A cars radiator contains 12 L of antifreeze at a 25 concentr

A car\'s radiator contains 12 L of antifreeze at a 25% concentration. How many liters must be drained and then replaced by pure antifreeze to bring the concentration to 50% 9the manufactor\'s \"safe\" level)?

Solution

let x=amount of 25% solution that needs to be drained and replaced with pure antifreeze
Then 25-x is the amount of 25% antifreeze left after the system was drained
Now we know that the amount of pure antifreeze left after the system was drained .25(12-x),plus the amount of pure antifreeze added back in (x) equals the amount of pure antifreeze in the final solution (12)(.50). So our equation to solve is:
.25(12-x)+x=(12)(.50) simplifying, we get:
3-.25x+x=6 subtract 3 from both sides:
.75x=3
x=4 liters needs to be drained and replaced by pure antifreeze----ans
ck
.25(12-4)+4=6
2+4=6
6=6