How many liters of a solution containing 22% sugar and how many liters of another solution containing 31% sugar must be mixed together to make 680 liters of solution containing 27% sugar

How many liters of a solution containing 22% sugar and how many liters of another solution containing 31% sugar must be mixed together to make 680 liters of solution containing 27% sugar?

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
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