Question 888095
A radiator can contain 6 gallons of liquid water and antifreeze. How many gallons of pure water would be needed to be mixed with 80% antifreeze to fill up the radiator with a 30% antifreeze solution?
***
let x=amt of pure water to mix
6-x=amt of 80% antifreeze to mix
..
80%(6-x)=30%*6
4.8-.8x=1.8
.8x=3
x=3.75
6-x=2.25
gallons of pure water would be needed to be mixed with 80% antifreeze=3.75