Question

Two towns $A$ and $B$ are connected by a regular bus service with a bus leaving in either direction every $T$ minutes. A man cycling with a speed of $20km{h}^{-1}$ in the direction $A$ to $B$ notices that a bus goes past him every $18$ min in the direction of his motion, and every $6$ min in the opposite direction. The period $T$ of the bus service is

• A. $4.5$ min
• B. $9$ min
• C. $12$ min
• D. $24$ min

Answer 1

Answer: (B)

Let $vkm{h}^{-1}$ be the constant speed with which the bus travel ply between the towns $A$ and $B$ .
Relative velocity of the bus from $A$ to $B$ with respect to the cyclist $=(v-20)km{h}^{-1}$
Relative velocity of the bus from $B$ to $A$ with respect to the cyclist $=(v+20)km{h}^{-1}$
Distance travelled by the bus in time $T$ (minutes) $=vT$
As per question
$\frac{vT}{v-20}=18$ or $vT=18v-18\times 20$ and $\frac{vT}{v+20}=6\dots \left(i\right)$
or $vT=6v+20\times 6\dots \left(ii\right)$
Equating $\left(i\right)$ and $\left(ii\right)$ , we get
$=18v-18\times 20=6v+20\times 6$
or $12v=20\times 6+18\times 20=480$ or $v=40km{h}^{-1}$
Putting this value of $v$ in $\left(i\right)$ , we get
$40T=18\times 40-18\times 20=18\times 20$
or $T=\frac{18\times 20}{40}=9$ min