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Divi and Dave run on a circular track. Divi completes 3 laps

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Divi and Dave run on a circular track. Divi completes 3 laps [#permalink] New post   03 Nov 2009, 23:11
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Divi and Dave run on a circular track. Divi completes 3 laps in every 4 mins, and Dave completes 2 laps in every 3 mins in opposite direction. Both start from points opposite to each other.Find the number of completed laps traveled by Divi when both will meet for the 3rd time at starting point of Dave.

A. 12
B. 13
C. 22
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Divi and Dave run on a circular track. Divi completes 3 laps [#permalink] New post   Updated on: 07 Nov 2023, 06:37
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virtualanimosity wrote:
Can any1 pls explain da concept behind such circular motion problem n meeting at startin pt problems
Q.Divi and Dave run on a circular track.Divi completes 3 laps in every 4 mins, and Dave completes 2 laps in every 3 mins in opposite direction.
Both start from points opposite to each other.Find the number of completed laps travelled by Divi when both will meet for the 3rd time at starting point of Dave.
1. 12
2. 13
3. 22


Responding to a pm:

Check out this video for a simple solution to such a question: https://youtu.be/BmWrx25SZq0


You can do it using the ratios approach. But first, look at the big picture.
Divi and Dave are at opposite points on a circle. They start running in opposite directions. They want to meet at the point from where Dave started. So basically, Dave wants to complete some number of full laps while Divi wants to complete half a lap and some number of full laps.
Speed of Divi = 3/4 laps/min
Speed of Dave = 2/3 laps/min
Ratio of speed of Divi:Dave = 3/4 : 2/3 = 9:8 (multiply the ratio by 12 to get integral numbers)
This means, for every 9 complete laps that Divi runs, Dave runs 8 laps i.e. 1 lap less. But we want Divi to run half a lap and some full laps. So for every 4.5 laps that Divi runs, Dave runs 4 laps (divide 9:8 by 2). That is when they meet for the first time! Both are at the starting point of Dave.
Now both start from the starting point of Dave. When will they come back again to starting point of Dave? When Divi completes 9 laps and Dave completes 8 laps.
They will come back to the starting point of Dave for the third time when Divi again does 9 more laps and Dave does 8 more.

Third time they meet at Dave's starting point is when Divi completes 4.5+9+9 = 22.5 i.e. 22 full laps (and a half)

Originally posted by KarishmaB on 09 Apr 2012, 05:19.
Last edited by KarishmaB on 07 Nov 2023, 06:37, edited 1 time in total.
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Re: time n Speed [#permalink] New post   04 Nov 2009, 23:09
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virtualanimosity wrote:
Can any1 pls explain da concept behind such circular motion problem n meeting at startin pt problems
Q.Divi and Dave run on a circular track.Divi completes 3 laps in every 4 mins, and Dave completes 2 laps in every 3 mins in opposite direction.
Both start from points opposite to each other.Find the number of completed laps travelled by Divi when both will meet for the 3rd time at starting point of Dave.
1. 12
2. 13
3. 22


Let's call Divi X and Dave Y, would be much easier. They start from the opposite points on the circular track. We need to find the # of completed laps by X when they meet 3rd time at the starting point of Y.

Well first of all it's obvious that X will run number of complete laps and a half of the lap, as they are going to meet at the starting point of Y, which is half of the lap away from the starting point of X.

Speed of X - \(\frac{3}{4}\) L/M;

Speed of Y - \(\frac{2}{3}\) L/M;

Let's calculate when are they going to meet first time at the starting point of Y: Y will run some # of complete laps =n, and X will run k times laps plus 1/2 lap (as he starts at the opposite point)=\(nk+0.5\) \((k>0)\).

Time needed for this will be equal: \(\frac{n}{(2/3)}=\frac{(nk+0.5)}{3/4}\)--> \(n=\frac{8}{(18-16k)}\).
As \(n\) is integer value of the laps Y should run, the only possible value for k is 1, so n=4.

Which means that they will meet for the first time when Y will complete 4 laps and X 4.5 laps.

Ratio of the speed of Y to the speed of X=\((2/3)/(3/4)=\frac{8}{9}\). Which means that after their first meeting they will meet again and again at the same point each time when Y will complete 9 laps and X will complete 8. So, for the time of third meeting X will complete \(4.5+8+8=22.5\) or \(22\) full laps.

Answer: 3 (22).
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Re: time n Speed [#permalink] New post   06 Nov 2009, 01:31
That was awesome solving...just one point is not clearly understood

Let's calculate when are they going to meet first time at the starting point of Y: Y will run some # of complete laps =n, and X will run k times laps plus 1/2 lap (as he starts at the opposite point)= nk+0.5.

Why do we take X runs nk+0.5 laps? Ok .5 is also understood, can you put some light on the 'nk' term.



Bunuel wrote:
virtualanimosity wrote:
Can any1 pls explain da concept behind such circular motion problem n meeting at startin pt problems
Q.Divi and Dave run on a circular track.Divi completes 3 laps in every 4 mins, and Dave completes 2 laps in every 3 mins in opposite direction.
Both start from points opposite to each other.Find the number of completed laps travelled by Divi when both will meet for the 3rd time at starting point of Dave.
1. 12
2. 13
3. 22


Let's call Divi X and Dave Y, would be much easier. They start from the opposite points on the circular track. We need to find the # of completed laps by X when they meet 3rd time at the starting point of Y.

Well first of all it's obvious that X will run number of complete laps and a half of the lap, as they are going to meet at the starting point of Y, which is half of the lap away from the starting point of X.

Speed of X - \(\frac{3}{4}\) L/M;

Speed of Y - \(\frac{2}{3}\) L/M;

Let's calculate when are they going to meet first time at the starting point of Y: Y will run some # of complete laps =n, and X will run k times laps plus 1/2 lap (as he starts at the opposite point)=\(nk+0.5\) \((k>0)\).

Time needed for this will be equal: \(\frac{n}{(2/3)}=\frac{(nk+0.5)}{3/4}\)--> \(n=\frac{8}{(18-16k)}\).
As \(n\) is integer value of the laps Y should run, the only possible value for k is 1, so n=4.

Which means that they will meet for the first time when Y will complete 4 laps and X 4.5 laps.

Ratio of the speed of Y to the speed of X=\((2/3)/(3/4)=\frac{8}{9}\). Which means that after their first meeting they will meet again and again at the same point each time when Y will complete 9 laps and X will complete 8. So, for the time of third meeting X will complete \(4.5+8+8=22.5\) or \(22\) full laps.

Answer: 3 (22).
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Re: time n Speed [#permalink] New post   06 Nov 2009, 02:42
Have the same doubt as virtualanimosity :

Quote:
Let's calculate when are they going to meet first time at the starting point of Y: Y will run some # of complete laps =n, and X will run k times laps plus 1/2 lap (as he starts at the opposite point)= nk+0.5.


Why do we take X runs nk+0.5 laps? Ok .5 is also understood, can you put some light on the 'nk' term.

Quote:
Ratio of the speed of Y to the speed of X=(2/3)/(3/4)=\frac{8}{9}. Which means that after their first meeting they will meet again and again at the same point each time when Y will complete 9 laps and X will complete 8


How does that ratio of speeds lead to that?

Thanks in advance.
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Re: time n Speed [#permalink] New post   06 Nov 2009, 11:05
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virtualanimosity wrote:
That was awesome solving...just one point is not clearly understood

Let's calculate when are they going to meet first time at the starting point of Y: Y will run some # of complete laps =n, and X will run k times laps plus 1/2 lap (as he starts at the opposite point)= nk+0.5.

Why do we take X runs nk+0.5 laps? Ok .5 is also understood, can you put some light on the 'nk' term.


Well, actually after calculation we get n+0.5=4+0.5=4.5, but when writing this equation we should take into account that generally we could have different values for the number of laps for X and Y, not necessarily n+0.5 and n, meaning that it's not necessary for X to run exactly the same # of laps as Y plus 0.5. It happened to be in our case that with given speeds it's 4 and 4.5.

4test1 wrote:
How does that ratio of speeds lead to that?


As for the 8/9. Consider this: we know that after 4.5 laps for X, they will be at the starting point of Y. X runs the lap in 4/3min=80sec and Y runs lap in 3/2 min=90sec. After some period of time they must be at the same exact point. In how many sec it will happen? LCM of 80 and 90 is 720, so in 720 sec. In 720 sec X will run 9 laps and Y will run 8 laps. So, their every meeting at this point will be every time when X runs 9 laps and Y 8.

Hope it's clear.
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Re: time n Speed [#permalink] New post   06 Nov 2009, 11:33
Thanks for explaining the logic behind the speed ratios - I like how you could skip through and use the lcm all in one step from dividing the speeds.

On virtualanimosity's question, I think the doubt is why take "nk" and not just say a variable "p"? It's very useful to take nk as it leads to being able to see that k=1 and hence n=4. But it's not apparent why you would assume that. I took a variable p instead, separate from n, which doesn't lead directly to 4 and 4.5 (could get 8 and 9 instead).

Trying to understand how and when one chooses the nk.
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Re: time n Speed [#permalink] New post   07 Nov 2009, 13:40
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Speed of Divi - 3/4 L/M

Speed of Dave - 2/3 L/M;

n: number of laps that Divi completes
k: number of laps that Dave completes

In order to meet @ the start of Dave

(n+0.5)/(3/4) = k/(2/3)
<=> k = 4(2n+1)/9

We can check the value of n to satisfy the above equation

n = 4, k = 4
n = 13, k = 12
n= 22, k = 20

So total number of laps Divi complete will be 22+0.5 = 22.5 or 22 full laps
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Re: time n Speed [#permalink] New post   12 May 2011, 04:41
pretty conceptual this one is.
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Re: time n Speed [#permalink] New post   11 Apr 2012, 04:10
VeritasPrepKarishma wrote:
virtualanimosity wrote:
Can any1 pls explain da concept behind such circular motion problem n meeting at startin pt problems
Q.Divi and Dave run on a circular track.Divi completes 3 laps in every 4 mins, and Dave completes 2 laps in every 3 mins in opposite direction.
Both start from points opposite to each other.Find the number of completed laps travelled by Divi when both will meet for the 3rd time at starting point of Dave.
1. 12
2. 13
3. 22


Responding to a pm:

You can do it using the ratios approach. But first, look at the big picture.
Divi and Dave are at opposite points on a circle. They start running in opposite directions. They want to meet at the point from where Dave started. So basically, Dave wants to complete some number of full laps while Divi wants to complete half a lap and some number of full laps.
Speed of Divi = 3/4 laps/min
Speed of Dave = 2/3 laps/min
Ratio of speed of Divi:Dave = 3/4 : 2/3 = 9:8 (multiply the ratio by 12 to get integral numbers)
This means, for every 9 complete laps that Divi runs, Dave runs 8 laps i.e. 1 lap less. But we want Divi to run half a lap and some full laps. So for every 4.5 laps that Divi runs, Dave runs 4 laps (divide 9:8 by 2). That is when they meet for the first time! Both are at the starting point of Dave.
Now both start from the starting point of Dave. When will they come back again to starting point of Dave? When Divi completes 9 laps and Dave completes 8 laps.
They will come back to the starting point of Dave for the third time when Divi again does 9 more laps and Dave does 8 more.

Third time they meet at Dave's starting point is when Divi completes 4.5+9+9 = 22.5 i.e. 22 full laps (and a half)


great explaination karishma!
will i ever be able to think like this?
this question is realy demotivating for me to attempt for GMAT.
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Re: time n Speed [#permalink] New post   11 Apr 2012, 06:11
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kashishh wrote:

will i ever be able to think like this?
this question is realy demotivating for me to attempt for GMAT.


It is certainly a tough question. I wouldn't expect anybody to see too many such questions. So cheer up and just try to wrap your head around the logic. Next time you will have an easier time solving such a question. Every new concept you come across adds to your preparation.
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Re: time n Speed [#permalink] New post   11 Apr 2012, 06:16
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VeritasPrepKarishma wrote:
kashishh wrote:

will i ever be able to think like this?
this question is realy demotivating for me to attempt for GMAT.


It is certainly a tough question. I wouldn't expect anybody to see too many such questions. So cheer up and just try to wrap your head around the logic. Next time you will have an easier time solving such a question. Every new concept you come across adds to your preparation.


Frankly speaking I wouldn't expect such question to appear on an actual test at all. The question is taken from a non-GMAT math book and is clearly out of the scope of the GMAT.
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Re: Divi and Dave run on a circular track.Divi completes 3 laps [#permalink] New post   29 Aug 2012, 07:04
Hi Bunuel & Karishma,

I want to ask has anyone encountered such kind of question in real GMAT.

Waiting for your valuable inputs.
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Re: Divi and Dave run on a circular track.Divi completes 3 laps [#permalink] New post   29 Aug 2012, 07:31
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fameatop wrote:
Hi Bunuel & Karishma,

I want to ask has anyone encountered such kind of question in real GMAT.

Waiting for your valuable inputs.


Check the post right above your's: divi-and-dave-run-on-a-circular-track-divi-completes-3-laps-86314.html#p1072602
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Re: Divi and Dave run on a circular track.Divi completes 3 laps [#permalink] New post   29 Aug 2012, 10:10
virtualanimosity wrote:
Divi and Dave run on a circular track.Divi completes 3 laps in every 4 mins, and Dave completes 2 laps in every 3 mins in opposite direction. Both start from points opposite to each other.Find the number of completed laps travelled by Divi when both will meet for the 3rd time at starting point of Dave.

1. 12
2. 13
3. 22


Since they start out when their positions are diametrically opposed, it means that each time they meet at Dave's starting point, the difference between the distance that Divi covered and that covered by Dave is a non-negative integer number of full laps + 0.5 lap.,

If \(T\) is the time lapsed until a meeting between the two, we can write \(\frac{3}{4}T-\frac{2}{3}T=n+0.5\) for a certain non-negative integer \(n\) where \(\frac{3}{4}\) is Divi's speed and \(\frac{2}{3}\) is Dave's speed in units of laps per minute.

Then \(T=12n+6.\)
The distance covered by Divi in time \(T\) is \(\frac{3}{4}(12n+6)=9n+4.5\) and this is valid for any meeting between the two under the given conditions.
For example, when \(n=0\), Divi will travel 4 full laps and an additional 0.5 lap, while Dave will travel 4 laps.
Any other meeting will occur after 9 more laps travelled by Divi.

From the above, we can see that the total number of full laps travelled by Divi is of the form \(9n+4\).

Answer 3, as \(22 = 9\cdot2+4.\)
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Re: Divi and Dave run on a circular track.Divi completes 3 laps [#permalink] New post   29 Aug 2012, 21:54
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fameatop wrote:
Hi Bunuel & Karishma,

I want to ask has anyone encountered such kind of question in real GMAT.

Waiting for your valuable inputs.


The question is not GMAT's style though the concepts you learn out of it can come in handy. Circular motion concepts are good to know and can be tested. Whether
they are actually tested at this point is hard to say since every person who takes GMAT signs a non disclosure agreement. The official guides include questions that were live years back so they hardly reflect the current scheme. So, to cover all bases, you should be comfortable with all your high school Math concepts.
As for this particular question, you come across such questions when preparing for CAT for Indian B schools. I am quite certain it was taken from CAT material so it doesn't have much relevance as far as GMAT is concerned.
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Re: time n Speed [#permalink] New post   01 Sep 2012, 11:46
Bunuel wrote:
virtualanimosity wrote:
Can any1 pls explain da concept behind such circular motion problem n meeting at startin pt problems
Q.Divi and Dave run on a circular track.Divi completes 3 laps in every 4 mins, and Dave completes 2 laps in every 3 mins in opposite direction.
Both start from points opposite to each other.Find the number of completed laps travelled by Divi when both will meet for the 3rd time at starting point of Dave.
1. 12
2. 13
3. 22


Let's call Divi X and Dave Y, would be much easier. They start from the opposite points on the circular track. We need to find the # of completed laps by X when they meet 3rd time at the starting point of Y.

Well first of all it's obvious that X will run number of complete laps and a half of the lap, as they are going to meet at the starting point of Y, which is half of the lap away from the starting point of X.

Speed of X - \(\frac{3}{4}\) L/M;

Speed of Y - \(\frac{2}{3}\) L/M;

Let's calculate when are they going to meet first time at the starting point of Y: Y will run some # of complete laps =n, and X will run k times laps plus 1/2 lap (as he starts at the opposite point)=\(nk+0.5\) \((k>0)\).

Time needed for this will be equal: \(\frac{n}{(2/3)}=\frac{(nk+0.5)}{3/4}\)--> \(n=\frac{8}{(18-16k)}\).
As \(n\) is integer value of the laps Y should run, the only possible value for k is 1, so n=4.

Which means that they will meet for the first time when Y will complete 4 laps and X 4.5 laps.

Ratio of the speed of Y to the speed of X=\((2/3)/(3/4)=\frac{8}{9}\). Which means that after their first meeting they will meet again and again at the same point each time when Y will complete 9 laps and X will complete 8. So, for the time of third meeting X will complete \(4.5+8+8=22.5\) or \(22\) full laps.

Answer: 3 (22).


Hi Bunuel,

When I solved it, I did same reasoning and got 4.5+8+8, which I calculated as 20 and this was not in the answer choice.
So then I took my original equation 9y-8x=4 and tried to substitute for values of x from the answer choices such that y is an integer and got the answer 22, which I think is wrong and 20.5 as correct.
How did you get 4.5+8+8 as 22.5? Sorry if this is something silly and I missed it :(
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Re: time n Speed [#permalink] New post   01 Sep 2012, 12:00
akashv wrote:
Bunuel wrote:
virtualanimosity wrote:
Can any1 pls explain da concept behind such circular motion problem n meeting at startin pt problems
Q.Divi and Dave run on a circular track.Divi completes 3 laps in every 4 mins, and Dave completes 2 laps in every 3 mins in opposite direction.
Both start from points opposite to each other.Find the number of completed laps travelled by Divi when both will meet for the 3rd time at starting point of Dave.
1. 12
2. 13
3. 22


Let's call Divi X and Dave Y, would be much easier. They start from the opposite points on the circular track. We need to find the # of completed laps by X when they meet 3rd time at the starting point of Y.

Well first of all it's obvious that X will run number of complete laps and a half of the lap, as they are going to meet at the starting point of Y, which is half of the lap away from the starting point of X.

Speed of X - \(\frac{3}{4}\) L/M;

Speed of Y - \(\frac{2}{3}\) L/M;

Let's calculate when are they going to meet first time at the starting point of Y: Y will run some # of complete laps =n, and X will run k times laps plus 1/2 lap (as he starts at the opposite point)=\(nk+0.5\) \((k>0)\).

Time needed for this will be equal: \(\frac{n}{(2/3)}=\frac{(nk+0.5)}{3/4}\)--> \(n=\frac{8}{(18-16k)}\).
As \(n\) is integer value of the laps Y should run, the only possible value for k is 1, so n=4.

Which means that they will meet for the first time when Y will complete 4 laps and X 4.5 laps.

Ratio of the speed of Y to the speed of X=\((2/3)/(3/4)=\frac{8}{9}\). Which means that after their first meeting they will meet again and again at the same point each time when Y will complete 9 laps and X will complete 8. So, for the time of third meeting X will complete \(4.5+8+8=22.5\) or \(22\) full laps.

Answer: 3 (22).


Hi Bunuel,

When I solved it, I did same reasoning and got 4.5+8+8, which I calculated as 20 and this was not in the answer choice.
So then I took my original equation 9y-8x=4 and tried to substitute for values of x from the answer choices such that y is an integer and got the answer 22, which I think is wrong and 20.5 as correct.
How did you get 4.5+8+8 as 22.5? Sorry if this is something silly and I missed it :(



Got the mistake. It is 4.5+9+9 and not 4.5+8+8. I hope I dont make such mistakes on the test
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Re: Divi and Dave run on a circular track. Divi completes 3 laps [#permalink] New post   26 Jun 2014, 03:14
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Since the ratio of speeds of Divi and Dave is 9:8 Divi has to run 9 laps to gain 1 lap i.e, they both meet at the same point when Divi runs 9 laps. To gain the initial 1/2 lap difference, Divi has to run 4.5 laps . That's when they first meet. The second and third time they meet at the same point is when Divi runs 9 laps both the times. So Divi totally runs 4.5+9+9= 22.5 laps i.e, completes 22 laps.
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Divi and Dave run on a circular track. Divi completes 3 laps [#permalink] New post   03 Sep 2014, 07:00
Question turns easy if we know some basics of circular speed and distance rule.
For example in this question as both are running in same direction, we will have to calculate aggregate speed by subtracting their respective speed.

Further as overall circular distance is not given, I will consider it 360degree.

Divi speed is 3*360/4 = 270 degree per min
Dave speed is 2*360/2 = 240 degree per min
Aggregate speed = 270-240=30 degree per min

At initial position distance between Divi and Dave is 180 degree.
Time taken to cover this 180 degree = 180/30 = 6min

Now their start point has changed and total distance to meet again between them will be 360 degree.
So time taken to cover this 360 = 360/30 = 12 min.

Similarly for 3rd meet they will take another 12min.

Total time = 6+12+12= 30min

Divi's revolution/min = 3/4
thus, in 30 min = 3/4*30 = 22.5

Ans = 22 (C)
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gmatclubot
Divi and Dave run on a circular track. Divi completes 3 laps [#permalink]
03 Sep 2014, 07:00
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