SOLUTION: How much water should be added to 40 liters of a 75% acid solution to dilute it to 60%? I know the current solution has 30 liters of acid (.75x40), but I'm having difficulty with t

75% of 40 liters is 30 liters of pure acid, which you got. 

Then you ask the question:

30 liters of acid is 60% of what?

      30 = .60x
       = x
      50 = x

So if you add enough water to make it 50 liters, 
those 30 liters of pure acid would make it 60% of
the 50.

Therefore to make the 40 liters 50 liters you need to add 10
liters of water.
-----------------------------------------------------------
However that method will only work when you are adding water.
Below is a method that will work when you add any percentage
solution to any other percentage solution.  In your case water
is considered a 0% solution but the method will work in all 
cases, not just for adding water.  Just put the percentage
of acid in what you're adding, which may be anywhere from 0% to 
100%, where the 0% is below. If you were adding pure acid, then
you would put 100% where the 0% is below


                  LITERS OF   PERCENT     LITERS OF
                    LIQUID    OF ACID     PURE ACID                 
---------------------------------------------------
Starting solution     40      75%         75%(40)
Added solution         x       0%          0%(x)
Final solution       40+x     60%         60%(40+x)

                   

                     75%(40) +  0%(x) = 60%(40+x)
                     .75(40) + .00(x) = .60(40+x)
                          30 +    0   = 24 + .60x
                                   30 = 24 + .60x
                                    6 = .60x
                                   10 = x

answer: 10 liters of water.

Edwin