Question 692246
I started to work on this yesterday and got sidetracked.  Hope this is not too late to be helpful.

Let x=percent of acid in the mixture
Then 100-x=percent of water in mixture
And let y= the original amount of mixture in vat
We know the following:
When we add acid, the amount of water remains the same before and after the acid is added and when we add water, the amount of acid remains the same before and after water is added, soooo:
We add 25 gal of acid:
Amount of water before=(100-x)*y=100y-xy
Amount of water after=20(y+25)
100y-xy=20(y+25)=20y+500
80y-xy=500--------------------eq1
When we add 25 gal of water:
Amount of acid before=xy
Amount of acid after=60(y+25)
xy=60y+1500 or
xy-60y=1500------------------------eq2

Add eq 1 and 2
20y=2000
y=100 gal ---amount of original mixture in vat
substitute y=100 in eq2
100x-6000=1500 or
x-60=15
x=75 percent-------------percent of acid in original mixture

Hope this helps---ptaylor