Cat Rat Dash

Let $t$ be the time in seconds, $x$ be the number of cat's rounds, and $y$ the number of rat's rounds.

Then $7t=36x$ , and $5t=24y$ .

Thus $\dfrac{x}{y} =\dfrac{7\times 24}{36\times 5} = \dfrac{14}{15}$

Since the cat would catch the rat immediately at the meeting point, $x = 14$ and $y=15$ .

Thus, $t = \dfrac{36\times 14}{7} =\dfrac{24\times 15}{5} = \boxed{72}$ .

This is my solution. The time traveled by the cat must be equal to the time traveled by the rat. Since they started at the same point, the total distance traveled by the cat must be divisible by 36 and the total distance traveled by the rat must be divisible by 24. So I took the LCM of 36 and 24 and that is 72. And then I computed the time traveled at an interval of 72 feet.

First trial. The cat traveled 72 feet in 72/7 seconds. The distance traveled by the rat in 72/7 seconds is 360/7 feet.

Second trial: The cat traveled 144 feet in 144/7 seconds. The distance traveled by the rat in 144/7 seconds is 720/7 feet.

Third trial: The cat traveled 216 feet in 216/7 seconds. The distance traveled by the rat in 216/7 seconds is 1080/7 feet.

Fourth trial: The cat traveled 288 feet in 288/7 seconds. The distance traveled by the rat in 288/7 seconds is 1440/7 feet.

Fifth trial: The cat traveled 360 feet in 360/7 seconds. The distance traveled by the rat in 360/7 seconds is 1800/7 feet.

Sixth trial: The cat traveled 432 feet in 432/7 seconds. The distance traveled by the rat in 432/7 seconds is 2160/7 feet.

Seventh trial: The cat traveled 504 feet in 72 seconds. The distance traveled by the rat in 72 seconds is 360 feet.

From here, I realized that 504 is divisible 36 and 360 is divisible by 24. So they meet again at point P after 72 seconds of running.