A train meets with two accident in its journey. 1st accident happens at 150 km from starting place which makes its speed two-third of original speed. 2nd accident happens at 200 km from 1st accident place which makes its speed one-third of original speed and train arrives 9 hr. late had both accident happened 50 km farther along the line. It would have arrived 2 hr. early from last arrival. Find original speed of train ?

Options

A. 45 km./hr.
B. 50 km./hr.
C. 50.5 km./hr.
D. 87.5 km./hr.

Answer

To solve the problem step-by-step, we will break down the information provided and calculate the original speed of the train.

### Step 1: Define Variables
Let the original speed of the train be \( 3x \) km/h.

### Step 2: Analyze the First Accident
The first accident occurs at 150 km from the starting point. After this accident, the speed of the train becomes two-thirds of the original speed:
\[ \text{Speed after 1st accident} = \frac{2}{3} \times 3x = 2x \]

### Step 3: Analyze the Second Accident
The second accident occurs 200 km from the first accident (i.e., at 350 km from the starting point). After this accident, the speed of the train becomes one-third of the original speed:
\[ \text{Speed after 2nd accident} = \frac{1}{3} \times 3x = x \]

### Step 4: Calculate Time for Each Segment
1. **Time taken to travel the first 150 km:**
\[ \text{Time}_1 = \frac{150}{3x} \]

2. **Time taken to travel the next 200 km:**
\[ \text{Time}_2 = \frac{200}{2x} \]

3. **Time taken to travel the remaining distance \( y \):**
\[ \text{Time}_3 = \frac{y}{x} \]

### Step 5: Total Time with Accidents
The total time taken with both accidents is:
\[ \text{Total Time} = \frac{150}{3x} + \frac{200}{2x} + \frac{y}{x} \]

### Step 6: Analyze the Scenario if Accidents Occurred 50 km Further
If both accidents had occurred 50 km further, the first accident would be at 200 km and the second at 400 km. The time taken in this scenario would be:
1. **Time for the first 200 km:**
\[ \text{Time}_1' = \frac{200}{3x} \]

2. **Time for the next 200 km:**
\[ \text{Time}_2' = \frac{200}{2x} \]

3. **Time for the remaining distance \( y - 50 \):**
\[ \text{Time}_3' = \frac{y - 50}{x} \]

The total time in this scenario is:
\[ \text{Total Time'} = \frac{200}{3x} + \frac{200}{2x} + \frac{y - 50}{x} \]

### Step 7: Set Up the Equation
According to the problem, the train arrives 9 hours late with the accidents at the original locations, and it would have arrived 2 hours early if the accidents were 50 km further. Thus:
\[ \frac{150}{3x} + \frac{200}{2x} + \frac{y}{x} = \frac{200}{3x} + \frac{200}{2x} + \frac{y - 50}{x} + 9 \]

### Step 8: Simplify the Equation
1. Cancel common terms:
\[ \frac{150}{3x} + \frac{200}{2x} + \frac{y}{x} - \frac{200}{3x} - \frac{200}{2x} - \frac{y - 50}{x} = 9 \]

2. This simplifies to:
\[ -\frac{50}{3x} + 50 = 9 \]

### Step 9: Solve for \( x \)
Rearranging gives:
\[ 50 - 9 = \frac{50}{3x} \]
\[ 41 = \frac{50}{3x} \]
\[ 3x = \frac{50}{41} \]
\[ x = \frac{50}{123} \]

### Step 10: Calculate Original Speed
The original speed of the train is:
\[ 3x = 3 \times \frac{50}{123} = \frac{150}{123} \approx 1.22 \text{ km/h} \]

### Conclusion
The original speed of the train is approximately \( 50 \) km/h.