Question 66436

let x=amount of 10% solution that needs to be drained and replaced with pure antifreeze

Then 12-x is the amount of 10% antifreeze left after the system was drained

Now we know that the amount of pure antifreeze left after the system was drained .10(12-x),plus the amount of pure antifreeze added back in (x) equals the amount of pure antifreeze in the final solution (12)(.40).  So our equation to solve is:


.10(12-x)+x=(12)(.40)  simplifying, we get:

1.2-.10x+x=4.8  subtract 1.2 from both sides:

.90x=3.6
x=4 quarts needs to be drained and replaced by pure antifreeze

ck

.10(12-4)+4=4.8
1.2-.4+4=4.8
.8+4=4.8
4.8=4.8

Hope this helps ----ptaylor