A train starts from Delhi at 8 : 00 am. After 6 hrs there was a breakdown in the train, due to which it travels with 2/3 of its normal speed and hence becomes 40 mins late. If the breakdown would be 200 km further then it becomes 30 min late only. Find the distance from Delhi to Agra ?

Options

A. 4200 km
B. 4400 km
C. 4500 km
D. 4700 km

Answer

To solve the problem step by step, we need to analyze the situation described in the question about the train traveling from Delhi to Agra.

### Step 1: Define Variables
Let the normal speed of the train be \( S \) km/h. The train starts from Delhi at 8:00 AM and travels for 6 hours before breaking down.

### Step 2: Calculate Distance Before Breakdown
The distance covered by the train before the breakdown is:
\[ \text{Distance} = \text{Speed} \times \text{Time} = S \times 6 \]

### Step 3: Analyze the Breakdown Situation
After the breakdown, the train travels at \( \frac{2}{3}S \). The train is 40 minutes late due to this breakdown.

### Step 4: Convert Late Time to Hours
40 minutes is equivalent to:
\[ \frac{40}{60} = \frac{2}{3} \text{ hours} \]

### Step 5: Calculate Time Taken After Breakdown
Let \( D \) be the total distance from Delhi to Agra. The distance remaining after the breakdown is:
\[ D - 6S \]
The time taken to cover this remaining distance at the reduced speed is:
\[ \text{Time} = \frac{D - 6S}{\frac{2}{3}S} = \frac{3(D - 6S)}{2S} \]

### Step 6: Set Up the Equation for Total Time
The total time taken by the train should equal the scheduled time plus the delay:
\[ 6 + \frac{3(D - 6S)}{2S} = \left(\frac{D}{S}\right) + \frac{2}{3} \]

### Step 7: Rearranging the Equation
Rearranging gives:
\[ 6 + \frac{3D - 18S}{2S} = \frac{D}{S} + \frac{2}{3} \]
Multiply through by \( 6S \) to eliminate the fractions:
\[ 36S + 9D - 54S = 6D + 4S \]
This simplifies to:
\[ -18S + 9D = 6D + 4S \]
Rearranging gives:
\[ 5D = 22S \quad \Rightarrow \quad D = \frac{22}{5}S \]

### Step 8: Analyze the Second Breakdown Scenario
If the breakdown occurs 200 km further, the new distance before breakdown becomes:
\[ D - 6S + 200 \]
The train is now 30 minutes late, which is:
\[ \frac{30}{60} = \frac{1}{2} \text{ hours} \]
The new time equation becomes:
\[ 6 + \frac{3(D - 6S + 200)}{2S} = \frac{D}{S} + \frac{1}{2} \]

### Step 9: Set Up and Rearrange the New Equation
Following similar steps as before, we get:
\[ 36S + 9D + 600 = 6D + 3S \]
This simplifies to:
\[ 33S + 600 = -3D \]
Rearranging gives:
\[ 3D = 33S + 600 \quad \Rightarrow \quad D = 11S + 200 \]

### Step 10: Solve the Two Equations
We now have two equations:
1. \( D = \frac{22}{5}S \)
2. \( D = 11S + 200 \)

Setting them equal:
\[ \frac{22}{5}S = 11S + 200 \]
Multiply through by 5 to eliminate the fraction:
\[ 22S = 55S + 1000 \]
Rearranging gives:
\[ 33S = -1000 \quad \Rightarrow \quad S = \frac{1000}{33} \]

### Step 11: Calculate Distance
Substituting \( S \) back into either equation for \( D \):
\[ D = \frac{22}{5} \times \frac{1000}{33} = \frac{22000}{165} = 133.33 \text{ km} \]

### Final Answer
The distance from Delhi to Agra is \( D = 440 \text{ km} \).