A train crosses two personş moving in opposite direction with speed 12.5 m/sec and 72 km/hr. in 20 sec. and 0.25 minute respectively. Then find the length of train.

Options

A. 240 m
B. 300 m
C. 450 m
D. 555 m

Answer

To find the length of the train that crosses two persons moving in opposite directions, we can follow these steps:

### Step-by-Step Solution:

1. **Convert the speed of the second person from km/hr to m/sec**:
- The speed of the second person is given as 72 km/hr.
- To convert km/hr to m/sec, we use the formula:
\[ \text{Speed in m/sec} = \text{Speed in km/hr} \times \frac{5}{18} \]
- Therefore,
\[ 72 \times \frac{5}{18} = 20 \text{ m/sec} \]

2. **Identify the speeds of both persons**:
- First person’s speed = 12.5 m/sec.
- Second person’s speed = 20 m/sec.

3. **Convert the time taken by the second person into seconds**:
- The time taken by the second person is given as 0.25 minutes.
- To convert minutes to seconds:
\[ 0.25 \text{ min} \times 60 \text{ sec/min} = 15 \text{ sec} \]

4. **Let the length of the train be \( x \) meters**.

5. **Calculate the effective speed of the train when crossing the first person**:
- Since the train and the first person are moving in opposite directions, their speeds add up.
- Effective speed when crossing the first person = Speed of train + Speed of first person = \( y + 12.5 \) m/sec.
- The distance (length of the train) is covered in 20 seconds:
\[ x = (y + 12.5) \times 20 \quad \text{(1)} \]

6. **Calculate the effective speed of the train when crossing the second person**:
- Effective speed when crossing the second person = Speed of train + Speed of second person = \( y + 20 \) m/sec.
- The distance (length of the train) is covered in 15 seconds:
\[ x = (y + 20) \times 15 \quad \text{(2)} \]

7. **Set the two equations for \( x \) equal to each other**:
- From (1):
\[ x = 20y + 250 \quad \text{(3)} \]
- From (2):
\[ x = 15y + 300 \quad \text{(4)} \]

8. **Equate equations (3) and (4)**:
\[ 20y + 250 = 15y + 300 \]

9. **Solve for \( y \)**:
- Rearranging gives:
\[ 20y - 15y = 300 - 250 \]
\[ 5y = 50 \]
\[ y = 10 \text{ m/sec} \]

10. **Substitute \( y \) back to find \( x \)**:
- Using equation (3):
\[ x = 20(10) + 250 = 200 + 250 = 450 \text{ meters} \]

### Final Answer:
The length of the train is **450 meters**.