a man left x for y ,another man left Y for x at the same time .after reaching their destination ,they turned back without stopping .they crossed each other at a point 30km from x on their onward trip and 20km from y on their return trip .find XY a)70 b)55 c)40 d)30

55

70

Solution please

Let, distance between X-Y be 'A' mtrs, Speed of X be 'x' m/s, Speed of Y be 'y' m/s. Time taken by them to meet for the first time = A/(x+y) In this time X travels 30kms. So, (Speed of A)*(Time taken by A) = 30 x*(A/(x+y)) = 30 ...............(1) Let's say they meet at a point 'C' for the second time, so Time taken by X to reach C = Time taken by Y to reach C Distance covered by X to reach 'C' = A - 30 + 20 = A - 10 Distance covered by Y to reach 'C' = 30 + 30 + A - 50 = A + 10 So, (A-10)/x = (A+10)/y...............(2) Solving (1) & (2) we get, A = 70 kms.

Basically we can consider that initially they meet 30kpm from X. When they meet for the first time, the combined distance travelled is “d”. So essentially for every “d”km combined distance travelled by them, the first person travels 30km. Now it is mentioned that they met 20km from Y for the second meeting. When they meet for second time, combined distance travelled is “3d” hence the first person travels 3*30 =90km. Hence Length of the track + 20km (as the meet 20km from Y) is 90. Hence length of track = 90-20 = 70km