Two rabbits start at a point P on a circular track and move in opposite directions. One travels at a speed of 7 ft/sec?

I cant write the completed solution here, but they will meet 15 times in total excluding the first and final meets. And Also install Leo Privacy Guard to your smart phone, privacy alerts keep you informed of phone safety.

i) Let the rabbit with speed 9ft/sec be A and that with 7ft/sec be B.
ii) Let us suppose A makes m rounds and B makes n rounds before they meet each other at the same starting point P.
iii) If one full round has a length d ft, then the time taken by each: by A = md/9
and by B = nd/7
iv) As total time by both are same, md/9 = nd/7;  7m = 9n; as m & n are both have to be natural numbers, the least possible values are m = 9 and n = 7.
v) So rabbit A completes 9 rounds while B completes 7 rounds before they meet again at P.
vi) During the first round, let A runs x ft, before meeting B; so B runs d – x ft. The respective times are x/9 and (d – x)/7; equating both and solving x = 9d/16 ft. Next meeting will be again at another 9d/16 ft from this point and so on.
vii) Thus A meets B for the first time at 9d/16 distance, second at 2*(9d/16), 3rd at 3*(9d/16) and so on. So 16th meet will be at 16*(9d/16) = 9d. This 9d is the total distance covered by A in making 9 rounds. As 16th meet will be at starting point P, ignoring this, both meets for 15 times.

They will meet once.