Question

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

Answer 1

Calucalte relative distance travelled by the bus in direction of motion of cyclist.
Let $V$ be the speed of bus running from town $A\text{to}B$.
Given: Speed of cyclist $v=20km/hr$
Every $18min$, the bus went past the cyclist, moving in direction of bus i.e., $18/60hr$
Distance covered by bus $=V-20\times 18/60$ .....$\left(1\right)$

Find the distance travelled by any bus untill the next bus starts.
Since one bus leaves after every $Tminutes$, the distance travelled by the bus will be
$=\frac{VT}{60}$ ....$\left(2\right)$
Both equations $\left(1\right)\text{and}\left(2\right)$ are equal i.e.,

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

Relative speed of the bus moving in the opposite direction of the cyclist$=(V+20)km/hr$
Similarly,
$\frac{(V+20)\times 60}{60}=\frac{VT}{60}$ ...$\left(3\right)$

From equations $\left(3\right)\text{and}\left(4\right)$, we get,
$\frac{(V+20)\times 18}{60}=\frac{(V+20)\times 6}{60}$
$\Rightarrow 3V-60=V+20$
$\Rightarrow 40km/hr$
Susbstituting the value of $V$ in equation $\left(3\right)$, we get,

$\frac{(40-20)\times 18}{60}=40\times \frac{T}{60}$

$\Rightarrow T=20\times \frac{18}{40}=\frac{36}{4}=9minutes$.