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|Milestone Muse (Posted on 2013-10-10)
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?
Comment 3 of 3 |
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 2nd and 3rd 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
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