The teachers were right.... we will need this one day!!! I didn't do so well in math class. Thank you for forums like this!

I am working with dilutions of a concentrate. (I will give both gallons and lbs.)

We have a diluted drum of 55 gallons/450lbs. Of that, 15 Gal/122.715 lbs is the concentrate. 40 Gal/327.285 lbs is the water. I think that is a 3:8 dilution ratio. When performing a solids/activity test on this diluted product, we find the finished drum has 11.65% solids/activity. We do not have the original concentrate and need to determine what the solids/activity of the original concentrate is.

(I hope someone will tell me it is as simple as multiplying 11.56 by 3)

On the flip side...... We often have a concentrate for which we know the % of solids/activity ie. 49%. Then I want to plug into a formula (which I don't have) to determine the resulting solids at a known ratio ie. 1:75, or 1:120 etc....

Again I hope I'm making this too complicated and there is some very simple formula that I missed that day I skipped class to go swimming!

humbly in appreciation.