Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of the bus service and the speed (assumed constant) of the buses.

The correct option is 15 min, 120 km/h.

Let us solve this problem step-by-step using the concept of relative velocity.

Let the constant speed of each bus be $v_b$ km/h.
Let the time interval after which a bus leaves in either direction be $T$ minutes.
The speed of the girl on the scooty is given as $v_s=60$ km/h.

The distance between any two consecutive buses moving in the same direction is constant and can be written as:

$d=v_b\times \frac{T}{60}$

Case 1: When buses move in the direction of the girl's motion (from X to Y)
The relative speed of the buses with respect to the girl is:

$v_{\mathrm{rel}}=v_b-v_s=v_b-60$

The girl observes that a bus passes her every 30 minutes ($30/60$ hours) in this direction. Therefore, the distance $d$ between consecutive buses is covered at this relative speed in 30 minutes:

$d=(v_b-60)\times \frac{30}{60}$

Equating the two expressions for $d$:

$v_b\times \frac{T}{60}=(v_b-60)\times \frac{30}{60}$

Simplifying this, we get Equation (1):

$v_{b}T=30(v_b-60)$

Case 2: When buses move in the opposite direction of the girl's motion
The relative speed of the buses with respect to the girl is:

$v_{\mathrm{rel}}=v_b+v_s=v_b+60$

The girl observes that a bus passes her every 10 minutes ($10/60$ hours) in the opposite direction. Therefore, the distance $d$ is covered at this relative speed in 10 minutes:

$d=(v_b+60)\times \frac{10}{60}$

Equating the two expressions for $d$:

$v_b\times \frac{T}{60}=(v_b+60)\times \frac{10}{60}$

Simplifying this, we get Equation (2):

$v_{b}T=10(v_b+60)$

Solving for Speed ($v_b$):
Equating Equation (1) and Equation (2):

$30(v_b-60)=10(v_b+60)$

Divide both sides by 10:

$3(v_b-60)=v_b+60$

$3v_b-180=v_b+60$

$2v_b=240$

$v_b=120$ km/h

Solving for Time Period ($T$):
Substitute the value of $v_b=120$ into Equation (1):

$120\times T=30(120-60)$

$120T=30\times 60$

$120T=1800$

$T=\frac{1800}{120}=15$ min

Thus, the time period $T$ of the bus service is 15 minutes, and the constant speed of the buses is 120 km/h.