$A$ train running at $25 \, km/hr$ takes $18 \, seconds$ to pass a platform. Next,it takes $12 \, seconds$ to pass a man walking at $5 \, km/hr$ in the opposite direction. Find the sum of the length of the train and that of the platform (in $m$).

$A$ car moving in the same direction passes a man walking at $4\, km/h$. The man can see the car for $3\, minutes$ and the visibility is up to a distance of $130\, m$. The speed of the car is: (in $km/h$)

$A$ 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:

$P$ and $Q$ are $27 \, km$ apart. Two trains with speeds of $24 \, km/h$ and $18 \, km/h$ respectively start simultaneously from $P$ and $Q$ and travel in the same direction. They meet at a point $R$ beyond $Q$. The distance $QR$ is (in $km$):

It takes $8 \, \text{hours}$ for a $600 \, \text{km}$ journey if $120 \, \text{km}$ is covered by train and the rest by car. It takes $20 \, \text{minutes}$ more if $200 \, \text{km}$ is covered by train and the rest by car. The ratio of the speed of the train to that of the car is:

$A$ 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 $\text{km}$)