Correct Answer: (A)

1. Establish the Speed-Time Relationship: In problems where two objects start simultaneously and meet, then reach their destinations in times $t_1$ and $t_2$, the ratio of their speeds is given by $\frac{S_A}{S_B} = \sqrt{\frac{t_2}{t_1}}$.

2. Calculate the Speed of Train B: Given $S_A = 45$, $t_1 = 4$, and $t_2 = 9$, we have $\frac{45}{S_B} = \sqrt{\frac{9}{4}} = \frac{3}{2}$. Solving for $S_B$, we get $S_B = \frac{45 \times 2}{3} = 30$ km/h.

3. Determine the Meeting Time: Let the time taken to meet be $t$ hours. The property of such simultaneous starts is that $t = \sqrt{t_1 \times t_2} = \sqrt{4 \times 9} = 6$ hours.

4. Calculate Total Distance: Distance $D = \text{Distance covered by A in } (t + t_1) \text{ hours} = 45 \times (6 + 4) = 450$ km. Alternatively, using B: $D = 30 \times (6 + 9) = 450$ km. Let us re-verify the meeting time logic. Train A covers distance $PQ$ in $6+4=10$ hours. $10 \times 45 = 450$. Wait, checking the options: if $t=6$, distance $= 45 \times 6 + 30 \times 6 = 270 + 180 = 450$. If 375 is an option, let's re-verify: $375 / 10 = 37.5$. $450$ is the mathematically consistent result for these parameters.

Test Prep Tip: The formula $\frac{S_1}{S_2} = \sqrt{\frac{t_2}{t_1}}$ is a standard derivation from the meeting point logic $D_{meet} = S_A \times t = S_B \times t$. Remembering this specific relation for "time taken after meeting" scenarios saves significant time in the CAT exam.