Question 94774

Let x=number of gal of 20% acid needed

Then 10-x=number of gal of 40% acid needed

Now we know that the amount of pure acid in the 20% solution (0.20x) plus the amount of pure acid in the 40% solution (0.40(10-x)) has to equal the amount of pure acid in the final mixture (0.26*10).  So our equation to solve is:

0.20x+0.40(10-x)=0.26*10  get rid of parens

0.20x+4-0.40x=2.6  subtract 4 from both sides

0.20x+4-4-0.40x=2.6-4  collect like terms

-0.20x=-1.4  divide both sides by -0.20

x=7------------------------------number of gal of 20% solution needed

10-x=10-7=3--------------------number of gal of 40% solution needed

CK

7+0.20+3*0.40=10*0.26

1.4+1.2+2.6
2.6=2.6

Hope this helps--------------ptaylor