$A$ $100\, m$ long train passes a bridge at the rate of $72\, km/h$ in $25\, s$. What is the length of the bridge (in $m$)?

$A$ train,after travelling $50 \, km$,meets with an accident and then proceeds at $\frac{3}{4}$ of its former speed and arrives at its destination $35 \, minutes$ late. Had the accident occurred $72 \, km$ further,it would have reached the destination only $15 \, minutes$ late. The normal speed of the train (in $km/hr$) is:

$A$ man with $\frac{3}{5}$ of his usual speed reaches the destination $2 \frac{1}{2}$ hours late. Find his usual time to reach the destination? (in $hours$)

If a distance of $50\, m$ is covered in $1\, minute$,$90\, m$ in $2\, minutes$,and $130\, m$ in $3\, minutes$,find the total distance covered in $15\, minutes$ (in $m$).

It takes $24 \, s$ for a train travelling at $93 \, km/h$ to cross entirely another train half its length travelling in the opposite direction at $51 \, km/h$. It passes a bridge in $66 \, s$. What is the length of the bridge in meters?

$A$ man covered a certain distance at some speed. Had he moved $3 \text{ km/hr}$ faster,he would have taken $40 \text{ minutes}$ less. If he had moved $2 \text{ km/hr}$ slower,he would have taken $40 \text{ minutes}$ more. The distance (in $\text{km}$) is: