SOLUTION: 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%
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: 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%
Log On
Question 103233: 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? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! 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
