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Two ants $\text{A}$ and $\text{B}$ start from a point $\text{P}$ on a circle at the same time, with $\text{A}$ moving clock-wise and $\text{B}$ moving anti-clockwise. They meet for the first time at $10:00$ am when $\text{A}$ has covered $60\%$ of the track. If $\text{A}$ returns to $\text{P}$ at $10:12$ am, then $\text{B}$ returns to $\text{P}$ at

- $10:25$am
- $10:18$am
- $10:27$am
- $10:45$am

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Ans is an option **(C)**

Let the diameter of the track be $d$ mtr. Then the length of the track is $\pi d$ mtr. Let the speeds of ant$_{1}$ and ant$_{2}$ be $v_{1},v_{2}$ mtr/min respectively.

Then, let in time $t$ minutes, ant$_{1}$ covers $60$% of the track i.e. distance of $0.6\pi d$ mtr.

Then in the same time $t$ minutes, ant$_{2}$ covers $40$% of the track i.e. distance of $0.4\pi d$ mtr.

For ant$_{1}$ we have following equations: $t=\frac{0.6\pi d}{v_{1}} \longrightarrow(1)$ and $t+12=\frac{\pi d}{v_{1}}\longrightarrow(1)$

Dividing $(2)$ by $(1),$ we get $t=18$ minutes

For ant$_{2}$ we have following equations: $t=\frac{0.4\pi d}{v_{2}}\longrightarrow(3)$ and $t+y=\frac{\pi d}{v_{2}}\longrightarrow(4) \; (t+y$ is the complete time taken by ant$_{2}$ to complete the track) .

Dividing $(4)$ by $(3),$ we get $y=27$ minutes

After $t$ minutes, it is $10:00$ A.M when the two ants meet for the first time. From there, it takes $27$ minutes for ant$_{2}$ to reach point P.

So, ant$_{2}$ reaches point P at $10:27\;\text{A.M.}$