Two cars leave simultaneously from points X and Y on the same road but in opposite directions. Their speeds are constant, and in the ratio 5 to 4 with the car leaving at X being the faster of the two.

The cars travel to and fro between X and Y. They meet for the second time at the 145th milestone and for the third time at the 201st milestone.

What are the respective milestones at X and Y?

Given the 5/4 ratio of speed, divide the distance between X and Y into nine equal segments "s"; then the distance between X and Y is 9*s.  Furthermore, (and temporarily) assign a position of zero to location X.  From there it is easy to see that the first meeting is at position 5*s.  Then if you think about it, the next meeting must occur after twice the length of time it took for the first to occur, same for all future meetings after that.  Accounting for the reversals in direction, the second meeting then occurs at position 3*s, and the third at 7*s. So the difference in distance between the second and third meeting points is 4*s, and by the problem statement that is equal to (201-145)=56, so s=14.  Now take either known meeting point (say 145).  It occurs at point 3*s=3*14=42, plus some offset to get to 145.  So that offset is 103
Therefore the answers are
X=103, Y=103+9*14=229

Edited on October 13, 2013, 10:30 am

Posted by Kenny M on 2013-10-13 10:29:56