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

Given that,

Let speed of bus $=V$

relative speed when cyclist and bus both move the same direction $=(V-20)km/h$

We know,

$Distance=speed\times time$

According to question ,

The bus went past cyclists every $18min$ when in the motion of his direction.

e.g distance covered by bus $=\frac{(V-20)\times 18}{60}km$

Every $T$ time bus travels distance $=VT$

e.g.$\frac{(V-20)\times 18}{60}=VT----------\left(1\right)$

Similarly,

The relative speed of bus when bus and cyclist move in opposite direction $=(v+20)km/h$

The bus went past every $6$ min in the opposite direction of his motion.

then,

$\frac{(V+20)\times 6}{60}=VT----------\left(2\right)$

Solve both equation ,

$\frac{(V-20)\times 18}{60}=\frac{(V+20)\times 6}{60}$

$\Rightarrow 3V-60=V+20$

$\Rightarrow 2V=80$

$\Rightarrow V=40Km/hputequation\left(1\right)$

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

$T=\frac{6}{40}$ hour $=9min$

Hence,

$V=40km/h$ and $T=9min$