Three persons P, Q and R start simultaneously from point A towards point B along a straight road. First, P reaches B, immediately turns back, and meets Q at a point which is 11 km away from B. After this, Q reaches B, turns back at once, and meets R at a point which is 9 km away from B. If the ratio of the speeds of P and R is 3 : 2, what is the distance between A and B in kilometres?

Introduction / Context:
This is a classic time, speed and distance puzzle where three people move along the same straight path and repeatedly turn back at the end point. The problem gives the meeting points between pairs of travellers and a ratio of speeds, and asks for the total distance between A and B.

Given Data / Assumptions:

• P, Q and R start together from A towards B.
• P reaches B first, then returns towards A and meets Q at a point 11 km from B (position D - 11 if D is the distance AB).
• Q then reaches B, turns back, and meets R at a point 9 km from B (position D - 9).
• Speed ratio of P to R is 3 : 2, so let speed of P = 3k km/h, speed of R = 2k km/h.
• Speed of Q is some constant q km/h.
• Total distance AB is D km, which we must find.

Concept / Approach:
We use positions at specific meeting times and the fact that all three start together. By writing equations for where P and Q meet, and where Q and R meet, in terms of D, k and q, we can eliminate q and solve for D. The key ideas are equal distances at meeting points and relationships between speeds and times.

Step-by-Step Solution:
Let D be the distance from A to B, and speeds be P = 3k, Q = q, R = 2k. Time for P to reach B first: t1 = D / (3k). At time t1, Q has travelled distance q * t1 and R has travelled 2k * t1. After reaching B, P turns back and meets Q after time t2. The meeting point is D - 11 from A. Position of P at meeting: D - 3k * t2. Position of Q at meeting: q * (t1 + t2). Since they meet at the same point D - 11, we have D - 3k * t2 = D - 11, so 3k * t2 = 11 and t2 = 11 / (3k). Also, q * (t1 + t2) = D - 11. Substitute t1 and t2 to get q * ((D / (3k)) + 11 / (3k)) = D - 11, so q = 3k (D - 11) / (D + 11). Next, from this meeting point at D - 11, Q takes time 11 / q to reach B. So Q reaches B at time T_QB = (D + 11) / (3k) + 11 / q. Q then turns back from B and meets R at point D - 9. Let t4 be the time after turning back until this meeting, so q * t4 = 9, hence t4 = 9 / q. Total time when Q meets R is T_QR = T_QB + t4 = (D + 11) / (3k) + 20 / q. R has been moving from A the whole time, so its distance at T_QR is 2k * T_QR, which must equal D - 9. So 2k * ((D + 11) / (3k) + 20 / q) = D - 9. Substitute q = 3k (D - 11) / (D + 11) and simplify. Solving the resulting equation gives possible values D = 1 km or D = 99 km. The distance cannot be 1 km because the meeting points are 11 km and 9 km from B, so D = 99 km is valid.

Verification / Alternative check:
With D = 99 km, we have t1 = 99 / (3k) = 33 / k, t2 = 11 / (3k). Meeting point for P and Q is 99 - 11 = 88 km from A. Using q from the earlier relation ensures that Q has also reached 88 km at the same time. Repeating the logic for the second meeting confirms that Q and R meet 9 km from B, which supports D = 99 km.

Why Other Options Are Wrong:
Distances such as 100 km or 89 km do not satisfy both meeting conditions when the speed ratio of P and R is fixed at 3 : 2. Distance 1 km is impossible because the meeting points are more than 1 km from B. The option 120 km similarly fails to produce the required 11 km and 9 km offsets from B when the movements are modelled.

Common Pitfalls:
Errors often arise from mishandling turning times or assuming symmetric motion without deriving equations. Learners may also forget that R starts at the same time as P and Q and never turns. Careful tracking of each segment of motion and checking physical plausibility of the final distance are essential steps.

Final Answer:
The distance between A and B is 99 km.