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Two ants A and B start from a point P on a circle at the same time, wi

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Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 09 Apr 2020, 06:58
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Question Stats:

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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. 10:45 am
B. 10:40 am
C. 10:27 am
D. 10:25 am
E. 10:18 am


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Re: Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 09 Apr 2020, 07:10
Bunuel wrote:
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. 10:45 am
B. 10:40 am
C. 10:27 am
D. 10:25 am
E. 10:18 am


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When A and B meet, A has covered 60%, so B has covered 40%, and their speeds are in ratio A:B=60:40=3:2..

Next A completes the remaining 40% in 1012-1000=12 minutes, while B has to cover 60% of the route..
B would cover 40% route in 3/2 more time => 12*3/2=18..
So B would cover 60% in 18*60/40=27.

Hence B reaches 27 minutes after the ants meet at 1000, so at 1000+27=1027

C
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Re: Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 09 Apr 2020, 07:38
Bunuel wrote:
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. 10:45 am
B. 10:40 am
C. 10:27 am
D. 10:25 am
E. 10:18 am


Attachment:
ants.png
ants.png [ 2.86 KiB | Viewed 754 times ]

A and B meet at point M at 10:00 AM
A travels the rest of the distance \(\frac{2D}{5}\) in 12 minutes, as it is given that A returns to P at 10:12 AM.
when A travels \(\frac{3D}{5}\), B travels \(\frac{2D}{5}\).
when A travels \(\frac{2D}{5}\), B will travel:
\(\frac{2D*5*2D}{5*3D*5}\)= \(\frac{4D}{15}\)
B travels \(\frac{4D}{15}\) in 12 minutes.
B will travel \(\frac{3D}{5}\) in \(\frac{12*15*3D}{4D*5}\) = 27 minutes.
So B will take 27 minutes from point M to reach point P.
Ans: C
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Re: Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 09 Apr 2020, 09:29
Bunuel wrote:
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. 10:45 am
B. 10:40 am
C. 10:27 am
D. 10:25 am
E. 10:18 am


Are You Up For the Challenge: 700 Level Questions



Given: 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.

Asked: If A returns to P at 10:12 am, then B returns to P at

When A.& B meet at 1000 hrs,
A has covered 60% & B has covered remaining 40% of the track
vA : vB =60:40 =3:2

A takes 12 mins to cover remaining 40% of the track
B will take 12*3/2 = 18 min to cover 40% of the track
B will take 18*60/40 = 27 mins to cover remaining 60% of the track

B will reach P at 1027 hrs

IMO C
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Re: Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 09 Apr 2020, 11:23
Bunuel wrote:
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. 10:45 am
B. 10:40 am
C. 10:27 am
D. 10:25 am
E. 10:18 am


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Ratio of speed of A and B = 3: 2
To cover remaining 40% ant A took 12 minutes. So for the total distance ant A took 12/.4 = 30 minutes and ant B took 45 minutes.
They started their journey (30-12)= 18 minutes before 10 am or, 9.42 am. So ant B will reach to P = 9.42 + 45 = 9.87 or 10.27 am.
C is the answer
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Re: Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 09 Apr 2020, 18:09
60% covered @10am, returns at 10:12am --> goes 40% in 12min
40 = r*1/5
r=200

60:40 --> ratio of person A to person B
3:2

3/2 = 200/x
x=400/3

Person B needs to go 60%
60 = 400/3*t
t = 180/400 = 18/40 = 9/20 = 27/60 = 27min

10:00am + 27min = 10:27am
C
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Re: Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 09 Apr 2020, 20:23
A covers 40 % of the distance is 12 minutes so will cover 60% of the distance in 18 minutes. Now B has to cover 60% of the distance and his speed is 2/3 of A's speed.
Therefore time taken by B =18×3/2 = 27 minutes
Option C is the answer

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Re: Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 13 Apr 2020, 03:39
Bunuel wrote:
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. 10:45 am
B. 10:40 am
C. 10:27 am
D. 10:25 am
E. 10:18 am




We see that ant A must cover 40% of the track in 12 minutes, which means it covers the entire track (or the circumference of the circle) in 12/0.4 = 30 minutes. This means both ants start moving from P at 10:12 am - 30 minutes = 9:42 am. Furthermore, we see that since it takes ant A 10:00 am - 9:42 am = 18 minutes to cover 60% of the track, it will take ant B the same (i.e., 18 minutes) to cover the remaining 40% of the track when they meet at 10:00 am. So it will take ant B 18/0.4 = 45 minutes to cover the entire track. Since ant B begins moving from P at 9:42 am, ant B will return to P at 9:42 am + 45 minutes = 10:27 am.

Answer: C
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Re: Two ants A and B start from a point P on a circle at the same time, wi  [#permalink]

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New post 16 Apr 2020, 07:22
Eq1- Speed of a / speed of b = 3/2

Speed of a = 40/12
Speed of b = 60/t

Put the values in eq1 and find t which comes out to be 27.

So time taken by B is 27 mins more after A and B meet at 10:00am
Hence 10:27 is the answer.

I tried solving by long method but then realized that placing speed in the ratios is more efficient and easy. Hope it helps.

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Re: Two ants A and B start from a point P on a circle at the same time, wi   [#permalink] 16 Apr 2020, 07:22

Two ants A and B start from a point P on a circle at the same time, wi

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