$A$ person has to cover a distance of $6 \, km$ in $45 \, minutes$. If he covers one-half of the distance in two-third of the total time,to cover the remaining distance in the remaining time,his speed (in $km/hr$) must be:

$A$ 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 $AM$):

$A$ car covers four successive $3 \, km$ stretches at speeds of $10 \, km/h$,$20 \, km/h$,$30 \, km/h$,and $60 \, km/h$ respectively. Its average speed over this total distance is (in $km/h$):

$A$ man covers a certain distance between his house and office on a scooter. With an average speed of $30 \, km/hr$,he is late by $10 \, min$. However,with a speed of $40 \, km/hr$,he reaches his office $5 \, min$ earlier. Find the distance between his house and office in $km$.

If a man walks at the rate of $5\, km/h$,he misses a train by $7\, minutes$. However,if he walks at the rate of $6\, km/h$,he reaches the station $5\, minutes$ before the arrival of the train. The distance covered by him to reach the station is (in $km$)

The average speed of a car is $1 \frac{4}{5}$ times the average speed of a bus. $A$ tractor covers $575 \text{ km}$ in $23 \text{ hours}$. How much distance will the car cover in $4 \text{ hours}$ if the speed of the bus is twice the speed of the tractor? (in $\text{km}$)