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

• A. 10: 25am
• B. 10 :18am
• C. 10: 27am
• D. 10: 45am

Answer 1

Answer: (C)

10: 27am

Answer 2

Let the track length be represented as 10x.
When they meet at 10 am, ant A has traveled 6x of the distance, and ant B has covered 4x of the distance.
Therefore, the speed of ant A to the speed of ant B is $\frac{6x}{4x}=\frac{3}{2}$​.
The ratio of the time taken by A and B to cover the same distance is $\frac{2}{3}$​.
The distance by ant A from the meeting point to point P is 4x, and similarly, the distance covered by ant B from the meeting point to point P is 6x.
Given that ant A took 12 minutes to reach point P, the time taken by ant B to cover a distance of 4x is $\frac{3}{2}\times12=18$ minutes.
However, ant B needs to cover a total distance of 6x.

Hence, the time required is $\frac{6x}{4x}\times18=27$ minutes.
Therefore, ant B reaches point P at 10:27 am.