After travelling 5 hours a train meets with an accident. Due to this it has to stop 2 hours. After this the train starts moving `55(5)/(9)`% of its speed , and reaches to its destination `12(2)/(9)` hours late . If the accident had occurred 150 Km ahead on the same line then the train would have reached the destination `10(8)/(9)` hours late . Find the original speed of the train ?

Answer

To find the original speed of the train, we will break down the problem step by step.

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

### Step 2: Calculate Distance Travelled Before Accident
The train travels for 5 hours before the accident, so the distance covered before the accident is:
\[ D_1 = 5S \text{ km} \]

### Step 3: Understand the Delay After the Accident
After the accident, the train stops for 2 hours. Then it continues at \( \frac{55}{9}\% \) of its original speed. To convert this percentage to a fraction:
\[ \frac{55}{9}\% = \frac{55}{900} = \frac{11}{180} \]
Thus, the new speed after the accident is:
\[ S' = S \times \frac{5}{9} = \frac{5S}{9} \text{ km/h} \]

### Step 4: Calculate Total Delay
The train reaches its destination \( 12 \frac{2}{9} \) hours late. Converting this to an improper fraction:
\[ 12 \frac{2}{9} = \frac{110}{9} \text{ hours} \]
Since the train stopped for 2 hours, the effective delay due to the speed reduction is:
\[ \text{Effective Delay} = \frac{110}{9} - 2 = \frac{110}{9} - \frac{18}{9} = \frac{92}{9} \text{ hours} \]

### Step 5: Relate Delay to Speed Ratio
The ratio of speeds before and after the accident is \( 9:5 \), which means the ratio of the time taken is \( 5:9 \). The difference in time (in units) is:
\[ \text{Difference in Time} = 4 \text{ units} \]
Thus, we can set up the equation:
\[ 4 \text{ units} = \frac{92}{9} \text{ hours} \]
From this, we can find the value of 1 unit:
\[ 1 \text{ unit} = \frac{92}{9} \div 4 = \frac{92}{36} = \frac{23}{9} \text{ hours} \]

### Step 6: Calculate Time for 5 Units
Now, we find the time for 5 units:
\[ 5 \text{ units} = 5 \times \frac{23}{9} = \frac{115}{9} \text{ hours} \]

### Step 7: Consider the Scenario of Accident 150 km Ahead
If the accident happened 150 km ahead, the total delay would be \( 10 \frac{8}{9} \) hours:
\[ 10 \frac{8}{9} = \frac{98}{9} \text{ hours} \]
Again, accounting for the 2-hour stop:
\[ \text{Effective Delay} = \frac{98}{9} - 2 = \frac{98}{9} - \frac{18}{9} = \frac{80}{9} \text{ hours} \]

### Step 8: Set Up the Equation for New Delay
Using the same speed ratio, we have:
\[ 4 \text{ units} = \frac{80}{9} \text{ hours} \]
Thus, the value of 1 unit in this case is:
\[ 1 \text{ unit} = \frac{80}{9} \div 4 = \frac{80}{36} = \frac{20}{9} \text{ hours} \]

### Step 9: Calculate Time for 5 Units Again
Now, we find the time for 5 units:
\[ 5 \text{ units} = 5 \times \frac{20}{9} = \frac{100}{9} \text{ hours} \]

### Step 10: Find the Difference in Time
The difference in time between the two scenarios (accident at C and accident 150 km ahead) is:
\[ \text{Difference} = \frac{115}{9} - \frac{100}{9} = \frac{15}{9} \text{ hours} \]

### Step 11: Calculate Original Speed
This time difference corresponds to the distance of 150 km:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{150 \text{ km}}{\frac{15}{9} \text{ hours}} = 150 \times \frac{9}{15} = 90 \text{ km/h} \]

### Final Answer
Thus, the original speed of the train is:
\[ \boxed{90 \text{ km/h}} \]