Water and Milk

Stage I:

Concentration of milk = $\frac{3}{(3+2)}$ x 100 = 60 %

Stage II:

Volume increased by 50 % after adding water. Thus, volume becomes $\frac{3}{2}$ times the original volume.

So the milk concentration becomes $\frac{2}{3}$ x 60 % = 40 %

Stage III:

30 L is withdrawn and replaced with water.

Final concentration of milk = $\frac{3}{(3+7)}$ x 100 = 30 %

Hence, $\frac{(40-30)}{40}$ th = $\frac{1}{4}$ th of milk has been removed from the mixture.

In other words, $\frac{1}{4}$ th of mixture = 30 L

So the volume in Stage II = 4 x 30 L = 120 L

120 L is $\frac{3}{2}$ times the original volume. Therefore, original volume = $\frac{2}{3}$ x 120 = 80 L