Xavier Aptitude Test 2020 Paper

For hare and tortoise speed is given. Tortoise will take $12\mathrm{hours}$ to complete $1\mathrm{round}$. And during this, hare will make $12\mathrm{rounds}$ of $\mathrm{OP}$. In the first round, both has started from point $\mathrm{O}$. After some time, hare will cross tortoise and distance between them will be $1\mathrm{km}$. After some more time, when hare is returning back from $\mathrm{P}$ to $\mathrm{O}$, before and after crossing tortoise, hare will be two more times $1\mathrm{km}$ apart from tortoise. So, in first round, there are three such occurrences.
In the second round, when the hare has started from point $\mathrm{O}$, while going and returning back, there will be four occurrences when before and after crossing the tortoise, the hare will be exactly $1\mathrm{km}$ 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 $\mathrm{P}$, there will be only $2\mathrm{cases}$.
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 $=20\times 2=40$.