In motion $BC$

$x = d _1+ d _2$

where $d _1 , d _2$ we the distance travelled with $10\,m / s$ and $15\,m / s$ respectively in equal time intervals

$\frac{' t^{\prime}}{2} \text { each }$

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

$d_1+d_2=x=\frac{t}{2}(10+15)=\frac{25 t }{2}$

$< v >=\frac{2 x }{\frac{ x }{5}+\frac{2 x }{25}}=\frac{2 \times 25}{5+2}=\frac{50}{7} m / s$

Ans. : $50$