Question 136750
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