Pls solve the first one

Dear student,
Average speed=total distance travelled/total time taken.
Let the distance between A and B is‘d’.
From A to B:
Let time taken = t₁
Then, $t_1=\frac{d}{60}$ (in hr)
From B to A:
Let time taken = t₂
Then, $t_2=\frac{d}{80}$ (in hr)
Now, total time taken, 
$t=t_1+t_2 =\frac{d}{60}+\frac{d}{80}=\frac{7d}{240} hr$
Total distance travelled = d + d = 2d
Then, Average speed is $= \frac{total dis\tan ce}{total time}=\frac{2d}{{\displaystyle \raisebox{1ex}{$7d$}\!\left/ \!\raisebox{-1ex}{$240$}\right.}}=\frac{480}{7}=68.57 km/hr$
$average velocity =\frac{total displacement}{total time}=\frac{0}{total time}= 0 km/hr$
(displcament is zero as the initial point = final point)
hence, 
Average speed=68.57 km/hr
Average velocity =0

Regards,