Babita travels a certain distance at a speed of 11 km/hr and double the earlier distance at 44 km/hr. She then returns to the starting point through the same route. If her average speed for the entire journey is 34 km/hr, then what is her speed (in km/hr) for the return journey?

Correct option is C

Given:

Outbound: distance d at 11 km/h, then 2d at 44 km/h.

Return: entire 3d at speed v km/h. Overall average speed =34 km/h.

Formula Used:
Average speed $$=\dfrac{\text{Total distance}}{\text{Total time}}$$​​
Time $$=\dfrac{\text{Distance}}{\text{Speed}}$$​​

Solution:
Outbound time =$$\dfrac{d}{11}+\dfrac{2d}{44}=\dfrac{d}{11}+\dfrac{d}{22}=\dfrac{3d}{22}$$​​

Return time $$=\dfrac{3d}{v}$$​​
Total distance = 6d

3$$4=\frac{6d}{\frac{3d}{22}+\frac{3d}{v}}$$​​

​$$=\frac{2}{\frac{1}{22}+\frac{1}{v}}$$​

​$$\frac{1}{22}+\frac{1}{v}=\frac{1}{17}$$​​

​$$\frac{1}{v}=\frac{1}{17}-\frac{1}{22}=\frac{5}{374}$$​​

v$$=\frac{374}{5}=74.8\ \text{km/h}$$​​