### Q.9:- 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. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?

**Answer:-**

Let *V* be the speed of the bus running between towns A and B.

Speed of the cyclist, *v* = 20 km/h

Relative speed of the bus moving in the direction of the cyclist

= *V* – *v* = (*V* – 20) km/h

The bus went past the cyclist every 18 min i.e., 18 / 60 h (when he moves in the direction of the bus).

Distance covered by the bus = (*V* – 20) × 18 / 60 km …. (i)

Since one bus leaves after every *T* minutes, the distance travelled by the bus will be equal to

*V* × *T* **/** 60 ….(ii)

Both equations (i) and (ii) are equal.

(V – 20) × 18 **/** 60 = *VT* / 60 ……(iii)

Relative speed of the bus moving in the opposite direction of the cyclist

= (*V* + 20) km/h

Time taken by the bus to go past the cyclist = 6 min = 6 / 60 h

∴ (*V* + 20) × 6 / 60 = *VT* / 60 ….(iv)

From equations (iii) and (iv), we get

(*V* + 20) × 6 **/** 60 = (V – 20) × 18 **/** 60

*V* + 20 = 3*V* – 60

2*V* = 80

*V* = 40 km/h

Substituting the value of *V* in equation (iv), we get

(40 + 20) × 6 **/** 60 = 40T **/** 60

T = 360 / 40 = 9 min