A train met with an accident 120 km from station A. It completed the remaining journey at 5/6 of its previous speed and reached 2 hours late at station B. Had the accident taken place 300 km further, it would have been only 1 hour late? What is the speed of the train?

Options

A. 100 km/h
B. 120 km/h
C. 60 km/h
D. 50 km/h

Answer

To solve the problem step by step, we will break down the information given and derive the speed of the train.

### Step 1: Define Variables
Let the original speed of the train be \( v \) km/h.

### Step 2: Understand the Journey
1. The train travels 120 km to point C before the accident.
2. After the accident, it travels the remaining distance to station B at a speed of \( \frac{5}{6}v \).

### Step 3: Calculate Time Taken in Each Scenario
1. **First Scenario (Accident at C)**:
- Distance from C to B = \( d \) km (unknown).
- Time taken from A to C = \( \frac{120}{v} \) hours.
- Time taken from C to B = \( \frac{d}{\frac{5}{6}v} = \frac{6d}{5v} \) hours.
- Total time = \( \frac{120}{v} + \frac{6d}{5v} \).
- The train is 2 hours late, so:
\[ \frac{120}{v} + \frac{6d}{5v} = T + 2 \]
where \( T \) is the normal time taken to reach B.

2. **Second Scenario (Accident at D, 300 km further)**
- Distance from C to D = 300 km.
- Time taken from A to C = \( \frac{120}{v} \).
- Time taken from C to D = \( \frac{300}{v} \).
- Time taken from D to B = \( \frac{d - 300}{\frac{5}{6}v} = \frac{6(d - 300)}{5v} \).
- Total time = \( \frac{120}{v} + \frac{300}{v} + \frac{6(d - 300)}{5v} \).
- The train is 1 hour late, so:
\[ \frac{120 + 300}{v} + \frac{6(d - 300)}{5v} = T + 1 \]

### Step 4: Set Up Equations
From the two scenarios, we can set up the following equations:

1. **First Scenario**:
\[ \frac{120}{v} + \frac{6d}{5v} = T + 2 \quad (1) \]

2. **Second Scenario**:
\[ \frac{420}{v} + \frac{6(d - 300)}{5v} = T + 1 \quad (2) \]

### Step 5: Eliminate T
Subtract equation (1) from equation (2):
\[ \left(\frac{420}{v} + \frac{6(d - 300)}{5v}\right) - \left(\frac{120}{v} + \frac{6d}{5v}\right) = 1 \]
This simplifies to:
\[ \frac{300}{v} + \frac{6(d - 300) - 6d}{5v} = 1 \]
\[ \frac{300}{v} - \frac{1800}{5v} = 1 \]
\[ \frac{300}{v} - \frac{360}{v} = 1 \]
\[ -\frac{60}{v} = 1 \quad \Rightarrow \quad v = 60 \text{ km/h} \]

### Conclusion
The speed of the train is **60 km/h**.