A supportive young hare and tortoise raced in opposite directions around a circular track that was 100 yards in diameter. They started at the same spot, but the hare did not move until the tortoise had a start of one eighth of the distance ( that is, the circumference of the circle). The hare held such a poor opinion of the other’s racing ability that he sauntered along, nibbling the grass until he met the tortoise. At this point the hare had gone one sixth of the distance. How many times faster than he went before must the hare now run in order to win the race ?
Total Answers and Comments: 11
Last Update: December 03, 2006 Asked by: sharad vasisht
RE: A supportive young hare and tortoise raced in oppo...
since the tortise std with 1/8th d circmfrnce of circle the tortise moved 1/6-1/8 when the hare moved 1/6.So ratio of their distances coverd is 1/6:1/24 which is nothin but ratio of their speeds since time taken by both to reach is same.Therefore the ratio is4:1
RE: A supportive young hare and tortoise raced in oppo...
since the hare didnt started till tortoise travelled 1/8th distance so the left over is 7/8th.after hare ran for 1/6th of the distance it met tortoise so tortoise walk for a distance of(7/8)-(1/6) 17/24 distance in the same time . so the ratio of distances ran in the same slot can gv the ratio of speed. so hare: tortoise 1/6 : 17/24 4/17 0.235.so its obvious that hare was slow wen compared to tortoise. so hare must be faster more than 4.25 times( 1/0.235) compared to tortoise.
RE: A supportive young hare and tortoise raced in oppo...
i think the answer is 5 times.Both are moving in opp dirns and they have met after hare covered 1/6 th dist.Now hare has to go 5/6 th and tortoise 1/5 th .
so hare should run more than 5 times faster than tortoise
