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distance time

by shashank.ism » Sat Feb 06, 2010 5:16 am
In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?

a.) 20 min
b.) 15 min
c.) 10 min
d.) 5 min
e.) None of these
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by harsh.champ » Sat Feb 06, 2010 5:22 am
shashank.ism wrote:In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?

a.) 20 min
b.) 15 min
c.) 10 min
d.) 5 min
e.) None of these
IMO,the ans. [spoiler]is (D)[/spoiler]
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by Brent@GMATPrepNow » Sat Feb 06, 2010 8:29 am
shashank.ism wrote:In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same starting point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?

a.) 20 min
b.) 15 min
c.) 10 min
d.) 5 min
e.) None of these
After 5 minutes the faster runner "laps" (passes) the slower runner. We are told that this is the first time this happened.
So, at this point, we know that the faster runner has run 1 lap more than the slower runner.
Since, we also know that the faster runner travels at twice the speed of the slower runner, we can conclude that, at this point, the faster runner has completed 2 laps and the slower runner has completed 1 lap.
The race is 4 laps.
If it took the faster runner 5 minutes to complete 2 laps, it will take a total of 10 minutes for the faster runner to complete the entire race (4 laps)
So, the answer is C
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by shashank.ism » Sat Feb 06, 2010 8:48 am
After 5 minutes the faster runner "laps" (passes) the slower runner. We are told that this is the first time this happened.
So, at this point, we know that the faster runner has run 1 lap more than the slower runner.
Since, we also know that the faster runner travels at twice the speed of the slower runner, we can conclude that, at this point, the faster runner has completed 2 laps and the slower runner has completed 1 lap.
The race is 4 laps.
If it took the faster runner 5 minutes to complete 2 laps, it will take a total of 10 minutes for the faster runner to complete the entire race (4 laps)
So, the answer is C

Brent I was actually confused regarding this problem.. how to take the right approach.. the faster runner has completed 2 laps and the slower runner has completed 1 lap. This line of yours actually solves the problem.and the OA is also C

I got your point a bit, but I am still confused how do you reached to this conclusion. why the first runner didn't met the second runner before the end of second lap.(although 2nd runner has completed 1 lap when 2nd runner has completed only half lap till that time.). will you explain it in more details.
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by anki_jain » Sat Feb 27, 2010 4:26 am
if faster completes 1000, slower would complete 500..likewise faster 1500, slower 750....faster 2000, slower 1000...so they will actually meet at the start point only when slower comes at it for the first time and faster comes for the second time. This is what Brent says.
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by cat er ina » Mon Mar 01, 2010 3:14 pm
what I don t get here is why the faster has already run two laps and not one when he meets the slowest. since it says that is the first time that happens I thought both were in their first lap.
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by cat er ina » Mon Mar 01, 2010 3:16 pm
what I don t get here is why the faster has already run two laps and not one when he meets the slowest. since it says that is the first time that happens I thought both were in their first lap.
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by Testluv » Mon Mar 01, 2010 5:26 pm
The moment after the race begins, the faster runner pulls ahead of the slower one. Because it is a circle, the faster runner will soon be far ahead of the slower one, until eventually he "catches up" and meets the slower runner again. At this "catch-up" point, regardless of the two runners' specific speeds, the faster runner has travelled exactly one more lap than the slower one.

The question tells us that they reach the same point for the first time at the five-minute mark. Because this is the first time they meet, we know that this is the first catch-up point, and that the faster runner has travelled one extra lap for the first time also.

If this is the first time the faster runner has travelled one extra lap, then the slower runner has ran one lap, while the faster runner has ran two. Since each lap is 1000 m, we know that the faster runner travels 2000 m in 5 min. So, that would be 10 min for 4000 m (c).
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by cat er ina » Tue Mar 02, 2010 12:29 pm
OK, I did not had clear in my mind the idea that the faster runner was in his second lap but it makes sense! thank you!
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by STEVEN SPIELBERG » Mon Mar 15, 2010 3:11 am
IMOC.
I want to win an OSCAR on the GMAT !!!
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