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 $20 \,km\,h^{-1}$ 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)

$9$ min

Answer 2

Let $v \,km \,h^{-1}$ 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 $vT=18v-18\times 2 0$ and $\frac{vT}{v+20}=6\quad\ldots\left(i\right)$ or $vT=6v+20\times6\quad\ldots\left(ii\right)$ Equating $\left(i\right)$ and $\left(ii\right)$ , we get $=18v-18\times20=6v+20\times 6$ or $12v=20\times6+18\times20=480$ or $v=40\,km\,h^{-1}$ Putting this value of $v$ in $\left(i\right)$ , we get $40T=18\times40-18\times20=18\times 20$ or $T=\frac{18\times20}{40}=9$ min