A person travels equal distances with speeds of $3 \mathrm{km}/\mathrm{hr}, 4 \mathrm{km}/\mathrm{hr}$ and $5 \mathrm{km}/\mathrm{hr}$ and takes a total time of $47$ minutes. The total distance (in $\mathrm{km}$) is:

A man travels $600$km by train at $80$km/hr, $800 \text{km}$ by ship at $40 \text{km/hr}$, $500 \text{km}$ by aeroplane at $400 \text{km/hr}$ and $100\text{ km}$ by car at $50 \text{km/hr}$. What is the average speed for the entire distance?

Walking $\frac{6}{7}$th of his usual speed, a man is $12$minutes too late. The usual time taken by him to cover that distance is:

In covering a certain distance, the speeds of $A$ and $B$ are in the ratio of $3:4$. $A$ takes $30$ minutes more than $B$ to reach the destination. The time taken by $A$ to reach the destination is:

A thief steals a car at $2.30 \mathrm{p}.\mathrm{m}.$ and drives it at $60 \mathrm{kmph}$. The theft is discovered at $3 \mathrm{p}.\mathrm{m}.$ and the owner sets-off in another car at $75 \mathrm{kmph}$. When will he overtake the thief?

The distance between two cities A and B is $330 \mathrm{km}$. A train starts from A at $8 \mathrm{a}.\mathrm{m}.$ and travels towards B at $60 \mathrm{km}/\mathrm{hr}$. Another train starts from B at $9 \mathrm{a}.\mathrm{m}.$ and travels towards A at $75 \mathrm{km}/\mathrm{hr}$. At what time do they meet?