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?

Let the milestones at X and Y show x and y, so that the distance from
X to Y is y – x miles. At any stage, the distances travelled by the cars
must be in the ratio 5:4. Thus, at the 2^nd and 3^rd meetings:

(y – x) + (y – 145)  = 5/4
(y – x) + (145 – x)                     giving    2x + y = 435   (1)

2(y – x) + (201 – x)  = 5/4
2(y – x) + (y – 201)                   giving    2x + 7y = 1809  (2)

Solving (1) and (2) simultaneously gives  x = 103 and y = 229.

Posted by Harry on 2013-10-14 20:24:34