Question 215910
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