A train met with an accident 120 km from station A it completed the remaining journey at 5/6 of its previous speed and reached 2 hours late at station B had the accident taken place 300 km further it would have been only 1 hour late? what is the total distance between A and B?

Options

A. 480 km
B. 520 km
C. 600 km
D. 720 km

Answer

To solve the problem, we need to analyze the situation step by step.

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

### Step 2: Analyze the First Scenario
The train meets with an accident 120 km from station A. After the accident, it travels the remaining distance \( D - 120 \) km at a speed of \( \frac{5}{6}S \).

- Time taken to travel the first 120 km:
\[ \text{Time}_1 = \frac{120}{S} \]

- Time taken to travel the remaining distance \( D - 120 \) km:
\[ \text{Time}_2 = \frac{D - 120}{\frac{5}{6}S} = \frac{6(D - 120)}{5S} \]

- Total time taken after the accident:
\[ \text{Total Time}_1 = \frac{120}{S} + \frac{6(D - 120)}{5S} \]

### Step 3: Calculate the Delay
According to the problem, the train is 2 hours late, so we can express this as:
\[ \text{Total Time}_1 = \text{Normal Time} + 2 \]
Where Normal Time to cover distance \( D \) at speed \( S \) is:
\[ \text{Normal Time} = \frac{D}{S} \]
Thus, we have:
\[ \frac{120}{S} + \frac{6(D - 120)}{5S} = \frac{D}{S} + 2 \]

### Step 4: Simplify the Equation
Multiplying through by \( S \) to eliminate the denominator:
\[ 120 + \frac{6(D - 120)}{5} = D + 2S \]
Simplifying gives:
\[ 120 + \frac{6D - 720}{5} = D + 2S \]
Multiplying through by 5 to eliminate the fraction:
\[ 600 + 6D - 720 = 5D + 10S \]
Rearranging gives:
\[ D - 10S = 120 \]
(Equation 1)

### Step 5: Analyze the Second Scenario
Now, if the accident had occurred 300 km further (i.e., at 420 km from A), the remaining distance would be \( D - 420 \) km, and the time taken would be:
\[ \text{Time}_3 = \frac{420}{S} + \frac{6(D - 420)}{5S} \]
This time, the train is only 1 hour late:
\[ \text{Total Time}_2 = \frac{D}{S} + 1 \]
Thus:
\[ \frac{420}{S} + \frac{6(D - 420)}{5S} = \frac{D}{S} + 1 \]

### Step 6: Simplify the Second Equation
Multiplying through by \( S \):
\[ 420 + \frac{6(D - 420)}{5} = D + S \]
Again, multiplying through by 5:
\[ 2100 + 6D - 2520 = 5D + 5S \]
Rearranging gives:
\[ D - 5S = 420 \]
(Equation 2)

### Step 7: Solve the System of Equations
Now we have two equations:
1. \( D - 10S = 120 \)
2. \( D - 5S = 420 \)

Subtracting Equation 1 from Equation 2:
\[ (D - 5S) - (D - 10S) = 420 - 120 \]
This simplifies to:
\[ 5S = 300 \implies S = 60 \text{ km/h} \]

### Step 8: Substitute Back to Find D
Using \( S = 60 \) in Equation 1:
\[ D - 10(60) = 120 \implies D - 600 = 120 \implies D = 720 \text{ km} \]

### Final Answer
The total distance between stations A and B is **720 km**.

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