By walking at $\frac{3}{4}$ of his usual speed,a man reaches his office $20 \text{ minutes}$ later than usual. His usual time is (in $\text{minutes}$):

$A$ train speeds past a pole in $15 \, \text{seconds}$ and speeds past a $100 \, \text{meters}$ long platform in $25 \, \text{seconds}$. Its length in meters is:

$A$ 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 $km/hr$):

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 $km$.

$A$ train leaves station $A$ at $5\, AM$ and reaches station $B$ at $9\, AM$ on the same day. Another train leaves station $B$ at $7\, AM$ and reaches station $A$ at $10.30\, AM$ on the same day. The time at which the two trains cross one another is (in $AM$):

$A$ and $B$ are two towns. $A$ car goes from $A$ to $B$ at a speed of $64 \, km/h$ and returns to $A$ at a slower speed. If its average speed for the whole journey is $56 \, km/h$,find its return speed (in $km/h$).