SOLUTION: A radiator contains 15 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 55% antifreeze?

Click here to see ALL problems on Mixture Word Problems

Question 662267: A radiator contains 15 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 55% antifreeze?
Found 2 solutions by ewatrrr, Edwin McCravy:
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi,
A radiator contains 15 quarts of fluid, 30% of which is antifreeze.
How much fluid should be drained and
replaced with pure antifreeze so that the new mixture is 55% antifreeze?
x + .30(15-x) = .55*15
.70x = .25*15
x = .25*15/.70

Answer by Edwin McCravy(20056) (Show Source):
You can put this solution on YOUR website!
A radiator contains 15 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 55% antifreeze?

When we drain out x quarts there will only be 15-x quarts left in there. 30% of those 15-x quarts is pure antifreeze. So that's .30(15-x) quarts of pure antifreeze in there after we have drained off x quarts. Now we replace those x quarts with pure antifreeze. Now the amount of pure antifreeze is .30(15-x)+x quarts. There are 15 quarts of liquid in the radiator, so the fraction of the 15 quarts that is pure antifreeze is this fraction: And that fraction is to be equal to 55% or .55 = .55 Multiply both sides by 15 .30(15-x)+x = 8.25 Solve that and get or quarts. Edwin