A and B left Kanpur simultaneously and travelled Lucknow, A's speed was 20 km/hr and that of B's was 12 km/ hr. 3 hour later,C left kanpur and travelled in the same direction.After some time he overtook 'B' and 120 minutes further he overtook 'A' . Find the speed of 'C'.

A and B have 3-hour head start. A travels 60 km (20×3), B travels 36 km (12×3). C overtakes B: let C's speed be s. Time to catch B = 36/(s-12). C then overtakes A: time to catch A = 60/(s-20). Given second catch is 120 minutes (2 hours) after first: 60/(s-20) - 36/(s-12) = 2. Solving: 60(s-12) - 36(s-20) = 2(s-20)(s-12). 60s-720-36s+720 = 2(s²-32s+240). 24s = 2s²-64s+480. s²-88s+480=0 doesn't give integer. Alternative approach: let C take t hours to catch B, then t+2 to catch A. Distance same: 12t + s(t+2) = 20t + s(t+4)? No. C catches B at time t: s×t = 12×(3+t). s = 12(3+t)/t. C catches A at time t+2: s×(t+2) = 20×(3+t+2) = 20×(5+t). So 12(3+t)(t+2)/t = 20(5+t). 12(3+t)(t+2) = 20t(5+t). This gives t=4, s=24 km/hr. Answer C is correct.