$A$ train goes from Ballygunge to Sealdah at an average speed of $20 \text{ km/h}$ and comes back at an average speed of $30 \text{ km/h}$. The average speed of the train for the whole journey is (in $\text{km/h}$)

$A, B$ and $C$ start together from the same place to walk around a circular path of length $12 \, km$. $A$ walks at the rate of $4 \, km/h$,$B$ at $3 \, km/h$,and $C$ at $\frac{3}{2} \, km/h$. They will meet together at the starting place at the end of (in $h$):

$A$ car driver leaves Bangalore at $8:30 \, am$ and expects to reach a place $300 \, km$ from Bangalore at $12:30 \, pm$. At $10:30 \, am$, he finds that he has covered only $40 \%$ of the distance. By how much does he have to increase the speed of the car in order to keep up with his schedule? (in $km/h$)

$A$ train running at $\frac{7}{11}$ of its normal speed reached a place in $22\, \text{hours}$. How much time could be saved if the train had run at its normal speed? (in $\text{hours}$)

$A$ train can travel $50\%$ faster than a car. Both start from point $A$ at the same time and reach point $B$,$75\, km$ away from $A$,at the same time. On the way,however,the train lost about $12.5\, \text{minutes}$ while stopping at the stations. The speed of the car is (in $km/hr$):

$A$ car is running at a speed of $108 \, km/hr$. What distance will it cover in $15 \, s$? (in $meters$)