After travelling 80 km, a train meets with an accident and then proceeds at 3/4 of its former speed and arrives at its destination 35 min late. Had the accident occurred 24 km further, it would have reached the destination only 25 min late. Find the speed of the train.

Options

A. 50km/h
B. 30 km/h
C. 48 km/h
D. 55 km/h

Answer

To solve the problem, let's break it down step by step.

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

### Step 2: Understand the Situation
1. After traveling 80 km, the train meets with an accident and then travels at \( \frac{3}{4}S \).
2. It arrives at its destination 35 minutes late.
3. If the accident occurred 24 km further (i.e., after traveling 104 km), it would arrive only 25 minutes late.

### Step 3: Calculate Time Taken
- **Time taken without accident**: Let \( T \) be the time taken to reach the destination without any accident.
- **Time taken after the accident at 80 km**:
- Distance from 80 km to the destination = \( D - 80 \) (where \( D \) is the total distance).
- Time taken after the accident = \( \frac{D - 80}{\frac{3}{4}S} = \frac{4(D - 80)}{3S} \).

- **Total time with accident**:
\[ T + \frac{4(D - 80)}{3S} = T + \frac{35}{60} \text{ hours} \]

### Step 4: Set Up the Equation
From the above, we can set up the equation:
\[ T + \frac{4(D - 80)}{3S} = T + \frac{7}{12} \]
This simplifies to:
\[ \frac{4(D - 80)}{3S} = \frac{7}{12} \]

### Step 5: Calculate for the Second Scenario
- **Time taken after the accident at 104 km**:
- Distance from 104 km to the destination = \( D - 104 \).
- Time taken after the accident = \( \frac{D - 104}{\frac{3}{4}S} = \frac{4(D - 104)}{3S} \).

- **Total time with accident**:
\[ T + \frac{4(D - 104)}{3S} = T + \frac{25}{60} \text{ hours} \]

### Step 6: Set Up the Second Equation
From the second scenario, we have:
\[ T + \frac{4(D - 104)}{3S} = T + \frac{5}{12} \]
This simplifies to:
\[ \frac{4(D - 104)}{3S} = \frac{5}{12} \]

### Step 7: Solve the Two Equations
Now we have two equations:
1. \( \frac{4(D - 80)}{3S} = \frac{7}{12} \)
2. \( \frac{4(D - 104)}{3S} = \frac{5}{12} \)

From the first equation:
\[ 4(D - 80) = \frac{7}{12} \cdot 3S \implies D - 80 = \frac{21S}{48} \implies D = \frac{21S}{48} + 80 \]

From the second equation:
\[ 4(D - 104) = \frac{5}{12} \cdot 3S \implies D - 104 = \frac{15S}{48} \implies D = \frac{15S}{48} + 104 \]

### Step 8: Set the Two Expressions for D Equal
\[ \frac{21S}{48} + 80 = \frac{15S}{48} + 104 \]

### Step 9: Solve for S
Rearranging gives:
\[ \frac{21S - 15S}{48} = 104 - 80 \]
\[ \frac{6S}{48} = 24 \implies S = 24 \cdot 8 = 192 \text{ km/h} \]

### Final Answer
The speed of the train is **192 km/h**.