A train X running at 72 km/h crosses another train Y running at 54 km/h in opposite direction in 15 seconds. If the length of Y is two-fifth that of X, then what is the length (in meter) of X?

Options

A. 375
B. 150
C. 300
D. 225

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step-by-Step Solution:

1. **Convert Speeds from km/h to m/s**:
- The speed of train X is given as 72 km/h. To convert this to meters per second:
\[ \text{Speed of train X} = 72 \times \frac{5}{18} = 20 \text{ m/s} \]
- The speed of train Y is given as 54 km/h. To convert this to meters per second:
\[ \text{Speed of train Y} = 54 \times \frac{5}{18} = 15 \text{ m/s} \]

2. **Determine the Relative Speed**:
- Since the trains are moving in opposite directions, their speeds will be added:
\[ \text{Relative Speed} = \text{Speed of train X} + \text{Speed of train Y} = 20 \text{ m/s} + 15 \text{ m/s} = 35 \text{ m/s} \]

3. **Set Up the Length Relationship**:
- Let the length of train X be \( L_X \) and the length of train Y be \( L_Y \).
- According to the problem, the length of train Y is two-fifths that of train X:
\[ L_Y = \frac{2}{5} L_X \]

4. **Use the Time Taken to Cross Each Other**:
- The time taken for the two trains to cross each other is given as 15 seconds. The total distance covered when they cross each other is the sum of their lengths:
\[ \text{Distance} = L_X + L_Y \]
- Using the relative speed and time, we can set up the equation:
\[ \text{Distance} = \text{Relative Speed} \times \text{Time} \]
\[ L_X + L_Y = 35 \text{ m/s} \times 15 \text{ s} = 525 \text{ m} \]

5. **Substituting the Length Relationship**:
- Substitute \( L_Y = \frac{2}{5} L_X \) into the equation:
\[ L_X + \frac{2}{5} L_X = 525 \]
- Combine like terms:
\[ \frac{5}{5} L_X + \frac{2}{5} L_X = 525 \]
\[ \frac{7}{5} L_X = 525 \]

6. **Solve for \( L_X \)**:
- Multiply both sides by \( \frac{5}{7} \):
\[ L_X = 525 \times \frac{5}{7} = 375 \text{ m} \]

### Final Answer:
The length of train X is **375 meters**.