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

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

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

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

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

Posted from my mobile device

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

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.

Posted from my mobile device

To make it easier let the distance of circle = 10

When they meet, A covers 60% of 10 or 6

B covers 4

(1st) when they meet, there is a constant travel time for both A and B. This means the Ratio of Distances covered will be directly proportional to the Ratio of Constant Speeds


Dist A : Dist B = 6 : 4 = 3 : 2

Speed A : Speed B = 3 : 2


(2nd) A travels another 4 distance to get back around to point P. How long will it take B to cover this 4 distance too? (but keep in mind that B still needs to cover 6 to get back to P)

Same Distance A and B covers is 4, thus:

Ratio of Speeds is inversely proportional to Ratio of Time traveled

Speed A : Speed B = 3x : 2x

Time for A to cover 4 : Time for B to cover 4 = 2x : 3x

It Takes A another 12 minute to cover this distance.

2x = 12 ——- x = 6

B will take 3x = 18 minutes to cover this distance of 4


(3rd)
Takes B ———-18 min———- to cover a distance of 4

Then how long will it take B to cover 6 distance?

T = (18 * 6) / 4 = 27 minutes

B will return to P at 10 : 27 AM

Posted from my mobile device

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