Correct Answer: B. 30 km Let the distance between P and Q be $D$ km. Let the speed of A be $S_A$ km/h and the speed of B be $S_B$ km/h. From the problem statement, the speed of A is 1.5 times the speed of B. So, $S_A = 1.5 S_B$, which means $\frac{S_A}{S_B} = \frac{1.5}{1} = \frac{3}{2}$. **Analysis of the First Meeting:** They start simultaneously and meet for the first time 12 km from Q. This means that B has covered a distance of 12 km. A has covered a distance of $(D - 12)$ km. Since they traveled for the same amount of time until they met, the ratio of the distances covered is equal to the ratio of their speeds. $\frac{\text{Distance covered by A}}{\text{Distance covered by B}} = \frac{S_A}{S_B}$ $\frac{D - 12}{12} = \frac{3}{2}$ Now, we solve for $D$: $2(D - 12) = 3 \times 12$ $2D - 24 = 36$ $2D = 36 + 24$ $2D = 60$ $D = 30$ km. This gives us a value for D. Now we need to check if this value is consistent with the information from the second meeting. **Analysis of the Second Meeting:** After the first meeting, A reaches Q and immediately reverses towards P. B reaches P and immediately reverses towards Q. They meet for the second time 6 km from P. Let's determine the total distance covered by A and B until the second meeting point (M2, 6 km from P). 1. **Distance covered by A ($D_A$):** A starts from P, travels to Q (distance $D$), then reverses from Q towards P and meets B at M2, which is 6 km from P. So, M2 is $(D-6)$ km from Q. $D_A = (\text{Distance P to Q}) + (\text{Distance Q to M2})$ $D_A = D + (D - 6) = 2D - 6$ 2. **Distance covered by B ($D_B$):** B starts from Q, travels to P (distance $D$), then reverses from P towards Q and meets A at M2, which is 6 km from P. $D_B = (\text{Distance Q to P}) + (\text{Distance P to M2})$ $D_B = D + 6$ Similar to the first meeting, since they travel for the same amount of time until the second meeting, the ratio of the total distances covered is equal to the ratio of their speeds. $\frac{D_A}{D_B} = \frac{S_A}{S_B}$ $\frac{2D - 6}{D + 6} = \frac{3}{2}$ Now, we solve for $D$: $2(2D - 6) = 3(D + 6)$ $4D - 12 = 3D + 18$ $4D - 3D = 18 + 12$ $D = 30$ km. Both the first and second meeting conditions, when combined with the given speed ratio, consistently yield a distance of 30 km between P and Q. Therefore, the distance between P and Q is 30 km. The concept that for two objects starting from opposite ends and moving back and forth, the total distance covered by both for the $n^{th}$ meeting is $(2n-1)D$ can also be used as a check: For the first meeting, total distance = D. For the second meeting, total distance = 3D. With $D=30$ km: $D_A + D_B = (2D-6) + (D+6) = 3D = 3 \times 30 = 90$ km. This is consistent with the general principle that the total distance covered by both for the second meeting is $3D$. The final answer is $\boxed{\text{30 km}}$.