Three bodies A, B and C start moving around a circular track

For this question , we will start with the body who is the slowest i.e. 3m/sec.

Since the questions asks us when all the three bodies are going to meet , so assume the time t is reqduired to do this
Distance travelled by I body: 3t
Distance travelled by II body: 5t
Distance travelled by III body: 9t

the distances should be equal to meet , and that is possible because of a circular track , length of the track: 60m

Keep this is mind , for circular track
The point on the circular track = n * length of the track + remaining distance
where n is a positive integer

Like if someone travels from point Z on the track 200 m then actually he is far from point Z by 20m .
As ,200 = 3*60 +20

Now i inserted the values :
A: 30 secs
I=3*30=90 = one length of track +30 ;
II=5*30=150 = 2 lenth of track + 30;
III= 9*30 =270 = 4 lenth of track +30;

so everyone is at 30 m after 30 secs.
Hence A is the answer.

thanks , probably this is the quickest way for this , but any algebric way to handle this?

Hi Karishma,

After calculating Relative speed of B & C over A. We can take LCM of time taken to complete one round by B & C to find out when all three will meet.
This shortcut is preferred once anyone has mastered the logic as suggested by karishma

If they all meet after T seconds, it means they covered the distances 3T, 5T, and 9T respectively.
Since they all arrive to the same spot, it means that the differences taken pairwise between the distances must be positive integer multiples of the length of the track, which is 60m. So, 2T, 4T, and 6T must all be multiples of 60. 2T multiple of 60 means T multiple of 30. The smallest T with this property is 30 and is on the list of answers.

Answer A.

Yes, you are right. You get that the time taken by B to complete one extra circle is 30 secs and time taken by C to complete one extra circle is 10 secs.
You take their LCM which is 30 secs. The theory explains why you should take the LCM.

Hi Karishma

I have lost control over my understanding though u mentioned very clearly.
Requesting you to again depicts the same for me.

Also the theory behind the logic and any other question of similar kind.


Rgds
Prasannajeet

i like to back solve it: (it takes at most 40 seconds) i explained my way below:
For circular distance of 60 meter, 60,120,180,240 all end in the same point

for body A , 30 * 3 = 90 m = 60 + 30 m (so 30m ahead from the starting point)
for body, B, 30 * 5 = 150 m = 120 + 30 m (so 30m ahead from the starting point)
for body C, 30 * 9 = 270 m = 240 + 30 m (so 30m ahead from the starting point)
Everyone 30m ahead of the starting point after 30 sec.

i am lucky enough that the 1st answer satisfies my findings.

Check out this post on circular motion: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/08 ... n-circles/
See if this helps else I will try to explain more in detail.

Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively. When will they meet for the first time after they started moving?

A. 30 seconds
B. 60 seconds
C. 15 seconds
D. 10 seconds
E. 25 seconds

From the onset I kind of figured you could solve with LCM's but I wasn't entirely sure why. I tried solving by figuring out how long it would take each of them to make one complete revolution and keep counting until their times aligned but I think that is incorrect because we're trying to figure out how long it takes for them to "meet up" we cannot solve that way. Is this correct?

If that is the case, then we need to figure out their relative speeds to determine when each body (let's call them A, B, C for the slow, medium and fast objects respectively) reaches the other.

B's relative rate to A is 5-3 = 2m/second so it takes B 30 seconds to move 60 meters away from A. In other words, at the 30 second mark, A and B are next to one another. A has traveled 90 meters and B has traveled 150 meters.

C's relative rate to A is 9-3 = 6m/second so it takes C 10 seconds to move 60 meters away from A. Every 10 seconds, it moves 60 meters (one full revolution) away from A. In 30 seconds (the time it takes A and B to meet up) it is 3 full revolutions ahead of A but is also next to it on the circle.

ANSWER A. 30 seconds.

LCM of time works for a question of a different type:
When will they meet for the first time AT THE STARTING POINT after they started moving?

Time take by A to cover a circle = 60/3 = 20 sec
Time taken by B to cover a circle = 60/5 = 12 sec
Time taken by C to cover a circle = 60/9 sec

So every 20 sec, A will be at the starting point.
Every 12 secs B will be at the starting point.
Every 60/9 sec, C will be at the starting point.

Taking their LCM, we get 60. So every 60 sec, all three will be at the starting point. All meet for the first time at the starting point after they start moving after 60 sec.

Let’s analyze the answer choices from the smallest to the largest.

D) 10 seconds

After 10 seconds, A, B, and C have moved 30 m, 50 m, and 90 m, respectively. While A and B haven’t completed one lap, C has completed one lap and 30 m more. We see that A and C meet at the same place on the track, but B doesn’t meet them at that place .

C) 15 seconds

After 15 seconds, A, B, and C have moved 45 m, 75 m, and 135 m, respectively. While A hasn’t completed one lap, B has completed one lap and 15 m more, and C has completed two laps and 15 m more. We see that B and C meet at the same place on the track, but A doesn’t meet them at that place.

E) 25 seconds

After 25 seconds, A, B and C have moved 75 m, 125 m, and 225 m, respectively. We see that A has completed one lap and 15 m more, B has completed two laps and 5 m more, and C has completed three laps and 45 m more. We see that each person is at a different place on the track.

A) 30 seconds

After 30 seconds, A, B, and C have moved 90 m, 150 m, and 270 m, respectively. We see that A has completed one lap and 30 m more, B has completed two laps and 30 m more and C has completed four laps and 30 m more. We see that all of them are at the same place on the track.

Alternate Solution:

Suppose that they meet after t seconds for the first time. In t seconds, they have covered a distance of 3t, 5t and 9t meters, respectively. Notice that the distance between A and B is 5t - 3t = 2t, A and C is 9t - 3t = 6t and B and C is 9t - 5t = 4t. If they are at the same spot after t seconds, all of 2t, 6t and 4t must be multiples of 60; say 2t = 60k, 6t = 60s and 4t = 60p. Then, t = 30k = 10s = 15p. Since t is a multiple of 30, 10 and 15; the smallest value of t is given by the LCM of these numbers. Therefore, the smallest value of t is LCM(30, 10, 15) = 30.

Answer: A

Given: Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively.
Asked: When will they meet for the first time after they started moving?

A, B, C will move 3k, 5k & 9k after moving for k seconds

A. 30 seconds
A, B, C will move 90, 150 & 270 m after moving for 30 seconds
They are all 30 m away from starting point on the circular track.

B. 60 seconds
A, B, C will move 180, 300 & 540 m after moving for 30 seconds
They are all at the starting point

C. 15 seconds
A, B, C will move 45, 75 & 135 m after moving for 15 seconds
They are not at the same point

D. 10 seconds
A, B, C will move 30, 50 & 90 m after moving for 10 seconds
They are not at the same point

E. 25 seconds
A, B, C will move 75, 125 & 175 m after moving for 15 seconds
They are not at the same point

IMO A

Given: Three bodies A, B and C start moving around a circular track of length 60m from the same point simultaneously in the same direction at speeds of 3 m/s, 5 m/s and 9 m/s respectively.
Asked: When will they meet for the first time after they started moving?

A, B, C will move 3k , 5k & 9k after moving for k seconds

5k - 3k = 60m = 2k
9k - 5k = 4k = 60(2m)
9k - 3k = 6k = 60(3m)

k = 30m seconds
Rule out answers C, D & E

If m = 1; k = 30
A, B, C will move 90m, 150m & 270m respectively after 30 seconds
They are all 30m away from the starting point and meet for the first time.

Since they meet for the first time after 30 seconds, there is no need to check for answer B. 60 seconds

IMO A

Hi VeritasKarishma
It is not necessary that they meet at the starting point for the first time. They can meet somewhere else.
Please check the correct solutions posted by me.

Please note the first line of my solution (highlighted below). This was an answer to the solution given by WholeLottaLove for a different kind of question.

Relative speed of A and B is (5-3)=2 than that of A. So 60/2= 30 sec .


Relative speed of B and C (9-5)=4 m/s. So,60/4=15 sec.


Relative speed pf A and C (9-3)=6 m/s. So, 60/6=10 sec.

the LCM of values 30, 15 and 10 is 30.


Therefore IMO A

Check this video for a faster method: https://youtube.com/shorts/CUmXZDOCXxw?feature=shared