A sportive young hare and a tortoise raced in opposite directions around a circular track that was 100 yards in diameter.
They started in the same spot but the hare did not move until the tortoise had a head start of 1/8 of the distance (that is,
the circumference of the circle).
The hare held such a poor opinion of the tortoise's racing ability that he sauntered along, nibbling the grass, until he met the tortoise.
At this point, the hare had gone 1/6 of the distance.
How many times faster than he went before must the hare now run in order to win the race?
Solution to the Problem:
The hare must go 85/4 faster than he did before.
The diameter of the track has no bearing on the problem.
When they meet, the hare has gone 1/6 of the way around the track in the time that it took the tortoise to go 17/24 (remember the hare did not start until the tortoise had gone 1/8 or 3/24). The hare has therefore been moving 17/4 times as fast as the tortoise (17/24 divided by 1/6).
The hare has 5/6 of the distance left to go compared to 1/6 of the distance for the tortoise, so the hare must go five times faster than the tortoise, or 85/4 faster than he went before.
Correctly solved by:
| 1. Aayan Shah | Lalitpur, Nepal |
| 2. Veena Mg | Bangalore, Karnataka, India |
| 3. Kelly Stubblefield | Mobile, Alabama |
| 4. Dr. Hari Kishan |
D.N. College, Meerut, Uttar Pradesh, India |
| 5. Davit Banana | Istanbul, Turkey |
| 6. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |