A train starts from a station and after traveling 100 km meets with an accident. And then the speed of the train reduces by 1/4th of its former speed. And travelling the remaining distance it reaches to its destination `1(7)/(8) ` hours late . If the accident occured 60 Km ahead then it reaches 15 min earlier . The find its original speed and the distance of its journey ?

Answer

To solve the problem, we need to find the original speed of the train and the total distance of its journey. Let's break down the problem step by step.

### Step 1: Define Variables
Let the original speed of the train be \( S \) km/h. The total distance of the journey is \( D \) km.

### Step 2: Understand the Journey
1. The train travels 100 km before the accident.
2. After the accident, the speed reduces to \( \frac{3}{4}S \) km/h.
3. The train is 1 hour and 7/8 hours (which is 1.875 hours) late when the accident occurs 60 km ahead of the 100 km mark.

### Step 3: Calculate Time Taken
- **Time taken to travel the first 100 km**:
\[ \text{Time} = \frac{100}{S} \]

- **Distance remaining after the accident**:
\[ \text{Remaining distance} = D - 100 \text{ km} \]

- **Time taken to travel the remaining distance at reduced speed**:
\[ \text{Time} = \frac{D - 100}{\frac{3}{4}S} = \frac{4(D - 100)}{3S} \]

### Step 4: Total Time Calculation
The total time taken by the train after the accident is:
\[ \text{Total time} = \frac{100}{S} + \frac{4(D - 100)}{3S} \]

### Step 5: Calculate the Late Time
The train is 1.875 hours late, so we can set up the equation:
\[ \frac{100}{S} + \frac{4(D - 100)}{3S} = \text{Original time} + 1.875 \]

### Step 6: Calculate the Early Time
When the accident occurs 60 km ahead (i.e., at 160 km), the time taken will be:
- **Time taken to travel 160 km**:
\[ \text{Time} = \frac{160}{S} \]
- **Remaining distance**:
\[ \text{Remaining distance} = D - 160 \]
- **Time taken after the accident**:
\[ \text{Time} = \frac{D - 160}{\frac{3}{4}S} = \frac{4(D - 160)}{3S} \]

The total time taken when the accident occurs 60 km ahead is:
\[ \frac{160}{S} + \frac{4(D - 160)}{3S} \]

### Step 7: Set Up the Early Time Equation
Since the train reaches 15 minutes (0.25 hours) earlier, we can set up the equation:
\[ \frac{160}{S} + \frac{4(D - 160)}{3S} = \text{Original time} - 0.25 \]

### Step 8: Solve the Equations
Now we have two equations:
1. \(\frac{100}{S} + \frac{4(D - 100)}{3S} = \text{Original time} + 1.875\)
2. \(\frac{160}{S} + \frac{4(D - 160)}{3S} = \text{Original time} - 0.25\)

We can solve these equations simultaneously to find \( S \) and \( D \).

### Step 9: Substitute and Simplify
To simplify the calculations, express the original time in terms of \( D \):
\[ \text{Original time} = \frac{D}{S} \]

Substituting this into both equations will allow us to isolate \( S \) and \( D \).

### Step 10: Conclusion
After solving the equations, we will find the values of \( S \) (original speed) and \( D \) (total distance).