If the first one - third of q journey is travelled at 20 km `h^(-1)` , Next one - third at `40 km h^(-1)` And the last one third at `60 km h^(-1)` then the average speed for the whole journey will be

Options

A. `32.7 km h^(-1)`
B. `35 km h^(-1)`
C. `40 km h ^(-1)`
D. `45 km h ^(-1)`

Answer

To find the average speed for the entire journey, we will follow these steps:

### Step 1: Define the total distance
Let the total distance of the journey be \( 3D \), where \( D \) is the distance for each of the three equal parts of the journey.

### Step 2: Calculate the time taken for each part of the journey
- For the first one-third of the journey (distance \( D \)) at a speed of \( 20 \, \text{km/h} \):
\[ t_1 = \frac{D}{20} \]

- For the second one-third of the journey (distance \( D \)) at a speed of \( 40 \, \text{km/h} \):
\[ t_2 = \frac{D}{40} \]

- For the last one-third of the journey (distance \( D \)) at a speed of \( 60 \, \text{km/h} \):
\[ t_3 = \frac{D}{60} \]

### Step 3: Calculate the total time taken for the journey
The total time \( T \) taken for the entire journey is the sum of the times for each part:
\[ T = t_1 + t_2 + t_3 = \frac{D}{20} + \frac{D}{40} + \frac{D}{60} \]

### Step 4: Find a common denominator and simplify the total time
The least common multiple of \( 20, 40, \) and \( 60 \) is \( 120 \). We can express each term with a denominator of \( 120 \):
\[ T = \frac{6D}{120} + \frac{3D}{120} + \frac{2D}{120} = \frac{(6 + 3 + 2)D}{120} = \frac{11D}{120} \]

### Step 5: Calculate the average speed
The average speed \( V_{avg} \) is defined as the total distance divided by the total time:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{3D}{T} = \frac{3D}{\frac{11D}{120}} = \frac{3D \cdot 120}{11D} \]
Here, \( D \) cancels out:
\[ V_{avg} = \frac{360}{11} \approx 32.727 \, \text{km/h} \]

### Step 6: Round the result
Thus, the average speed for the whole journey is approximately:
\[ V_{avg} \approx 32.7 \, \text{km/h} \]

### Final Answer
The average speed for the whole journey is approximately \( 32.7 \, \text{km/h} \).

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