# CAT2019-2: 81

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

1. $10:25$am
2. $10:18$am
3. $10:27$am
4. $10:45$am

edited

Ans is 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:   $t=\frac{0.6\pi d}{v_{1}}$  -----(1)    and     $t+12=\frac{\pi d}{v_{1}}$  ------(2)

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

For ant$_{2}$ we have following equations:   $t=\frac{0.4\pi d}{v_{2}}$  -----(3)    and     $t+y=\frac{\pi d}{v_{2}}$  ------(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  $10:27$ A.M.

238 points 2 2 4

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