Two towns $\text{A}$ and $\text{B}$ are connected by a regular bus service with a bus leaving in either direction every $\text{T}\text{minutes}$. A man cycling with a speed $20\text{km/hr}$ in the direction $A$ to $B$ notices that a bus goes past him every $18\text{minutes}$ in the direction of his motion, and every $6\text{minutes}$ in the opposite direction. What is the period $T$ of the bus service and with what speed do the buses ply on the road. Assume the buses ply with constant speed on the road.

The correct option is C $V=40\text{km/hr}$
Let $V$ be the speed of the bus running between towns $\text{A}$ and $\text{B}$.

Given, speed of the cyclist $=20\text{Km/h}$

Relative speed of bus moving in the direction of cyclist with respect to cyclist $=(\text{V}-20)\text{km/h}$

Every $18\text{mins}$, the bus went past the cyclist.

So, the distance covered by the bus $d=(V-20)\frac{18}{60}\text{km}.......\left(1\right)$

As, one bus leaves every $T\text{mins}$

$\text{So},d=V\left(\frac{T}{60}\right)......\left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$ we have,

$(V-20)\frac{18}{60}=\frac{VT}{60}......\left(3\right)$

Relative speed of the bus moving in the opposite direction of the cyclist $=(V+20)\text{km/h}$

Time taken by the bus to go past the cyclist $=\frac{6}{60}\text{h}$

$\therefore (V+20)\left(\frac{6}{60}\right)=\frac{VT}{60}.....\left(4\right)$

Using equations $\left(3\right)$ and $\left(4\right)$ we have:

$(V+20)\frac{6}{60}=(V-20)\frac{18}{60}$

$\Rightarrow 2V=80$

$\Rightarrow V=40\text{km/h}$

Putting the value of $V$ in equation $\left(4\right)$ we have,

$T=\frac{360}{40}=9\text{min}$

Hence, option $\left(B\right)$ and $\left(C\right)$ are the correct answers.