If a man walks at the rate of 5 km/hour he misses a train by 7 minutes. However if he walks at the rate of 6km/hour he reaches the station 5 minutes before the arrival of the train. The distance covered by him to reach the station is

Options

A. 6km
B. 7km
C. 6.25km
D. 4km

Answer

To solve the problem, we will denote the distance to the station as \( D \) kilometers and the time taken to reach the station when he walks at different speeds.

1. **Understanding the Problem**:
- When the man walks at 5 km/h, he misses the train by 7 minutes. This means he arrives 7 minutes late.
- When he walks at 6 km/h, he arrives 5 minutes early.

2. **Convert Time to Hours**:
- 7 minutes = \( \frac{7}{60} \) hours
- 5 minutes = \( \frac{5}{60} \) hours

3. **Setting Up the Equations**:
- Let \( T \) be the time (in hours) when the train arrives at the station.
- When walking at 5 km/h:
\[ \text{Time taken} = \frac{D}{5} \]
Since he misses the train by 7 minutes:
\[ \frac{D}{5} + \frac{7}{60} = T \quad \text{(1)} \]

- When walking at 6 km/h:
\[ \text{Time taken} = \frac{D}{6} \]
Since he arrives 5 minutes early:
\[ \frac{D}{6} - \frac{5}{60} = T \quad \text{(2)} \]

4. **Equating the Two Expressions for T**:
- From equation (1):
\[ T = \frac{D}{5} + \frac{7}{60} \]
- From equation (2):
\[ T = \frac{D}{6} - \frac{5}{60} \]
- Setting them equal to each other:
\[ \frac{D}{5} + \frac{7}{60} = \frac{D}{6} - \frac{5}{60} \]

5. **Clearing the Fractions**:
- Multiply through by 60 to eliminate the denominators:
\[ 60 \left(\frac{D}{5}\right) + 7 = 60 \left(\frac{D}{6}\right) - 5 \]
\[ 12D + 7 = 10D - 5 \]

6. **Solving for D**:
- Rearranging gives:
\[ 12D - 10D = -5 - 7 \]
\[ 2D = -12 \]
\[ D = 6 \text{ km} \]

Thus, the distance covered by him to reach the station is **6 kilometers**.