You can put this solution on YOUR website! From: "Intially, a tank contains 20 gallons of a 30% antifreeze solution." We know the amount of antifreeze is:
.30(20) = 6 gallons
.
Let x = amount of 80% solution added
.
"inital amt of antifreeze" + "added antifreeze" = "50% concentration"
.30(20) + .80x = .50(20+x)
6 + .80x = 10 + .50x
.30x = 4
x = 4/.30
x = 13.333 gallons
or 13 1/3 gallons
You can put this solution on YOUR website! Let x=amount of 80% solution that needs to be added
Now we know that the amount of pure antifreeze in the 30% solution (0.30*20) plus the amount of pure antifreeze in the 80% solution that's added(0.80x) has to equal the amount of pure antifreeze in the final mixture(0.50(20+x)). So, our equation to solve is:
0.30*20+0.80x=0.50(20+x) get rid of parens and simplify
6+0.80x=10+0.50x subtract 6 and also 0.50x from each side
6-6+0.80x-0.50x=10-6+0.50x-0.50x collect like terms
0.30x=4 divide each side by 0.30
x=13.333 gal ---------------------------amount of 80% antifreeze needed
CK
0.30*20+0.80*13.333=0.50(33.333)
6+10.667=16.667
16.667=16.667
Hope this helps---ptaylor
