Two trains start simultaneously from A and B 180 kms apart, towards B and A respectively at speeds in the ratio 5 : 4. They cross each other at M and after reaching B and A turn back immediately and meet again at N. Find the distance MN.

Let speeds be 5x and 4x. First meeting at M: distances covered are in ratio 5:4, so train from A travels (5/9)×180 = 100 km to reach M. Second meeting at N: after first meeting, trains continue to destinations (train from A travels 80 km more to B, train from B travels 100 km more to A), then turn back. Total distance covered by both trains between first and second meeting is 2×180 = 360 km. Train from A covers (5/9)×360 = 200 km in this period. But from M to B is 80 km, and B to A is 180 km, so from M, it goes 80 + 180 = 260 km to reach A, but it only covers 200 km total. So from M towards A after the turn, it covers 200 - 80 = 120 km. Distance from A = 180 - 120 = 60 km. M was at 100 km from A, so MN = 100 - 60 = 40 km. Alternatively, using the formula: For two trains starting from opposite ends, MN = (difference in speeds)/(sum of speeds) × distance between starting points = (5x-4x)/(5x+4x) × 180 = (1/9) × 180 = 20 km from one end. But M is at 100 km from A, so N is at 100 - 40 = 60 km from A, giving MN = 40 km.