A boy walks to his school at a distance of $6km$ with a speed of $2.5km/h$ and walks back with a constant speed by $4km/h$. His average speed for trip expressed in $km/h$ is

The correct option is B $\frac{40}{13}$

$\text{Step 1: Time calculation}$

$\text{Forward Journey:}$

$d=6km;v_1=2.5km/h$

Time $t_1=\frac{d}{v_1}=\frac{6}{2.5}$ $=2.4hr$

$\text{Backward Journey:}$

$d=6km;v_2=4km/hr$

Time $t_2=\frac{d}{v_2}=\frac{6}{4}$ $=1.5hr$

$\text{Step 2: Average speed calculation}$

We know that, Average speed $V_{av}=\frac{Totaldistance}{totaltime}$

$\therefore V_{av}=\frac{(6+6)km}{t_1+t_2}=\frac{12km}{(2.4+1.4)hr}$ $=\frac{40}{13}km/hr$

Hence average speed for trip is $\frac{40}{13}km/hr$

$\therefore$ Option B is correct