# CAT 2019 Quant Question with Solution 49

Question:
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. $10: 25 \mathrm{am}$
2. $10: 18 \mathrm{am}$
3. $10: 27 \mathrm{am}$
4. $10: 45 \mathrm{am}$

Let the track length be 10x.

When they meet at 10 am, ant A travelled 6x of the distance and ant B travelled 4x of the distance.

Therefore, $\frac{\text{Speed of ant A}}{\text{Speed of ant A}}=\frac{6x}{4x}=\frac{3}{2}$

And, the ratio of time taken by A and B to cover the same distance = $\frac{2}{3}$

The distance by ant A from meeting point to point P was 4x. Similarly, the distance covered by ant B from meeting point to point P was 6x.

Given, ant A took 12 min to reach P.

Therefore, to cover a distance of 4x, time taken by ant B = $\frac{3}{2}\times 12=18$min.

But, ant B has to cover a total of 6x distance.

Hence, the time required =$\frac{6x}{4x}\times 18=27$ min.

Therefore, ant B reaches P at 10:27 am.

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