Replacement of solution

Final volume of A and B does not change because the volume is being replaced.

Suppose final % of milk in A and B is 100a% and 100b%. 

Total quantity of milk in solution A = Volume of A*(100a)/100 = 100*(100a)/100 = 100a

Total quantity of milk in solution B = Volume of B*(100b)/100 = 200*(100b)/100 = 200b

So, the total quantity of milk = 100a+200b = 100
After removing 40 litre from A now, the amount of milk in A = 60a

After removing 40 litre from A now, the amount of milk in B = 200b+40a

Now, after adding 40 litre from B, the amount of milk in A = 60a + $$\ \frac{\ 40}{240}$$*(200*b+40*a)

Since, the total quantity of milk will remain unchanged even after mixing,

60*a + $$\ \frac{\ 40}{240}$$*(200*b+40*a) = 100a.

On solving we get 200a=200b i.e. a=b. Hence, 300a =100. Thus, a=b=33.33%

Let x be the initial volume of the solution. Hence, 5x/8 is the amount of milk and 3x/8 is the amount of water. 25% of x=2x/8 water is added.
Hence, the solution has 5x/8 milk and 5x/8 water now. Hence, the solution has equal parts of milk and water.
Removing 100 liters of solution and adding water is the same as removing 50 liters of milk and adding 50 liters of water.
So,
(5x/8 -50)/(5x/8+50) = 3/7. So, x = 200