A person travels certain distance at `4(1)/(6)m//sec` and returns to the starting point at 12km/h. His average speed (in km/h) is :

Options

A. `12_(2)^(1)`
B. `13_(3)^(1)`
C. `13_(2)^(1)`
D. `10_(3)^(1)`

Answer

To find the average speed of the person who travels a certain distance at two different speeds, we can use the formula for average speed when the distance is the same for both parts of the journey.

### Step-by-Step Solution:

1. **Convert the first speed from m/s to km/h**:
- The first speed is given as \(4 \frac{1}{6} \text{ m/s}\).
- Convert \(4 \frac{1}{6}\) to an improper fraction:
\[ 4 \frac{1}{6} = \frac{25}{6} \text{ m/s} \]
- To convert m/s to km/h, we multiply by \(18/5\):
\[ \text{Speed in km/h} = \frac{25}{6} \times \frac{18}{5} = \frac{25 \times 18}{6 \times 5} = \frac{450}{30} = 15 \text{ km/h} \]

2. **Identify the second speed**:
- The second speed is given as \(12 \text{ km/h}\).

3. **Use the average speed formula**:
- The formula for average speed when the distance is the same is:
\[ \text{Average Speed} = \frac{2 \times S_1 \times S_2}{S_1 + S_2} \]
- Here, \(S_1 = 15 \text{ km/h}\) and \(S_2 = 12 \text{ km/h}\).

4. **Plug in the values**:
- Calculate the average speed:
\[ \text{Average Speed} = \frac{2 \times 15 \times 12}{15 + 12} = \frac{360}{27} = \frac{120}{9} = 13 \frac{1}{3} \text{ km/h} \]

### Final Answer:
The average speed of the person is \(13 \frac{1}{3} \text{ km/h}\).

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