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

The Correct Option is C

Let the total length of the track be \( 10x \).

When the ants meet at 10:00 AM:

• Ant A has traveled \( 6x \)
• Ant B has traveled \( 4x \)

So, the ratio of their speeds is: \[ \frac{6x}{4x} = \frac{3}{2} \]

This means the ratio of the time taken (to cover the same distance) is the inverse: \[ \frac{2}{3} \]

From the meeting point to point P:

• Ant A needs to cover \( 4x \)
• Ant B needs to cover \( 6x \)

Given that Ant A takes 12 minutes to cover \( 4x \), the time Ant B would take to cover the same distance of \( 4x \) is: \[ \frac{3}{2} \times 12 = 18 \text{ minutes} \]

Now, since Ant B must travel \( 6x \), the total time will be: \[ \frac{6x}{4x} \times 18 = \frac{3}{2} \times 18 = 27 \text{ minutes} \]

Therefore, Ant B will reach point P at:

10:27 AM