CAT 2019 Shift 2 Paper

Question : 92 of 100

Marks: +1, -0

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
10:45 am
10:27 am
10:18 am

Validate

Solution:

Let the track length be

10 {x}

.
When they meet at

10 {am}

, ant A travelled

6 {x}

of the distance and ant

{B}

travelled

4 {x}

of the distance.
Therefore,

\;{\;\text " Speed of ant "\; A}/{\;\text " Speed of ant "\; A}=\;{6 x}/{4 x}=\;{3}/{2}

And, the ratio of time taken by

{A}

and

{B}

to cover the same distance

=\;{2}/{3}

The distance by ant A from meeting point to point P was

4 x

. Similarly, the distance covered by ant B from meeting point to point

{P}

was

6 {x}

.
Given, ant A took 12 min to reach P.
Therefore, to cover a distance of

4 x

, time taken by ant

B=\;{3}/{2} × 12=18

min.
But, ant

{B}

has to cover a total of

6 {x}

distance.
Hence, the time required

=\;{6 x}/{4 x} × 18=27 {min}

.
Therefore, ant

{B}

reaches

{P}

10: 27 {am} .

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