Question

A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train's departure. The distance (in km) from his home to the railway station is

• A. 26
• B. 18
• C. 16
• D. 20

Answer 1

The Correct Option is D

Step 1: Define variables

Let the distance from the man’s home to the station be: \[ d \ \text{km} \] Let \( t \) (in hours) be the time of the train’s departure after he leaves home.

Step 2: First scenario — walking at 12 km/h

He arrives 10 minutes late (that is, \( \frac{1}{6} \) hour after the train departs). Thus: \[ t + \frac{1}{6} = \frac{d}{12} \]

Step 3: Second scenario — walking at 15 km/h

He arrives 10 minutes early (that is, \( \frac{1}{6} \) hour before the train departs). Thus: \[ t - \frac{1}{6} = \frac{d}{15} \]

Step 4: Form the system of equations

We now have: \[ \begin{cases} t + \frac{1}{6} = \frac{d}{12} \quad (1) \\ t - \frac{1}{6} = \frac{d}{15} \quad (2) \end{cases} \]

Step 5: Subtract equation (2) from equation (1)

\[ \left(t + \frac{1}{6}\right) - \left(t - \frac{1}{6}\right) = \frac{d}{12} - \frac{d}{15} \] \[ \frac{1}{3} = \frac{5d - 4d}{60} \] \[ \frac{1}{3} = \frac{d}{60} \]

Step 6: Solve for \( d \)

Multiply through by 60: \[ 20 = d \]

Final Answer:

\[ \boxed{\text{The distance is 20 km.}} \]