To solve this problem, we first need to calculate the speeds of the two trains and then determine the time they meet. Let the distance between Station A and Station B be $D$ (in kilometers). Here’s the step-by-step process:

Step 1: Calculate the speed of the first train

The first train travels from A to B in 4 hours (from 5:00 am to 9:00 am). Its speed is: $\text{Speed}_1 = \frac{D}{4} \, \text{km/h}$

Step 2: Calculate the speed of the second train

The second train travels from B to A in $3.5$ hours (from 7:00 am to 10:30 am). Its speed is: $\text{Speed}_2 = \frac{D}{3.5} = \frac{2D}{7} \, \text{km/h}$

Step 3: Write the relative positions as functions of time

Let $t$ represent the time (in hours) after 7:00 am when the two trains meet. At this moment:

1. The first train has already been traveling for $t + 2$ hours (it started at 5:00 am).
2. The second train has been traveling for $t$ hours (it started at 7:00 am).

The distance covered by the first train is: $\text{Distance}_1 = \text{Speed}_1 \times (t + 2) = \frac{D}{4} \times (t + 2)$

The distance covered by the second train is: $\text{Distance}_2 = \text{Speed}_2 \times t = \frac{2D}{7} \times t$

Since the two trains meet, the sum of these distances equals $D$: $\text{Distance}_1 + \text{Distance}_2 = D$

Step 4: Solve the equation

Substitute the expressions for $\text{Distance}_1$ and $\text{Distance}_2$: $\frac{D}{4} \times (t + 2) + \frac{2D}{7} \times t = D$

Divide through by $D$ (since $D \neq 0$): $\frac{t + 2}{4} + \frac{2t}{7} = 1$

Eliminate the fractions by multiplying through by 28 (the least common multiple of 4 and 7): $7(t + 2) + 8(2t) = 28$

Simplify: $7t + 14 + 16t = 28$

Combine terms: $23t + 14 = 28$

Solve for $t$: $23t = 14 \quad \Rightarrow \quad t = \frac{14}{23} \, \text{hours}$

Step 5: Convert $t$ to minutes

$t = \frac{14}{23} \times 60 \approx 36.52 \, \text{minutes}$

Thus, the two trains meet approximately $36.52 \, \text{minutes}$ after 7:00 am. This is around 7:37 am.

Correct Answer:

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