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$\text{A}$ and $\text{B}$ are two railway stations $90 \; \text{km}$ apart. A train leaves $\text{A}$ at $9:00 \; \text{am},$ heading towards $\text{B}$ at a speed of $40 \; \text{km/hr}.$ Another train leaves $\text{B}$ at $10:30 \; \text{am},$ heading towards $\text{A}$ at a speed of $20 \; \text{km/hr}.$ The trains meet each other at

1. $11:45 \; \text{am}$
2. $10:45 \; \text{am}$
3. $11:20 \; \text{am}$
4. $11:00 \; \text{am}$

We can draw the diagram for better understanding.

We know that, $\boxed{\text{Speed} = \frac{\text{Distance}}{\text{Time}}}$

Distance travelled by train $1$ in $\text{1:30 hr} = 40 \times \frac{3}{2} = 60 \; \text{km.}$

Time taken  by train $1$ and train $2$  to meet each other $= \frac{30}{60} = \frac{1}{2} \; \text{hr} = 30 \; \text{minutes}$

$\therefore$ The time when trains meet each other $= \text{10:30 am} + \text{30 minutes} = \text{11:00 am.}$

Correct Answer $: \text{D}$

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