You can
put this solution on YOUR website!Let the unknown amount of pollution be "x".
Then Plant II puts out the pollution at 1/x amount per hour.
And Plant I puts out the pollution at 2/x amount per hour.
Together they put out the pollution at 1/x+2/x = 3/x amt. per hour
Working together for 3 hours they put out (3/x)(3)=9/x amount
The slower Plant (Pl II) would require 9 hrs. to put out 9/x amount.
Cheers,
Stan H.