a radiator holds 5 gallons. how much of the 30% antifreeze solution should be drained and replaced with pure water to reduce the solution to 9% antifreeze?

Answer 1

If x gallons are drained, then 30% of what remains is antifreeze. So, since water is 0% antifreeze,

.3(5-x) + 0x = .09*5

Answer 2

0.3