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 use it to find the speed of the train.

### Step 1: Define Variables
Let the speed of the train before the accident be \( S \) km/h.

### Step 2: Determine the Distance and Time from Point C to B
The train meets with an accident 120 km from station A, which we will call point C. The remaining distance from C to B is \( D \) km.

The train travels from C to B at \( \frac{5}{6}S \) km/h after the accident.

### Step 3: Calculate the Time Taken from C to B
At normal speed \( S \), the time taken to travel from C to B would be:
\[ \text{Time}_{normal} = \frac{D}{S} \]
At reduced speed \( \frac{5}{6}S \), the time taken is:
\[ \text{Time}_{reduced} = \frac{D}{\frac{5}{6}S} = \frac{6D}{5S} \]

### Step 4: Set Up the Equation for Being 2 Hours Late
According to the problem, the train is 2 hours late when it travels from C to B. Therefore, we can set up the equation:
\[ \frac{6D}{5S} = \frac{D}{S} + 2 \]

### Step 5: Simplify the Equation
Multiply through by \( 5S \) to eliminate the denominators:
\[ 6D = 5D + 10S \]
This simplifies to:
\[ D = 10S \]

### Step 6: Determine the Scenario with the Accident at Point D
If the accident had occurred 300 km further (at point D), the distance from D to B would be \( D - 300 \) km. The time taken at normal speed would be:
\[ \text{Time}_{normal} = \frac{D - 300}{S} \]
At reduced speed \( \frac{5}{6}S \):
\[ \text{Time}_{reduced} = \frac{D - 300}{\frac{5}{6}S} = \frac{6(D - 300)}{5S} \]

### Step 7: Set Up the Equation for Being 1 Hour Late
The train is only 1 hour late in this scenario:
\[ \frac{6(D - 300)}{5S} = \frac{D - 300}{S} + 1 \]

### Step 8: Simplify the Second Equation
Multiply through by \( 5S \):
\[ 6(D - 300) = 5(D - 300) + 5S \]
This simplifies to:
\[ 6D - 1800 = 5D - 1500 + 5S \]
\[ D - 1800 + 1500 = 5S \]
\[ D - 300 = 5S \]

### Step 9: Substitute \( D \) from Step 5
From Step 5, we know \( D = 10S \). Substitute this into the equation:
\[ 10S - 300 = 5S \]
\[ 5S = 300 \]
\[ S = 60 \]

### Conclusion
The speed of the train is \( 60 \) km/h.