SOLUTION: There are two containers of lemonade; one contains 6% of sugar, and the other is with 12%. To make 800oz of lemonade with 10.5% of sugar, how do you need to mix the two?

Question 348570: There are two containers of lemonade; one contains 6% of sugar, and the other is with 12%. To make 800oz of lemonade with 10.5% of sugar, how do you need to mix the two?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Actually, we can do this with two equations and two unknowns or with one equation and one unknown---I prefer one equation and one unknown:
Let x=amount of lemonade with 12% sugar that is needed
Then 800-x=amount of lemonade with 6% sugar that is needed
Now we know that the amount of sugar in the 12% lemonade (0.12x) plus the amount of sugar in the 6% lemonade (0.06(800-x)) has to equal the amount of pure lemonade in the final mixture (0.105*800). So our equation to solve is:
0.12x+0.06(800-x)=0.105*800 get rid of parens
0.12x+48-0.06x=84 subtract 48 from each side and collect like terms
0.06x=36
x=600 oz ----amount of 12% lemonade needed
800-x=800-600=200 oz----amount of 6% needed
CK
600*0.12 +200*0.06=800*0.105
72+12=84
84=84
Hope this helps--ptaylor
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