Two trains of equal length are running on parallel tracks in the same direction at the speeds of $46 \, km/h$ and $36 \, km/h$. The faster train passes the slower train in $36 \, seconds$. The length of each train is (in $m$):

$A$ car travels $80 \, km$ in $2 \, hours$ and a train travels $80 \, km$ in $3 \, hours$. The ratio of the speed of the car to that of the train is

$A$ train covers a distance of $10 \text{ km}$ in $12 \text{ minutes}$. If its speed is decreased by $5 \text{ km/hr}$,the time taken by it to cover the same distance will be:

$A$ man performs $\frac{2}{15}$ of the total journey by rail,$\frac{9}{20}$ by tonga,and the remaining $10 \ km$ on foot. The total journey is (in $km$):

$A$ 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:

Two men start together to walk a certain distance,one at $4 \, km/h$ and another at $3 \, km/h$. The former arrives half an hour before the latter. Find out the distance in $km$.