Let the speed of each bus $=v_{b} \text{ km } h^{-1}$ and speed of cyclist $=v_{c}=20 \text{ km } h^{-1}$ Case I : Relative speed of the buses plying in the direction of motion ofcyclist $\text{i. e.}$ from $A$ to $B=v_{b}-v_{c}= (v_{b}-20) \text{ km } h^{-1}$ since the bus goes past the cyclist every 18 minutes $\left( = \frac{18}{60} \text{ h} \right)$, $\therefore\;$ Distance covered by the bus w.r.t. the cyclist $= (v_{b}-20) \times \frac{18}{60} \text{ km } \ldots$ (i) Since the bus leaves after every T minutes, so the distance covered by thebus in T min $\left( = \frac{T}{60} \text{ h} \right)$ is given by $=v_{b} \times \frac{T}{60} \ldots$(ii) ∴ From (i) and (ii), we get or $(v_{b}-20) \times \frac{18}{60} = v_{b} \times \frac{T}{60}$or $v_{b}-20 = v_{b} \times \frac{T}{18} \ldots$(iii)Case II : Relative speed of the bus coming from town B to A w.r.t. cyclist$(v_b+20) \text{ km } h^{-1}$ (∵ cyclist is moving from A to B)Since the bus goes past the cyclist after every 6 minutes∴ Distance covered by the bus w.r.t. the cyclist$= (v_{b}+20) \times \frac{6}{60} \text{ km } \ldots$(iv)Also distance covered by bus in T minutes is$=v_{b} \times \frac{T}{60} \text{ km } \ldots$(v)∴ Equating (iv) and (v), we get$(v_{b}+20) \times \frac{6}{60} = v_{b} \times \frac{T}{60}$or $v_{b}+20 = v_{b} \times \frac{T}{6} \ldots$(vi)Dividing (vi) by (iii), we get$\frac{v_{b}+20}{v_{b}-20}=3$or $v_{b}+20=3v_{b}-60$ or $2v_{b}=80$ or $v_{b}=40 \text{ km } h^{-1}$Putting the value of $v_{b}$ in equation (iii), we get$40-20=40 \times \frac{T}{18}$ or $20=40 \times \frac{T}{18}$or $T=20 \times \frac{18}{40}$$=9$ minutes$\therefore\; v_{b}=40 \text{ km } h^{-1},$$T=9$ minutes.