SOLUTION: A chemist needs to mix a solution containing 30% insecticede with a solution containing 50% insecticede to make 200 L of a solution that is 42% insecticide. How much of each soluti

Question 215910This question is from textbook Algebra 1
: A chemist needs to mix a solution containing 30% insecticede with a solution containing 50% insecticede to make 200 L of a solution that is 42% insecticide. How much of each solution should she use? This question is from textbook Algebra 1

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 30% insecticide needed
Then (200-x)=amount of 50% insecticide needed
Now we know that the amount of pure insecticide that exists in the 30% solution(0.30x) plus the amount of pure insecticide that exists in the 50% solution (0.50(200-x)) has to equal the amount of pure isecticide that exists in the final mixture(0.42*200). So, our equation to solve is:
0.30x+0.50(200-x)=0.42*200 get rid of parens and simplify
0.30x+100-0.50x=84 subtract 100 from each side
0.30x+100-100-0.50x=84-100 collect like terms
-0.20x=-16 divide each side by -0.20
x=80 L----------------------amount of 30% solution needed
200-x=200-80=120 L---------------amount of 50% solution
CK
0.30*80+0.50*120=0.42*200
24+60=84
84=84

Hope this helps---ptaylor
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