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DS - time speed

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anuramm
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DS - time speed [#permalink] New post   30 Aug 2004, 10:12
00:00
A
B
C
D
E

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(N/A)

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Can anyone explain what the first stem means.



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Re: DS - time speed [#permalink] New post   30 Aug 2004, 10:55
The first stem basically means, X crossed Y for the second time. This stem does not indicate anything about TIMES, hence is insufficient.

The answer to this problem would be B.

Total miles = 10,000
Track size = 400
No. of laps = 10,000 / 400 = 25

Now X takes 60 seconds/lap, for the whole race (25 laps) = 25 x 60
Now Y takes 62.4 seconds/lap, for the whole race (25 laps) = 25 x 62.4

Total race time for X : Total race time for Y
=> (25 x 60) / (25 x 62.4) = 25/26
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Re: DS - time speed [#permalink] New post   30 Aug 2004, 11:12
A is insufficient.

My choice is B
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Re: DS - time speed [#permalink] New post   30 Aug 2004, 11:17
I would choose B too.

St.1 doesnt indicate anything about the time for X and Y
St.2 gives us the average of both x and y, while we know they both had to run 25 laps.

Regards,

Alex
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Re: DS - time speed [#permalink] New post   30 Aug 2004, 11:20
The first stem means that the first runner crossed the second two times. The net distance between them is therefore 400x2=800.
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Re: DS - time speed [#permalink] New post   30 Aug 2004, 12:20
assuming average times over the race, doesn't A) indicate that X was two laps ahead of Y? thus, meaning that X had finished 10,000 meters, whereas Y had only finished 9,200 meters? so isn't the ratio of time the same as the ratio of distance? thus the ratio of Y to X was 9200 to 10000, or 23/25?
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Re: DS - time speed [#permalink] New post   30 Aug 2004, 16:33
well, we know that while X (first runner) completed 25 laps, Y (second runner) completed 23 laps. Hence, in the same time, one completed 10,000, while the other completed 9,200. I think, we can pick any time for T here and get the relationship of t1 to t2.
hence I'd go for D
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Re: DS - time speed [#permalink] New post   30 Aug 2004, 23:49
I'll deal with the simple one first.

From the question, we know they need to do 10,000/400=25 laps

From (2), X takes 60*25 = 1500 seconds, and Y takes 62.4*25 = 1560 seconds
With this in hand, (2) is sufficient.

Now we need to consider (1).
When X finishes 25 laps, he lapped Y for the second time. This means he is 800 meters ahead.
All we have is information about the distance X and Y have already ran, but nothing about their speed, and so we can conclude the time they took and hence the ratio of their time.
Therefore (1) is not sufficient.

(B) should be the answer.
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Re: DS - time speed [#permalink] New post   31 Aug 2004, 01:22
The OA is B.
I couldn't understand what A means exactly. Thanks guys for explaining the meaning.
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Re: DS - time speed [#permalink] New post   01 Sep 2004, 07:48
I think what is tricky whith statement 1 is that we do not know how much time will need Y to complete the 2 remaining laps after he was crossed by X.

I would also choose B
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Re: DS - time speed [#permalink] New post   01 Sep 2004, 08:57
twixt wrote:
I think what is tricky whith statement 1 is that we do not know how much time will need Y to complete the 2 remaining laps after he was crossed by X.

I would also choose B


Wow! I choose D considering that Y travels at a constant speed. Good catch.

Just for discussion, If the question is:
Two Runners, X and Y, competed in a 10000 m run. The race took place in a oval track 400m long. What is tha ration od X's time to Y's time, if they run at constant speed?
(1) Just a X crossed the finish line, he lapped Y for the second time.
(2) X averaged 60s per lap, and Y averaged 62.4 s per lap.

the answer would be (D).



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Re: DS - time speed [#permalink]
01 Sep 2004, 08:57
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