Let the total journeybe x km. Then $\frac{\mathrm{x}}{3}$km at the speed of 25 km/hr and $\frac{\mathrm{x}}{4}$km at 30 km/hr and the rest distance $\left(\mathrm{x}-\frac{\mathrm{x}}{3}-\frac{\mathrm{x}}{4}\right)=\frac{5}{12}\mathrm{x}$

at the speed of 50 km/hr.

Total time taken during the journey of x km

$$\begin{gathered} =\frac{\mathrm{x}}{3\times 25}\mathrm{hrs}+\frac{\mathrm{x}}{4\times 30}\mathrm{hrs}+\frac{5\mathrm{x}}{12\times 50}\mathrm{hrs} \\ =\frac{\mathrm{x}}{75}\mathrm{hrs}+\frac{\mathrm{x}}{120}\mathrm{hrs}+\frac{\mathrm{x}}{12\times 10}\mathrm{hrs} \\ =\frac{\mathrm{hrsx}}{75}+\frac{\mathrm{hrsx}}{120}+\frac{\mathrm{hrsx}}{120} \end{gathered}$$

(Expand the fraction to get the least common denominator)

$$\begin{gathered} =\frac{8\mathrm{hrsx}}{8\times 75}+\frac{5\mathrm{hrsx}}{5\times 120}+\frac{5\mathrm{hrs}}{5\times 120} \\ =\frac{8\mathrm{hrsx}+5\mathrm{hrsx}+5\mathrm{hrsx}}{600} \\ =\frac{18\mathrm{hrsx}}{600}=\frac{3}{100}\mathrm{hrsx} \end{gathered}$$

∴ Average speed =$\frac{\mathrm{x}}{{\displaystyle \frac{18\mathrm{x}}{600}}}=\frac{100}{3}=33\frac{1}{3}$ km/hr

Hence, Option A is correct.