Question

A man covers half of his journey by train at 90 km/hr, one-third of the remainder by bus at 30 km/hr and the rest by cycle at 10 km/hr. The average speed during the entire journey is ______

• A. 22.5 km/hr
• B. 28.5 km/hr
• C. 30.0 km/hr
• D. 32.5 km/hr

Hint:

Assume a convenient total distance, find the time for each leg, then divide total distance by total time.

Answer 1

The Correct Option is A

Step 1: Represent the journey with a variable distance.
Let the total distance be $D$. Half the journey, $\frac{D}{2}$, is covered by train at 90 km/hr, so this leg takes $\frac{D}{180}$ hours.

Step 2: Handle the bus leg symbolically.
The remaining half is $\frac{D}{2}$. One-third of that is $\frac{D}{6}$, covered by bus at 30 km/hr, taking $\frac{D}{180}$ hours as well, since $\frac{D/6}{30} = \frac{D}{180}$.

Step 3: Handle the cycle leg symbolically.
What's left after the train and bus legs is $\frac{D}{2} - \frac{D}{6} = \frac{D}{3}$, covered by cycle at 10 km/hr, taking $\frac{D/3}{10} = \frac{D}{30}$ hours.

Step 4: Add the times and form the average speed.
Total time $= \frac{D}{180} + \frac{D}{180} + \frac{D}{30} = \frac{D}{180} + \frac{D}{180} + \frac{6D}{180} = \frac{8D}{180} = \frac{2D}{45}$. Average speed $= \frac{D}{\text{total time}} = \frac{D}{2D/45} = \frac{45}{2} = 22.5$ km/hr.

Final Answer:
Since $D$ cancels out completely, the average speed is always 22.5 km/hr no matter the actual distance. \[ \boxed{22.5 \text{ km/hr}} \]