SOLUTION: A radiator contains 15 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 15 gal

Question 219776: A radiator contains 15 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 15 gal of a 40% antifreeze solution?
Found 2 solutions by ptaylor, checkley77:
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of the 20% solution that needs to be drained and replaced with pure antifreeze
Now we know that the amount of pure antifreeze in the initial radiator after x amount is drained (0.20(15-x)) plus the amount of pure antifreeze that is added(x) has to equal the amount of pure antifreeze in the final radiator (0.40*15), so our equation to solve is:
0.20(15-x)+x=0.40*15 get rid of parens
3-0.20x+x=6 subtract 3 from each side
3-3-0.20x+x=6-3 collect like terms
0.80x=3 divide each side by 0.80
x=3.75 gal----amount of the 20% solution that needs to be drained and replaced with pure antifreeze
CK
0.20*(15-3.75)+3.75=0.40*15
2.25+3.75=6
6=6
Hope this helps---ptaylor
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
(15*.20)-(x*.20)+x=15*.40
3-.2x+x=6
.8x=6-3
.8x=3
x=3/.8
x=3.75 gal. of pure antifreeze is used.
Proof:
3-3.75*.20+3.75=6
3-.75+3.75=6
6=6
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