From 9:00 AM to 10:30 AM = 1.5 hours. Speed of A = 40 km/h. So, distance covered by A: \[ = \frac{3}{2} \times 40 = 60 \text{ km} \]
Since total distance between them is 90 km, Remaining distance between A and B = \[ 90 - 60 = 30 \text{ km} \]
Relative distance covered: \[ 40t + 20t = 30 \Rightarrow 60t = 30 \Rightarrow t = \frac{1}{2} \text{ hour} \]
\[ \frac{1}{2} \text{ hour} = 30 \text{ minutes} \] So they meet at: \[ 10:30 \text{ AM} + 30 \text{ min} = \boxed{11:00 \text{ AM}} \]
The correct option is (D): \[ \boxed{11:00 \text{ AM}} \]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: