The average of the speed of the boat upstream and the speed of the boat downstream is equal to the:

The sum of the times taken by train $P$ and train $Q$ to cross their own lengths is twice the time taken by them to cross each other when they are travelling in opposite direction to each other. If $P$ takes $20 \mathrm{s}$ to cross a stationary pole, then find the time taken by $Q$ to do the same (in $\mathrm{seconds}$).

A man can row $20 \mathrm{kmph}$ in still water. It takes him thrice as long to row up as to row down the river. Find the speed of the stream.

A person saves $5 \min$ in covering a certain distance by increasing his speed by $20\%$. What is the time taken to cover the distance at his usual speed?

Two trains $T_1$ and $T_2$ are travelling from station $A$ and station $B$ respectively in the opposite direction and they are $600 \mathrm{km}$ apart. Train $T_1$ starts from station $A$ with the speed $90 \mathrm{kmph}$ at $9:30 \mathrm{a}.\mathrm{m}.$ and train $T_2$ starts station $B$ with the speed $50 \mathrm{kmph}$ at $11:30 \mathrm{a}.\mathrm{m}.$ At what time will they cross each other?

In a $100 \mathrm{m}$ race, $A$ beats $B$ by $30 \mathrm{m}$ and in the same race, $B$ beats $C$ by $20 \mathrm{m}$. By what distance does $A$ beat $C$?

The following steps are involved in solving the above problem. Arrange them in sequential order.

$\left(A\right)$ The ratio of distance covered by $A$ and $B=100:70$ and the ratio of distances covered by $B$ and $C=100:80$.

$\left(B\right)$ When $A$ is at $100 \mathrm{m}$ from the starting point, $C$ will be at $56 \mathrm{m}$ from the starting point.

$\left(C\right)$ $A$ beats $C$ by $100-56$, i.e., $44 \mathrm{m}$.

$\left(D\right)$ The ratio of distances covered by $A$ and $C$ $=\frac{100}{70}\times \frac{100}{80}=\frac{100}{56}$