Ncert Exercises & Solutions

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?
1. $9$ min
2. $6$ min
3. $18 $ min
4. $2 $ min

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 (18/60h)

Distance covered by the bus = $(V - 20) \frac{18}{60} k m . . . . (1)$

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

$V \times \frac{T}{60} . . . . . (2)$

Both equations (1) and (2) are equal.

(V - 20) \times \frac{18}{60} = \frac{V T}{60} . . . . . (3)

The relative speed of the bus moving in the opposite direction of the cyclist = (V + 20) km/h

Time is taken by the bus to go past the cyclist= $6 m i n = \frac{6}{60} h$

$\Rightarrow (V + 20) \frac{6}{60} = \frac{V T}{60} . . . . . (4)$

From equations (3) and (4), we get

$(V + 20) \times \frac{6}{60} = (V - 20) \times \frac{18}{60}$
$V + 20 = 3 V - 60$
$2 V = 80$
$V = 40 k m / h$

Substituting the value of V in equation (4), we get

$(40 + 20) \times \frac{6}{60} = \frac{40 T}{60}$
$T = \frac{360}{40} = 9 m i n$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions