$A$ person covers half of his journey at $30 \text{ km/h}$ and the remaining half at $20 \text{ km/h}$. The average speed for the whole journey is (in $\text{km/h}$)

$A$ truck covers a distance of $550 \, m$ in $1 \, min$,whereas a bus covers a distance of $33 \, km$ in $45 \, min$. The ratio of their speeds is

$A$ train leaves the station at $5 \, am$ at $60 \, km/h$. Another train leaves the same station at $6.30 \, am$ at $75 \, km/h$ and travels in the same direction as the first train. At what time and at what distance from the station will they meet?

$A$ 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 time taken by them is $14 \, hours$. Then $A$'s speed is equal to (in $km/hr$)

$A$ 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 $\text{km}$)

$A$ man starts from a place $P$ and reaches the place $Q$ in $7 \, hours$. He travels $\frac{1}{4}^{th}$ of the distance at $10 \, km/h$ and the remaining distance at $12 \, km/h$. The distance,in $km$,between $P$ and $Q$ is