Question

There is a regular bus service between towns A and B, with a bus leaving towns A and B every $T$ minutes. A cyclist moving with a speed of $20km{h}^{-1}$ in the direction A to B notices that a bus goes past him every $18mins$ in his direction and every $6mins$ in the opposite direction. What is the period $T$ of bus service?

• A. $9$ $minutes$
• B. $12$ $minutes$
• C. $15$ $minutes$
• D. $18$ $minutes$

Answer 1

Answer: (A)

$9$ $minutes$

Let velocity of bus is $V$. Then separation between two nearest buses is $VT$
A bus moving in the same direction travels the distance $VT$ in 18 minutes to meet the cyclist.
So, $t=\frac{S_{rel}}{V_{rel}}$

or, $\frac{18}{60}=\frac{VT}{V-20}......\left(1\right)\{Vrel=V-20\}$
A bus moving in opposite direction travels the distance $VT$ in 6 minutes to meet the cyclist.
So, $t=\frac{S_{rel}}{V_{rel}}\Rightarrow \frac{6}{60}=\frac{VT}{V+20}......\left(2\right)\{Vrel=V+20\}$
From $1$ and $2$, we get:

$3=\frac{V+20}{V-20}$

$\Rightarrow V=40$ kmph
From $1$

$\frac{18}{60}=\frac{40T}{40-20}\Rightarrow T=\frac{9}{60}$ hrs. $=\frac{9}{60}\times 60$ min$=9$ min