Solution:
For hare and tortoise speed is given. Tortoise will take 12hours to complete 1round. And during this, hare will make 12rounds of OP. In the first round, both has started from point O. After some time, hare will cross tortoise and distance between them will be 1km. After some more time, when hare is returning back from P to O, before and after crossing tortoise, hare will be two more times 1km apart from tortoise. So, in first round, there are three such occurrences.
In the second round, when the hare has started from point O, while going and returning back, there will be four occurrences when before and after crossing the tortoise, the hare will be exactly 1km apart. But the first occurrence of round 2 is already counted in round 1. So, in second round as well, there will be total 3 occurrences.
In the third, fourth and fifth rounds, there will be 4 such occurrences.
In the sixth round, because the tortoise will be at point P, there will be only 2cases.
Now, till round 6 there are 20 such occurrences. And from round 7 to 12, it will be exactly the same but in reverse order of 2,4,4,4,3,3
Hence, total such occurrences =20×2=40.
© examsnet.com