Question 285515: how many gallons of a 25% solution must be mixed with 60 gallons of a 50% solution to get a 30% solution? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=number of gal of 25% solution needed
Now we know that the amount of "pure stuff" in the 25% solution (0.25x) plus the amount of "pure stuff" in the 60 gal of the 50% solution ((0.50*60) has to equal the amount of "pure stuff" in the final mixture (0.30(60+x)). So, our equation to solve is:
0.25x+0.50*60=0.30(60+x) get rid of parens and simplify
0.25x+30=18+0.30x subtract 0.25x and also 18 from each side
30-18=0.30x-0.25x
0.05x=12 divide each side by 0.05
x=240 gal of 25% solution is needed
CK
240*0.25 +0.50*60=0.30*300
60+30=90
90=90
