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

$10:25$ am

$10:18$ am

$10:27$ am

$10:45$ am

✅ Correct Option: 3

Two ants start from point P simultaneously:

Ant A moves clockwise
Ant B moves anti-clockwise
They meet at 10:00 am (A has covered 60% of track)
A returns to P at 10:12 am
We need to find when B returns to P

When two objects move in opposite directions on a circular track, their first meeting tells us about their relative speeds.

At 10:00 am (first meeting):

A covered 60% of the track
B covered 40% of the track (since together they completed one full circle)

When moving in opposite directions, they meet after covering exactly one complete circle between them.

Since both ants moved for the same time:

$\dfrac{\text{Speed of A}}{\text{Speed of B}} = \dfrac{60\%}{40\%} = \dfrac{3}{2}$

From 10:00 am to 10:12 am = 12 minutes

In these 12 minutes, A covered: 100% - 60% = 40% of the track

So A covers:

40% of track in 12 minutes
100% of track in $12 \times \dfrac{100}{40} = 30$ minutes

Since $\dfrac{\text{Speed of A}}{\text{Speed of B}} = \dfrac{3}{2}$ :

$\text{Speed of B} = \dfrac{2}{3} \times \text{Speed of A}$

Time is inversely proportional to speed:

$\text{Time for B} = \dfrac{3}{2} \times \text{Time for A} = \dfrac{3}{2} \times 30 = 45 \text{ minutes}$

At 10:00 am, B had covered 40% of the track.

B still needs to cover: 100% - 40% = 60% of the track

Time for B to cover 60%: $45 \times \dfrac{60}{100} = 27$ minutes

Therefore, B returns to P at: 10:00 am + 27 minutes = 10:27 am

In circular motion problems with opposite directions, the first meeting always occurs when the faster object has covered more than 50% of the track. This immediately tells us which object is faster and by how much!

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