SOLUTION: Michelle had two containers, each containing a mixture of alcohol and disinfectant. One was 40% disinfectant and the other was 80% disinfectant. How much of each should be used t

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Question 67414This question is from textbook Advanced mathematics
: Michelle had two containers, each containing a mixture of alcohol and disinfectant. One was 40% disinfectant and the other was 80% disinfectant. How much of each should be used to get 600 milliliters of a solution that is 50% disinfectant? This question is from textbook Advanced mathematics

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x= amount of 80%
Then 600-x= amount of 40%
Now we know that the amount of pure disinfectant in the 80% mixture (.8x) plus the amount of pure disinfectant in the 40% misture .4(600-x) must equal the amount of pure disinfectant in the final solution .5(600). So our equation to solve is:

.8x+.4(600-x)=.5(600) simplifying we get
.8x+240-.4x=300 subtract 240 from both sides and collect like terms:
.4x=60
x=150 ml -----------------amount of 80% solution
600-x=600-150=450 ml-----------amount of 40% solution
CK
450(.4)+150(.8)=.5(600)
180+120=300
300=300

Hope this helps----ptaylor