with 90 gallons of 10% antifreeze to get a mixture that is 50% difference?
Log On
Question 507450: How many gallons of a 60% antifreeze solution must be mixed with 90 gallons of 10% antifreeze to get a mixture that is 50% difference? Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! How many gallons of a 60% antifreeze solution must be mixed with 90 gallons of 10% antifreeze to get a mixture that is 50% difference?
.
Let x = amount (gallons) of 60% antifreeze
then
.60x + .10(90) = .50(x+90)
.60x + 9 = .50x+45
.10x + 9 = 45
.10x = 36
x = 360 gallons