Question 573777
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