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Relative speed on circular path

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Relative speed on circular path

by akpareek » Mon Feb 11, 2013 7:21 am
John, Vicky and David starts with 2 m/s, 2.5 m/s and 5 m/s at same point at same time in 100 meters circular track. John and David start in anti-clock direction whereas Vicky starts in clock-wise direction. After how many seconds will they meet second time?

A) 100/3 B) 200/3 C) 100 D) 400/3 E) 200
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by vishalbpr » Tue Feb 12, 2013 2:16 am
akpareek wrote:After how many seconds will they meet second time?
This is not clear... There are 3 people in calculations, so who 2 should we reference here?
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by GMATGuruNY » Tue Feb 12, 2013 7:13 am
This seems to be the intent of the problem:
At 12pm, John, Vicky and David all start running from the same point around a 100-meter circular track. John and David run in one direction, while Vicky runs in the opposite direction. John runs at a speed of 2 meters per second, Vicky at a speed of 2.5 meters per second, and David at a speed of 5 meters per second. How many seconds will have passed when the three runners meet for the second time?

A) 100/3 B) 200/3 C) 100 D) 400/3 E) 200
David and John:
When people travel in the same direction, SUBTRACT THEIR RATES.
David's rate - John's rate = 5-2 = 3 meters per second.
Thus, every second, David SURPASSES John by 3 meters.

To meet John, David must surpass John by the ENTIRE LENGTH OF THE TRACK (100 meters).
Time required = (track length)/(rate difference) = 100/3 seconds.
Thus, every 100/3 seconds, David will surpass John by the entire length of the track.
In other words, David and John will MEET every 100/3 seconds.

David and Vicky:
When people travel in opposite directions, ADD THEIR RATES.
The reason is that the people are WORKING TOGETHER to cover the distance between them.
David's rate + Vicky's rate = 5 + 2.5 = 15/2 meters per second.
Thus, every second, David and Vicky cover 15/2 meters between them.

Since the track is 100 meters, David and Vicky will MEET every time they work together to cover 100 meters between them.
In other words, David and Vicky will meet each time they travel a MULTIPLE of 100 meters.

Since David and John meet every 100/3 seconds, we need to determine how many intervals of 100/3 seconds must pass for David and Vicky to travel a multiple of 100 meters.
Distance traveled by David and Vicky in 100/3 seconds = r*t = (100/3)(15/2) = 250 meters.
Since 100/3 seconds = 250 meters, 200/3 seconds = 500 meters, which is a multiple of 100.

Thus, all 3 runners will meet every 200/3 seconds, implying that they will meet for the SECOND TIME after 400/3 seconds.

The correct answer is D.
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by ziko » Wed Feb 13, 2013 9:27 pm
GMATGuruNY wrote:This seems to be the intent of the problem:
At 12pm, John, Vicky and David all start running from the same point around a 100-meter circular track. John and David run in one direction, while Vicky runs in the opposite direction. John runs at a speed of 2 meters per second, Vicky at a speed of 2.5 meters per second, and David at a speed of 5 meters per second. How many seconds will have passed when the three runners meet for the second time?

A) 100/3 B) 200/3 C) 100 D) 400/3 E) 200
David and John:
When people travel in the same direction, SUBTRACT THEIR RATES.
David's rate - John's rate = 5-2 = 3 meters per second.
Thus, every second, David SURPASSES John by 3 meters.

To meet John, David must surpass John by the ENTIRE LENGTH OF THE TRACK (100 meters).
Time required = (track length)/(rate difference) = 100/3 seconds.
Thus, every 100/3 seconds, David will surpass John by the entire length of the track.
In other words, David and John will MEET every 100/3 seconds.

David and Vicky:
When people travel in opposite directions, ADD THEIR RATES.
The reason is that the people are WORKING TOGETHER to cover the distance between them.
David's rate + Vicky's rate = 5 + 2.5 = 15/2 meters per second.
Thus, every second, David and Vicky cover 15/2 meters between them.

Since the track is 100 meters, David and Vicky will MEET every time they work together to cover 100 meters between them.
In other words, David and Vicky will meet each time they travel a MULTIPLE of 100 meters.

Since David and John meet every 100/3 seconds, we need to determine how many intervals of 100/3 seconds must pass for David and Vicky to travel a multiple of 100 meters.
Distance traveled by David and Vicky in 100/3 seconds = r*t = (100/3)(15/2) = 250 meters.
Since 100/3 seconds = 250 meters, 200/3 seconds = 500 meters, which is a multiple of 100.

Thus, all 3 runners will meet every 200/3 seconds, implying that they will meet for the SECOND TIME after 400/3 seconds.

The correct answer is D.
Well explained! But i have one question to clarify, is it GMAT type of question, because i have not seen such question so far, i have seen other similar type, when we consider only two people.

Would be glad to know our opinion.
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by GMATGuruNY » Fri Feb 15, 2013 8:26 am
ziko wrote:
GMATGuruNY wrote:This seems to be the intent of the problem:
At 12pm, John, Vicky and David all start running from the same point around a 100-meter circular track. John and David run in one direction, while Vicky runs in the opposite direction. John runs at a speed of 2 meters per second, Vicky at a speed of 2.5 meters per second, and David at a speed of 5 meters per second. How many seconds will have passed when the three runners meet for the second time?

A) 100/3 B) 200/3 C) 100 D) 400/3 E) 200
David and John:
When people travel in the same direction, SUBTRACT THEIR RATES.
David's rate - John's rate = 5-2 = 3 meters per second.
Thus, every second, David SURPASSES John by 3 meters.

To meet John, David must surpass John by the ENTIRE LENGTH OF THE TRACK (100 meters).
Time required = (track length)/(rate difference) = 100/3 seconds.
Thus, every 100/3 seconds, David will surpass John by the entire length of the track.
In other words, David and John will MEET every 100/3 seconds.

David and Vicky:
When people travel in opposite directions, ADD THEIR RATES.
The reason is that the people are WORKING TOGETHER to cover the distance between them.
David's rate + Vicky's rate = 5 + 2.5 = 15/2 meters per second.
Thus, every second, David and Vicky cover 15/2 meters between them.

Since the track is 100 meters, David and Vicky will MEET every time they work together to cover 100 meters between them.
In other words, David and Vicky will meet each time they travel a MULTIPLE of 100 meters.

Since David and John meet every 100/3 seconds, we need to determine how many intervals of 100/3 seconds must pass for David and Vicky to travel a multiple of 100 meters.
Distance traveled by David and Vicky in 100/3 seconds = r*t = (100/3)(15/2) = 250 meters.
Since 100/3 seconds = 250 meters, 200/3 seconds = 500 meters, which is a multiple of 100.

Thus, all 3 runners will meet every 200/3 seconds, implying that they will meet for the SECOND TIME after 400/3 seconds.

The correct answer is D.
Well explained! But i have one question to clarify, is it GMAT type of question, because i have not seen such question so far, i have seen other similar type, when we consider only two people.

Would be glad to know your opinion.
Rate problems about circular travel are rare.
The one here -- which involves three people, two moving in one direction, with the other moving in the opposite direction -- seems especially complex.
An actual GMAT question would probably involve only two people moving either in the same direction or in opposite directions.
That being said, the solution here requires very little calculating; a test-taker who understands the reasoning should be able to determine the correct answer within two minutes.
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