Let ,in the mixture $x$ litre water and $y$ litre milk was present at first.

Now we added 1 litre of water in the mixture.

Now ,the in the mixture $(x+1)$ litre water and $y$ litre milk present .

Now percentage of milk is $\large =\frac{y}{x+1+y}$

Now it is given as $25$% .

$\large \frac{y}{x+1+y}=\frac{1}{4}$

or, $\large x-3y=-1$ ……………………….(1)

Now we added again 1 litre of milk in this mixture.

Now ,the in the mixture $(x+1)$ litre water and $(y+1)$ litre milk present .

Now percentage of milk is $\large =\frac{y+1}{x+1+y+1}$

Now it is given as $40$% .

$\large \frac{y+1}{x+1+y+1}=\frac{2}{5}$

or, $\large 2x-3y=1$ ……………………….(2)

Now $(2)-(1)$

$\large 2x-3y-x+3y=1-(-1)$

so , $x=2$

Putting value of $x$ in $(2)$ we get,

$y=1$

So, initially in the mixture 2 litre water and 1 litre milk was present .

So percentage of milk initialy is=$\large \frac{1}{2+1}*100%$=$\large 33\frac{1}{3}%$%