Question 103233
Let x=number of liters of the 22% sugar solution needed
Then 680-x=number of liters of the 31% solution needed

Now we know that the pure sugar in the 22% solution (0.22x) plus the pure sugar in the 31% solution (0.31(680-x)) has to equal the pure sugar in the final mixture(0.27*680).  So our equation to solve is:

0.22x+0.31(680-x)=0.27*680  get rid of parens

0.22x+210.8-0.31x=183.6  subtract 210.8 from both sides

0.22x+210.8-210.8-0.31x=183.6-210.8  collect like terms
-0.09x=-27.2  divide both sides by -0.09

x=302.222222222---- liters of 22% solution

680-x=680-302.222222222222222222---=377.77777777777777----liters of 31% solution

CK
302.2222*0.0.22+377.7777*0.31=680*0.27
66.4888888888888+117.11111111111111=183.6
183.599999999----~183.6


Hope this helps----ptaylor