$A$ man leaves a point $P$ and reaches the point $Q$ in $4 \, hours$. Another man leaves the point $Q$,$2 \, hours$ earlier and reaches the point $P$ in $4 \, hours$. Find the time in which the first man meets the second man (in $hours$).

$A$ man driving at $\frac{3}{4}$ of his original speed reaches his destination $20 \text{ minutes}$ later than the usual time. Then the usual time is (in $\text{minutes}$)

$A$ man travelled a distance of $80\, km$ in $7\, hours$,partly on foot at the rate of $8\, km/h$ and partly on bicycle at $16\, km/h$. The distance travelled on foot is (in $km$):

Buses start from a bus terminal with a speed of $20 \text{ km/h}$ at intervals of $10 \text{ mins}$. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of $8 \text{ mins}$? (in $\text{km/h}$)

$A$ man takes $6\, h \, 15\, min$ in walking a distance and riding back to the starting place. He could walk both ways in $7\, h \, 45\, min$. The time taken by him to ride both ways is:

$A$ man travelled a distance of $61 \, km$ in $9 \, h$,partly on foot at the rate of $4 \, km/h$ and partly on bicycle at the rate of $9 \, km/h$. The distance travelled on foot was (in $km$):