Question 573777: How many gallons of 60% antifreeze solution must be mixed with 80 gallons of 30% antifreeze to get a mixture that is 50% antifreeze? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of 60% antifreeze needed
Now we know that the amount of pure antifreeze that exists before the mixture takes place has to equal the amount of pure antifreeze that exists after the mixture takes place, sooooo:
0.60x+0.30*80=0.50(80+x)
0.60x+24=40+0.50x subtract 0.50x and also 24 from each side
0.10x=16
x=160 gal-----amount of 60% antifreeze needed
CK
0.60*160+24=0.50*240
96+24=120
120=120
Hope this helps---ptaylor
