SOLUTION: i am having a hard time on this. any help will be appreciated. a chemist needs 140 milliliters of a 33% solution but has only 28% and 42% solutions available. find how many mill

Question 159874: i am having a hard time on this. any help will be appreciated.
a chemist needs 140 milliliters of a 33% solution but has only 28% and 42% solutions available. find how many milliliters of each that should be mixed to get the desired solution.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
OK
Let x=amount of 28% solution needed
Then 140-x=amount of 42% solution needed
Now we know that the amount of pure stuff in the 28% solution(0.28x) plus the amount of pure stuff in the 42% solution (0.42(140-x)) has to equal the amount of pure stuff in the final solution (0.33*140). So, our equation to solve is:
0.28x+0.42(140-x)=0.33*140 get rid of parens
0.28x+58.8-0.42x=46.2 subtract 58.8 from each side
0.28x+58.8-58.8-0.42x=46.2-58.8 collect like terms
-0.14x=-12.6 divide each side by -0.14
x=90 ml------------------------------------amount of 28% solution needed
140-x=140-90=50 ml --------------------------amount of 42% solution needed
CK
0.28*90+0.42*50=0.33*140
25.2+21=46.2
46.2=46.2

Hope this helps---ptaylor
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