Question 384814: A 30 L solution is 80% antifreeze. How much water must be added to produce a solution that is 60% antifreeze? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of water needed
Now we know that the amount of pure antifreeze before the mixture takes place (0.80*30) is equal to the amount of pure antifreeze after the mixture takes place (0.60(30+x)). Sooo our equation to solve is:
0.80*30=0.60(30+x) simplify
24 =18+0.60x subtract 18 from each side
0.60x=6
x=10 L amount of water needed
CK
0.80*30=0.60*40
24=24
Hope this helps---ptaylor
