One third part of a certain journey is covered at the speed of 24 km/hr, one fourth part at the speed of 36 km/hr and the rest part at the speed of 60 km/hr. What will be the average speed for the whole journey?

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

The journey is completed in three segments:
 • First segment: $$\frac{1}{3}D$$ km at 24 km h−1
 • Second segment: $$\frac{1}{4}D$$ km at 36 km h−1
 • Remaining segment: $$1-\frac{1}{3}-\frac{1}{4} = \frac{12-4-3}{12} = \frac{5}{12}$$ of the journey, so $$\frac{5}{12}D$$ km at 60 km h−1

Time taken for each segment (time = distance ÷ speed):
$$t_1 = \frac{\frac{1}{3}D}{24} = \frac{D}{72} \text{ h}$$
$$t_2 = \frac{\frac{1}{4}D}{36} = \frac{D}{144} \text{ h}$$
$$t_3 = \frac{\frac{5}{12}D}{60} = \frac{5D}{720} = \frac{D}{144} \text{ h}$$

Total time for the whole journey:
$$T = t_1 + t_2 + t_3 = \frac{D}{72} + \frac{D}{144} + \frac{D}{144}$$
Combine the two equal terms:
$$T = \frac{D}{72} + \frac{2D}{144} = \frac{D}{72} + \frac{D}{72} = \frac{2D}{72} = \frac{D}{36} \text{ h}$$

Average speed $$V_{\text{avg}}$$ is defined as total distance divided by total time:
$$V_{\text{avg}} = \frac{D}{T} = \frac{D}{\frac{D}{36}} = 36 \text{ km h}^{-1}$$

Thus the average speed for the entire journey is 36 km/hr.

Option D which is: 36 km/hr