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 speed of 20km/h 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 do the buses ply on the road?

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Speed of cyclist, v = 20 km/h

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

Relative speed of bus in the direction of cyclist   = V - v = (V - 20) km/h

Distance covered by bus = (V - 20) 18/60 ….. (i)  (Given)

Distance traveled by the bus =  V × T/60 ......(ii)

Solving eq (i) and (ii), we get

(V - 20) × 18/60 = VT/60 ....... (iii)

Relative speed of bus in opposite direction of cyclist = (V + 20) km/h

Time taken by the bus to go past the cyclist = 6/60 h  (Given)

⇒ (V - 20) × 6/60 = VT/60 .... (iv)

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

⇒  (V - 20) × 6/60 = (V - 20) × 18/60  

⇒  V - 20 = 3V - 60

⇒  2V = 80

⇒ V = 80/2

⇒ V = 40 km/h

Hence, the speed of of bus is 40 km/h.

Putting V's value in equation (iv), we get

⇒  (40 + 20) × 6/60  = 40T/60

⇒  60 × 6/60  = 40T/60

⇒  6 = 40T/60

⇒ 40T = 6 × 60

⇒ 40T = 360

⇒ T = 360/40

⇒ T = 9 minutes

Hence, the period T of the bus service is 9 minutes.

Answer 2

Explanation:

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 = 3V – 60

2V = 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

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