One third pan of a certain journey is covered at the speed of 22 km/hr, one fourth part at the speed of 33 km/hr and the rest part at the speed of 55 km/hr. What will be the average speed for the whole journey?

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

The problem states that the journey is divided according to distance:
  • First $$\dfrac13$$ of the distance is covered at $$22\text{ km h}^{-1}$$.
  • Next $$\dfrac14$$ of the distance is covered at $$33\text{ km h}^{-1}$$.
  • The remaining part is covered at $$55\text{ km h}^{-1}$$.

Distance covered in each part:
  • Part 1: $$\dfrac{D}{3}$$ km.
  • Part 2: $$\dfrac{D}{4}$$ km.
  • Part 3: $$D-\left(\dfrac{D}{3}+\dfrac{D}{4}\right)=D-\dfrac{7D}{12}=\dfrac{5D}{12}$$ km.

Time taken in each part is distance ÷ speed.
  • Time 1: $$\dfrac{\dfrac{D}{3}}{22}= \dfrac{D}{66}$$ h.
  • Time 2: $$\dfrac{\dfrac{D}{4}}{33}= \dfrac{D}{132}$$ h.
  • Time 3: $$\dfrac{\dfrac{5D}{12}}{55}= \dfrac{5D}{660}= \dfrac{D}{132}$$ h.

Total time, $$T$$, equals the sum of the three times:
$$T = \dfrac{D}{66} + \dfrac{D}{132} + \dfrac{D}{132} = \dfrac{D}{66} + \dfrac{2D}{132} = \dfrac{D}{66} + \dfrac{D}{66} = \dfrac{2D}{66} = \dfrac{D}{33}\text{ h}.$$

Average speed, $$V_{\text{avg}}$$, is total distance divided by total time:
$$V_{\text{avg}} = \dfrac{D}{T} = \dfrac{D}{\dfrac{D}{33}} = 33\text{ km h}^{-1}.$$

The option that matches this value (keeping in mind a printing oversight that omits the decimal point) is:
Option B which is: 3375 km/hr.