SOLUTION: A solution containing 30% insecticide is to be mixed with a solution containing 50% insecticide to make 200 L of a solutions containing 42% insecticide. How much of each solution s

Question 136750This question is from textbook Algebra 1
: A solution containing 30% insecticide is to be mixed with a solution containing 50% insecticide to make 200 L of a solutions containing 42% insecticide. How much of each solution should be used? 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 in the 30% solution(0.30x) plus the amount of pure insecticide in the 50% solution(0.50(200-x)) has to equal the amount of pure insecticide 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
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 both sides by -0.20
x=80 L----------------------amount of 30% insecticide needed
200-x=200-80=120 L---------------------amount of 50% insecticide needed
CK
0.30*80+0.50*120=0.42*200
24+60=84
84=84

Hope this helps---ptaylor
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