A hare and a tortoise run between points O and P located exactly 6 km from each other on a straight line. They start together at O, go straight to P and then return to O along the same line. They run at constant speeds of 12 km/hr and 1 km/hr respectively. Since the tortoise is slower than the hare, the hare shuttles between O and P until the tortoise goes once to P and returns to O. During the run, how many times are the hare and the tortoise separated by an exact distance of 1 km from each other?

The total distance to be travelled is 6+6 = 12km

We know the speeds of the hare and the tortoise.

Time taken by the Tortoise to finish one round = $$\frac{12}{1}=12hr$$. 

We know the speed of the Hare is 12 km/h. Let's find the distance travelled by the hare in 12 hrs. = $$12\cdot12=144km$$

This is nothing but $$\frac{144}{12}=12\ \text{rounds}\ $$. So, Hare will make 12 rounds of OP.

In the first round, both have started from point O. After some time, the distance between them will be 1 km. 

After some more time, when the hare is returning from P to O, before and after crossing the tortoise, the hare will be two more times 1km apart from the tortoise. So, in the first round, there are three such occurrences.

In the second round, when the hare starts from point O, while going and returning, there will be four occurrences when, before and after crossing the tortoise, the hare will be exactly 1 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 P, there will be only 2 cases.

Now, till around 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.