A train leaves station A at 5 AM and reaches station B at 9 AM on the same day. Another train leaves station B at 7 AM and reaches station A at 10 : 30 AM on the same day. The time at which the two train cross each other is :

Options

A. `8 : 26` AM
B. `7 : 36` AM
C. `7 : 56` AM
D. 0.33333333333333

Answer

To solve the problem, we need to determine the time at which the two trains cross each other. We will first calculate the speeds of both trains and then find the time of crossing.

### Step-by-Step Solution:

1. **Determine the time taken by Train A:**
- Train A leaves Station A at 5 AM and reaches Station B at 9 AM.
- Time taken = 9 AM - 5 AM = 4 hours.

2. **Determine the time taken by Train B:**
- Train B leaves Station B at 7 AM and reaches Station A at 10:30 AM.
- Time taken = 10:30 AM - 7 AM = 3.5 hours.

3. **Assume the distance between Station A and Station B is D.**
- Speed of Train A = Distance / Time = D / 4 hours.
- Speed of Train B = Distance / Time = D / 3.5 hours.

4. **Convert the speeds to a common format:**
- Speed of Train A = D / 4 = 0.25D km/h.
- Speed of Train B = D / 3.5 = (2/7)D km/h.

5. **Calculate the distance covered by each train until they meet:**
- Let the time taken until they meet be t hours after 5 AM.
- Distance covered by Train A in t hours = Speed of Train A × t = (0.25D) × t.
- Distance covered by Train B in (t - 2) hours (since Train B starts at 7 AM) = Speed of Train B × (t - 2) = (2/7)D × (t - 2).

6. **Set up the equation based on the total distance:**
- The total distance covered by both trains when they meet is equal to D.
- Therefore, we have:
\[ (0.25D) \cdot t + \left(\frac{2}{7}D\right) \cdot (t - 2) = D \]

7. **Simplify the equation:**
- Dividing through by D (assuming D ≠ 0):
\[ 0.25t + \frac{2}{7}(t - 2) = 1 \]
- Expanding the equation:
\[ 0.25t + \frac{2}{7}t - \frac{4}{7} = 1 \]

8. **Combine like terms:**
- To combine the terms, convert 0.25 to a fraction:
\[ \frac{1}{4} = \frac{7}{28} \]
- Convert \(\frac{2}{7}\) to a fraction with a denominator of 28:
\[ \frac{2}{7} = \frac{8}{28} \]
- Now the equation becomes:
\[ \frac{7}{28}t + \frac{8}{28}t - \frac{4}{7} = 1 \]
- Combine the t terms:
\[ \frac{15}{28}t - \frac{4}{7} = 1 \]

9. **Solve for t:**
- Convert \(-\frac{4}{7}\) to a fraction with a denominator of 28:
\[ -\frac{4}{7} = -\frac{16}{28} \]
- Now the equation is:
\[ \frac{15}{28}t + \frac{16}{28} = 1 \]
- Rearranging gives:
\[ \frac{15}{28}t = 1 - \frac{16}{28} = \frac{12}{28} \]
- Therefore:
\[ t = \frac{12}{28} \cdot \frac{28}{15} = \frac{12}{15} = \frac{4}{5} \text{ hours} = 48 \text{ minutes} \]

10. **Determine the crossing time:**
- Since Train A leaves at 5 AM, they cross at:
\[ 5:00 AM + 0:48 = 5:48 AM \]

### Final Answer:
The two trains cross each other at **5:48 AM**.