A train after travelling 50 km meets with an accident and then proceeds at `(3)/(4)` of its former speed and arrives as its destination 35 minutes late. Had the accident occurred 24km further, it would have reached the destination only 25 minutes late. Find the speed of the train and the distance which the train traveles ?

Answer

To solve the problem step by step, let's denote the initial speed of the train as \( S \) km/h.

### Step 1: Define the Variables
Let:
- \( S \) = initial speed of the train (in km/h)
- The speed after the accident = \( \frac{3}{4}S \) km/h
- Distance covered before the accident = 50 km
- Distance after the accident = \( D - 50 \) km (where \( D \) is the total distance to the destination)

### Step 2: Time Taken in Two Scenarios
1. **First Scenario (Accident after 50 km)**:
- Time taken to travel the first 50 km = \( \frac{50}{S} \) hours
- Time taken to travel the remaining distance \( D - 50 \) km at \( \frac{3}{4}S \) = \( \frac{D - 50}{\frac{3}{4}S} = \frac{4(D - 50)}{3S} \) hours
- Total time taken = \( \frac{50}{S} + \frac{4(D - 50)}{3S} \)
- Since the train is 35 minutes late, we convert 35 minutes to hours: \( \frac{35}{60} = \frac{7}{12} \) hours.
- Therefore, the equation becomes:
\[ \frac{50}{S} + \frac{4(D - 50)}{3S} = T + \frac{7}{12} \]

2. **Second Scenario (Accident after 74 km)**:
- Time taken to travel the first 74 km = \( \frac{74}{S} \) hours
- Time taken to travel the remaining distance \( D - 74 \) km at \( \frac{3}{4}S \) = \( \frac{4(D - 74)}{3S} \) hours
- Total time taken = \( \frac{74}{S} + \frac{4(D - 74)}{3S} \)
- Since the train is 25 minutes late, we convert 25 minutes to hours: \( \frac{25}{60} = \frac{5}{12} \) hours.
- Therefore, the equation becomes:
\[ \frac{74}{S} + \frac{4(D - 74)}{3S} = T + \frac{5}{12} \]

### Step 3: Set Up the Equations
From the two scenarios, we can set up the following equations:

1. \( \frac{50}{S} + \frac{4(D - 50)}{3S} = T + \frac{7}{12} \)
2. \( \frac{74}{S} + \frac{4(D - 74)}{3S} = T + \frac{5}{12} \)

### Step 4: Eliminate \( T \)
Subtract the second equation from the first:
\[ \left(\frac{50}{S} - \frac{74}{S}\right) + \left(\frac{4(D - 50)}{3S} - \frac{4(D - 74)}{3S}\right) = \frac{7}{12} - \frac{5}{12} \]
This simplifies to:
\[ -\frac{24}{S} + \frac{4(74 - 50)}{3S} = \frac{2}{12} \]
\[ -\frac{24}{S} + \frac{96}{3S} = \frac{1}{6} \]
\[ -\frac{24}{S} + \frac{32}{S} = \frac{1}{6} \]
\[ \frac{8}{S} = \frac{1}{6} \]
Thus, solving for \( S \):
\[ S = 48 \text{ km/h} \]

### Step 5: Find Total Distance \( D \)
Now, substituting \( S \) back into one of the original equations to find \( D \):
Using the first scenario:
\[ \frac{50}{48} + \frac{4(D - 50)}{3 \times 48} = T + \frac{7}{12} \]
Let’s assume \( T = \frac{D}{S} \):
\[ \frac{50}{48} + \frac{4(D - 50)}{144} = \frac{D}{48} + \frac{7}{12} \]
Solving this will give us the total distance \( D \).

### Final Calculation
After solving, we find:
- The total distance \( D = 134 \) km.

### Conclusion
The speed of the train is \( 48 \) km/h, and the total distance it travels is \( 134 \) km.

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