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 $20 kmh^{-1}$ 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 $v\, km\, h^{-1}$ be the constant speed with which the buses 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
$\frac{v T}{v-20}=18$
or $v T=18 v-18 \times 20$
and $\frac{v T}{v+20}=6$
or $v T=6 v+20 \times 6$
Equating (i) and (ii), we get
$18 v-18 \times 20=6 v+20 \times 6$
or $12 v=20 \times 6+18 \times 20=480$
or $v =40 km h ^{-1}$
Putting this value of $v$ in (i), we get
$40 T =18 \times 40-18 \times 20=18 \times 20$
or $T =\frac{18 \times 20}{40}=9 min$