1. Let the total distance be $D$.

2. First half of the distance:

   • Distance covered $= \dfrac{D}{2}$
   • Speed $= 5\ \text{m/s}$
   • Time taken: $t_1 = \frac{D/2}{5} = \frac{D}{10}$

3. Remaining half of the distance: This part is travelled at two speeds, $10\ \text{m/s}$ and $15\ \text{m/s}$, for equal times.

   Let each of these times be $t$.

   Then distance covered in this remaining half is: $10t + 15t = 25t$

   But this equals $\dfrac{D}{2}$, so $25t = \frac{D}{2}$ $t = \frac{D}{50}$

   Hence time for the second half: $t_2 = 2t = \frac{2D}{50} = \frac{D}{25}$

4. Total time: $T = t_1 + t_2 = \frac{D}{10} + \frac{D}{25}$ Taking LCM $= 50$, $T = \frac{5D + 2D}{50} = \frac{7D}{50}$

5. Mean speed: $v_{\text{avg}} = \frac{\text{total distance}}{\text{total time}} = \frac{D}{7D/50} = \frac{50}{7}\ \text{m/s}$

6. Given mean speed is $\dfrac{x}{7}\ \text{m/s}$, so $\frac{x}{7} = \frac{50}{7}$ Therefore, $x = 50$

Final Answer: $50$