At `60%` out its usual speed, a train of length L metres crosses a platform 240 metres long in 15 seconds. At its usual speed, the train crosses a pole in 6 seconds. What is the value of L?( in metres)

Options

A. 270
B. 225
C. 220
D. 480

Answer

To solve the problem, we need to find the length of the train (L) based on the information given about its speed and the time taken to cross a platform and a pole.

### Step-by-Step Solution:

1. **Determine the speed at 60% of its usual speed:**
Let the usual speed of the train be \( V \) m/s. At 60% of its usual speed, the speed becomes:
\[ \text{Speed at 60%} = 0.6V \]

2. **Calculate the total distance covered while crossing the platform:**
When the train crosses a platform of length 240 meters, the total distance covered is the length of the train plus the length of the platform:
\[ \text{Total distance} = L + 240 \text{ meters} \]

3. **Use the time taken to find the speed at 60%:**
The train crosses the platform in 15 seconds. Using the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
we can write:
\[ L + 240 = 0.6V \times 15 \]
Simplifying this gives:
\[ L + 240 = 9V \quad \text{(Equation 1)} \]

4. **Determine the speed at usual speed:**
The train crosses a pole in 6 seconds at its usual speed \( V \):
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
Therefore:
\[ L = V \times 6 \]
Simplifying this gives:
\[ L = 6V \quad \text{(Equation 2)} \]

5. **Substitute Equation 2 into Equation 1:**
Now, substitute \( L = 6V \) into Equation 1:
\[ 6V + 240 = 9V \]

6. **Solve for V:**
Rearranging the equation gives:
\[ 240 = 9V - 6V \]
\[ 240 = 3V \]
\[ V = \frac{240}{3} = 80 \text{ m/s} \]

7. **Find the length of the train (L):**
Substitute \( V \) back into Equation 2:
\[ L = 6V = 6 \times 80 = 480 \text{ meters} \]

### Final Answer:
The length of the train \( L \) is **480 meters**.