$A$ man travelled a certain distance by train at the rate of $25 \text{ km/h}$ and walked back at the rate of $4 \text{ km/h}$. If the whole journey took $5 \text{ h } 48 \text{ min}$,what is the distance? (in $\text{km}$)

$A$ person has to reach a certain place at a certain time. He finds that he will be $15 \text{ minutes}$ late if he walks at $4 \text{ km/h}$,and $10 \text{ minutes}$ earlier if he walks at $6 \text{ km/h}$. How far has he to walk? (in $\text{km}$)

$A$ man travels first $50 \, km$ at $25 \, kmph$, next $40 \, km$ at $20 \, kmph$, and then $90 \, km$ at $15 \, kmph$. His average speed for the whole journey (in $kmph$) is:

Two men $A$ and $B$ start from a place $P$ walking at $3 \, km/h$ and $3.5 \, km/h$ respectively. How many $km$ will they be apart at the end of $3 \, h$,if they walk in opposite directions (in $.5$)?

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

If a person walks at $14 \, km/h$ instead of $10 \, km/h$,he would have walked $20 \, km$ more. The actual distance travelled by him is (in $km$):