$A$ train passes two persons walking in the same direction at a speed of $3 \, km/h$ and $5 \, km/h$ respectively in $10 \, s$ and $11 \, s$ respectively. The speed of the train is (in $km/h$):

Walking $\frac{6}{7}$ of his usual speed, a man is $12 \, \text{minutes}$ late. The usual time taken by him to cover that distance is

Three men step off together from the same spot. Their steps measure $63 \, cm$,$70 \, cm$,and $77 \, cm$ respectively. The minimum distance each should cover so that all can cover the distance in complete steps is (in $cm$):

$A$ student goes to school at the rate of $2 \frac{1}{2} \text{ km/h}$ and reaches $6 \text{ min}$ late. If he travels at the speed of $3 \text{ km/h}$,he is $10 \text{ min}$ early. The distance (in $\text{km}$) between the school and his house is:

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 $m$).

$A$ thief is spotted by a policeman from a distance of $200 \text{ meters}$. When the policeman starts to chase the thief,the thief also starts running. Assuming the speed of the thief is $10 \text{ km/h}$ and that of the policeman is $12 \text{ km/h}$,how far will the thief have run before he is overtaken by the policeman? (in $\text{km}$)