Question 146833
Let x=amount of 80% antifreeze needed

Now we know that the amount of pure antifreeze in the 40% barrel(0.40*30) plus the amount of pure antifreeze in the amount that was added (0.80*x) has to equal the amount of pure antifreeze in the final solution (0.50(30+x)).  So our equation to solve is:

0.40*30+0.80x=0.50(30+x) get rid of parens and simplify

12+0.80x=15+0.50x  subtract 0.50x and also 12 from each side

12-12+0.80x-0.50x=15-12+0.50x-0.50x  collect like terms

0.30x=3  divide both sides by 0.30

x=10 gal-----------------------------amount of 80% antifreeze needed

CK

0.40*30+0.80*10=0.50*40

12+8=20
20=20


Hope this helps---ptaylor