The other tutor's answer is correct, but his method will work ONLY
when the percentage desired is exactly half-way between the two
percentages.  Here is a more general way to do the problem, which will
work in ALL mixture problems: 

Make this chart:

              amt of liquid     % acid   amt of pure acid contained 
1st solution       
2nd solution        
final solution

Let x be the answer, so fill that in first     




              amt of liquid     % acid   amt of pure acid contained 
1st solution        x           
2nd solution      
final solution     

We are told that there are 5 L of the second solution, so

              amt of liquid     % acid   amt of pure acid contained 
1st solution        x           
2nd solution        5             
final solution     

Now we add x PLUS 5 to get the amt of final liquid, so we fill
x+5 in

              amt of liquid     % acid   amt of pure acid contained 
1st solution        x           
2nd solution        5         
final solution     x+5          

Next we fill in the three percentages as decimals:

              amt of liquid     % acid   amt of pure acid contained 
1st solution        x             .25           
2nd solution        5             .15            
final solution     x+5            .20          

Then we multiply the three amts of liquid by the percentages expressed
as decimals to get the amt of pure acid in each amount:

              amt of liquid     % acid   amt of pure acid contained 
1st solution        x             .25            .23x
2nd solution        5             .15            .15(5)
final solution     x+5            .20            .20(x+5)

The equation comes from:

                 


                 .25x + .15(5) = .20(x+5)

Multiply through by 100  

                   25x + 15(5) = 20(x+5)
                      25x + 75 = 20x + 100
                            5x = 25
                             x = 5              
Edwin