Swimming pool circular strip

Water current $r = 3\text{ km/h} = 50\text{ m/min}$ (clockwise). Rana's speed in still water $S_r = 3\text{ km/h} = 50\text{ m/min}$. Rana swims downstream (clockwise), so his net speed $V_R = 50 + 50 = 100\text{ m/min}$. Ayub swims upstream (counterclockwise), so his net speed $V_A = S_a - 50\text{ m/min}$.

For first meeting, Ayub covers $60\text{ m}$ in time $t_1 = \frac{60}{S_a - 50}$. Rana covers distance $V_R t_1 = 100 \times \frac{60}{S_a - 50} = \frac{6000}{S_a - 50}$. Since they start from diametrically opposite points, $\pi r = 60 + \frac{6000}{S_a - 50}$.

For second meeting, Rana covers $180\text{ m}$ in time $t_2 = \frac{180}{100} = 1.8\text{ min}$. In $t_2$, total relative distance covered equals full circumference $2\pi r$. $2\pi r = 180 + (S_a - 50) \times 1.8$. Substituting $\pi r$: $2\left(60 + \frac{6000}{S_a - 50}\right) = 180 + 1.8(S_a - 50)$. Let $t = S_a - 50$: $120 + \frac{12000}{t} = 180 + 1.8t \implies 3t^2 + 100t - 20000 = 0 \implies t = \frac{200}{3}$.

Thus $S_a = 50 + \frac{200}{3} = \frac{350}{3}\text{ m/min}$. Circumference $2\pi r = 180 + 1.8 \times \frac{200}{3} = 300\text{ m}$.

If Ayub swims clockwise (downstream), his net speed $= S_a + 50 = \frac{350}{3} + 50 = \frac{500}{3}\text{ m/min}$. Time for one round $= \frac{300}{500/3} = \frac{9}{5}\text{ min} = 1\text{ min } 48\text{ sec}$.