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

To find the average speed over the entire journey, we need to consider the distances covered and the speeds for each segment of the journey accordingly. 

1. Let's assume the total distance of the journey is \(D\).
2. The first part of the journey is covered by train. The man travels half the total journey by train, that is, \(\frac{D}{2}\), at a speed of 90 km/hr. 
   The time taken for this part is: \(\text{Time}_\text{train} = \frac{\frac{D}{2}}{90} = \frac{D}{180}\)
3. The remaining distance after the train journey is \(\frac{D}{2}\). The man covers one-third of this remaining distance by bus at 30 km/hr. 
   The distance by bus is: \(\frac{\frac{D}{2}}{3} = \frac{D}{6}\)
   The time taken by bus is: \(\text{Time}_\text{bus} = \frac{\frac{D}{6}}{30} = \frac{D}{180}\)
4. The distance covered by cycle is the rest of the journey. 
   The remaining distance is: \(\frac{D}{2} - \frac{D}{6} = \frac{3D - D}{6} = \frac{D}{3}\)
   This is covered by cycle at 10 km/hr. 
   The time taken by cycle is: \(\text{Time}_\text{cycle} = \frac{\frac{D}{3}}{10} = \frac{D}{30}\)
5. Total time taken for the entire journey is the sum of all individual times: \(\text{Total Time} = \frac{D}{180} + \frac{D}{180} + \frac{D}{30} = \frac{D}{180} + \frac{D}{180} + \frac{6D}{180} = \frac{8D}{180} = \frac{4D}{90}\)
6. The average speed for the entire journey is given by: \(\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{D}{\frac{4D}{90}} = \frac{90}{4} = 22.5 \text{ km/hr}\)

Thus, the average speed during the entire journey is 22.5 km/hr.

Answer 2

Step 1: Pick a convenient total distance.
Since the actual distance is not given, assume the total journey is \(D = 180\) km (a number that divides evenly at each stage). Average speed is total distance divided by total time, so whatever value of \(D\) we pick will not change the final answer.

Step 2: Find the train leg.
Half the journey by train \(= 90\) km at 90 km/hr, so time \(t_1 = \frac{90}{90} = 1\) hr. Remaining distance \(= 90\) km.

Step 3: Find the bus leg.
One-third of the remaining 90 km \(= 30\) km, covered by bus at 30 km/hr, so time \(t_2 = \frac{30}{30} = 1\) hr. Remaining distance \(= 90 - 30 = 60\) km.

Step 4: Find the cycle leg.
The rest, 60 km, is covered by cycle at 10 km/hr, so time \(t_3 = \frac{60}{10} = 6\) hr.

Step 5: Compute the average speed.
Total time \(= t_1 + t_2 + t_3 = 1 + 1 + 6 = 8\) hr. Average speed \(= \frac{180}{8} = 22.5\) km/hr.

Final Answer:
The average speed for the whole journey is 22.5 km/hr. \[ \boxed{22.5 \text{ km/hr}} \]