SOLUTION : $40 {m} / {sec}$.
Total distance $={L}$ and
Total time taken $=t$
A $\frac{x \quad C \quad(L-x)}{t_1=t / 2 \quad t_2=t / 2} B$
Let $x$ and $({L}-x)$ be the distance travelled in time $t_1$ and $t_2$, then
$\qquad x=20 t_1=20 t / 2=10 t$
$\text { And }(L-x)=60 t_2=60 t / 2=30 t$
$\text { Average speed }=\frac{\text { Total distance }}{\text { Total time }}=\frac{10 t+30 t}{t}=40 {m} / {sec} \text {. }$