Let total distance travelled by horse rider be $2x$.

Now, the time taken to cover first half distance is $t_{AB}=\frac{x}{5}$

Second half distance $x$ between $BC$ is covered in two parts, $x=d_1+d_2$

here $d_1$ is the distance travelled with $10 \mathrm{m} {\mathrm{s}}^{-1}$ and $d_2$ is the distance travelled with $15 \mathrm{m} {\mathrm{s}}^{-1}$ in equal time intervals $\frac{t}{2}$ each.

Then, $d_1=\frac{10t}{2}, d_2=\frac{15t}{2}$

Second half distance $x=\frac{\mathrm{t}}{2}(10+15)=\frac{25t}{2}$

Now, the mean speed of rider is $v=\frac{2x}{\frac{x}{5}+\frac{2x}{25}}=\frac{2\times 25}{5+2}=\frac{50}{7} \mathrm{m} {\mathrm{s}}^{-1}$

Thus, the value of $x=50$.