Question 94760: How many gallons of a 40% solution must be mixed with 96 gallons of 60% solution to obtain a 52% solution? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=number of gal of 40% solution needed
Now we know that the amount of pure liquid in the 40% solution (0.40x) plus the amount of pure liquid in the 60% solution (0.60*96) has to equal the amount of pure liquid in the final mixture(96+x)*0.52. So our equation to solve is:
0.40x+(0.60*96)=0.52(96+x) get rid of parens
0.40x+57.6=49.92+0.52x subtract 0.52x and also 57.6 from both sides
0.40x-0.52x+57.6-57.6=49.92-57.6+0.52x-0.52x collect like terms
-0.12x=-7.68 divide both sides by -0.12
x=64-------------------------number of gal of 40% solution needed
CK
0.40*64+0.60*96=0.52(96+64)
25.6+57.6=83.2
83.2=83.2
