One litre of 0.15 M HCI and one litre of 0.3 M HCI is given. What is the maximum volume of 0.2 M HCI which one can make from these two solutions? No water is added.

Since no water added, so volume of solution cannot exceed 2L. So, less concentrated solution should be taken in its total volume. Only the small portion of more concentrated solution is to be mixed, so that the total concentrated is less than (0.3 M HCI). Let x L of 0.3 M solution is mixed. 

Total volume = (x+1) L

$\begin{array}{l}{\mathrm{M}}_1{\mathrm{V}}_1+{\mathrm{M}}_2{\mathrm{V}}_2={\mathrm{M}}_3{\mathrm{V}}_3 \left[{\mathrm{V}}_3=(1+\mathrm{x})\mathrm{L}\right.\\ 0.3\times \mathrm{xL}+0.15\times 1\mathrm{L}={\mathrm{M}}_3(1+\mathrm{x})\mathrm{L}\\ \text{ Final molarity }=0.2\mathrm{M}\\ 0.3\mathrm{x}+0.15=0.2(1+\mathrm{x})\\ \mathrm{x}=0.5\mathrm{L}\\ \text{ Maximum volume }=1+0.5=1.5\mathrm{L}\end{array}$