SOLUTION: A 12 quart cooling system is filled with a solution that is 10% antifreeze. The desired strength of the solution is 40% antifreeze. How many quarts of solution need to be drained

Question 66436: A 12 quart cooling system is filled with a solution that is 10% antifreeze. The desired strength of the solution is 40% antifreeze. How many quarts of solution need to be drained and replaced with pure antifreeze to reach the desired strength?
Found 2 solutions by ptaylor, josmiceli:
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!

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

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let x = the amount of mixture I want to remove
Then .1*x = the amount of antifreeze in the mixture
Then x = the amount of antifreeze I want to put back in,
making 12 = the amount of mixture once again
stated in a formula,

I say that 4 quarts of 10% mixture must be removed and
replaced with pure antifreeze. Check this
4 + 8 = 12, so

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