Question 64

Let the total distance of the journey be $D$ km.

1. First leg (Train):

   • Distance $d_1 = \dfrac{D}{2}$
   • Speed $v_1 = 60\text{ km/h}$
   • Time $t_1 = \dfrac{D/2}{60} = \dfrac{D}{120}\text{ hours}$

2. Second leg (Bus):

   • Remaining distance $= D - \dfrac{D}{2} = \dfrac{D}{2}$
   • Half of the remainder: $d_2 = \dfrac{1}{2} \times \dfrac{D}{2} = \dfrac{D}{4}$
   • Speed $v_2 = 30\text{ km/h}$
   • Time $t_2 = \dfrac{D/4}{30} = \dfrac{D}{120}\text{ hours}$

3. Third leg (Cycle):

   • Remaining distance: $d_3 = D - d_1 - d_2 = D - \dfrac{D}{2} - \dfrac{D}{4} = \dfrac{D}{4}$
   • Speed $v_3 = 10\text{ km/h}$
   • Time $t_3 = \dfrac{D/4}{10} = \dfrac{D}{40} = \dfrac{3D}{120}\text{ hours}$

4. Total time and average speed:

   $$\text{Total time } T = t_1 + t_2 + t_3 = \dfrac{D}{120} + \dfrac{D}{120} + \dfrac{3D}{120} = \dfrac{5D}{120} = \dfrac{D}{24}\text{ hours}$$
   $$\text{Average speed} = \dfrac{\text{Total Distance}}{\text{Total Time}} = \dfrac{D}{\frac{D}{24}} = 24\text{ km/h}$$

Thus, the average speed is 24 km/h, which is option (C).