Answer –

Let the speed of each bus be denoted by V_b speed of the cyclist be V_c. 

And according to the question, V_c = 20 km/h

Buses travelling in the same direction as cyclists have a relative velocity = V_b – V_c
Buses travelling in the same direction as the cyclist, pass the cyclist every 18 minutes, i.e. (18/60) seconds.

Then distance covered is simply the product of speed and time taken, so

Distance = (V_b – V_c ) x 18/60

Also, we can write that the distance is equal to V_b x (T/60) due to the fact that the buses are leaving every T minutes.

Therefore, by equating two values of distance we get –

(V_b – V_c ) x 18/60 = V_b x (T/60)                                                      —— (1)

The relative velocity of buses travelling in the opposite direction as cyclists is V_b + V_c.
And the buses pass the cyclist every 6 minutes, or (6/60) seconds.

Similar to the above case, Distance = (V_b + V_c ) x 6/60

Therefore we can write,

(V_b +V_c ) x 6/60 = V_b x (T/60)                                                         ——(2)

Dividing equation (2) by equation (1), we get

\[\frac{\left[ ({{V}_{b}}-{{V}_{c}})\times \frac{18}{60} \right]}{\left[ ({{V}_{b}}+{{V}_{c}})\times \frac{6}{60} \right]}=\frac{{{V}_{b}}\times \frac{T}{60}}{{{V}_{b}}\times \frac{T}{60}}\]

$\frac{\left( {{V}_{b}}-{{V}_{c}} \right)18}{\left( {{V}_{b}}+{{V}_{c}} \right)3}=1$

Or, (V_b – V_c )3 = (V_b +V_c )

Upon putting the value of V_{c,  We get –}

(V_b – 20 )3= (V_b + 20 )

3V_b – 60 = V_b + 20

2V_b = 80

V_b = 80/2

V_b = 40 km/h

In order to find the value of T, we put the values of V_b and V_c in equation (1)

We get =>  (40 – 20) x (18/60) = 40 x (T/60)

T = (20 x 18) /40

Therefore, T = 9 minutes