Question 94760
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


Hope this helps------ptaylor