SOLUTION: Story Problem: If I want to make the perfect 12 ounce cup of chocolate milk, it requires that the mixture is 42% syrup. What i have now is 8 ounces of a milk/syrup mixture that I

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Question 88945: Story Problem:
If I want to make the perfect 12 ounce cup of chocolate milk, it requires that the mixture is 42% syrup. What i have now is 8 ounces of a milk/syrup mixture that I know contains 30% syrup. What must the syrup concentration be in the remaining mixture that I must add in order to achieve perfection?
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=syrup concentration in the remaining mixture that must be added which is 12-8 or 4 oz
Now we know that the amount of pure syrup in the 8oz mixture (0.30(8)) plus the amount of pure syrup in the 4oz mixture (x(4)) must equal the amount of pure syrup in the final mixture (0.42(12)).
So our equation to solve is:
0.30*8+4x=0.42*12
2.40+4x=5.04 subtract 2.40 from both sides
2.40-2.40+4x=5.04-2.40 collect like terms
4x=2.64 divide both sides by 4
x=0.66--or 66%--------syrup concentration in the remaining mixture that must be added
CK
0.30*8+0.66*4=0.42*12
2.40+2.64=5.04
5.04=5.04
Hope this helps---ptaylor