$A$ $280\,m$ long train is moving at a speed of $60\,km/h$. The time taken by the train to cross a platform $220\,m$ long is (in $seconds$):

$A$ is faster than $B$. $A$ and $B$ each walk $24 \, km$. The sum of their speeds is $7 \, km/hr$ and the sum of the times taken by them is $14 \, hours$. Then $A$'s speed is equal to (in $km/hr$):

$A$ car travels the first one-third of a certain distance with a speed of $10\, km/hr$,the next one-third distance with a speed of $20\, km/hr$,and the last one-third distance with a speed of $60\, km/hr$. The average speed of the car for the whole journey is (in $km/hr$):

$A$ train leaves the station $1 \, hr$ before the scheduled time. The driver decreases its speed by $4 \, km/hr$. At the next station $120 \, km$ away,the train reaches on time. Find the original speed of the train in $km/hr$.

$A$ $280 \, m$ 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 $m/s$?

$A$ train covers a distance between station $A$ and station $B$ in $45 \, min$. If the speed of the train is reduced by $5 \, km/h$,then the same distance is covered in $48 \, min$. The distance between stations $A$ and $B$ is? (in $km$)