$A$ train $50\, m$ long passes a platform $100\, m$ long in $10\, s$. The speed of the train in $m/s$ is:

$A$ car travels $20\%$ slower than a train. Both start from point $A$ at the same time and reach point $B$ $240\, km$ away at the same time. On the way,the train takes $48\, minutes$ for stopping at stations. What is the speed (in $km/hr$) of the car?

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

$A$ 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:

One of the two buses completes a journey of $300 \, km$ in $7 \frac{1}{2} \, hours$ and the other a journey of $450 \, km$ in $9 \, hours.$ The ratio of their average speed is:

$A$ $280 \, m$ long train moving with an average speed of $108 \, km/h$ crosses a platform in $12 \, s$. $A$ man crosses the same platform in $10 \, s$. What is the speed of the man in $m/s$?