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At 6 PM, Renee and Eva begin cycling up a 10 kilometre slope

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At 6 PM, Renee and Eva begin cycling up a 10 kilometre slope [#permalink] New post 09 Oct 2006, 03:47
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At 6 PM, Renee and Eva begin cycling up a 10 kilometre slope at different but constant speeds. On reaching the top of the slope, each turns around and cycles down to the bottom of the slope at a constant speed that is twice the speed at which she ascended. How far apart will the two girls be when Renee reaches the bottom of the slope?

(1) When Renee has cycled 14 kilometres, she cycles by Eva.
(2) Renee reaches the top of the slope 40 minutes before Eva does.

Last edited by kevincan on 09 Oct 2006, 08:30, edited 1 time in total.
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 [#permalink] New post 09 Oct 2006, 06:53
How could choice (1) even happen if they are cycling at constant speeds the entire time?
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Re: DS: Eva and Renee [#permalink] New post 09 Oct 2006, 07:34
1) IF rene and eva move up at different speeds and their speeds are constant.so,if eva was ahead of rene for the first 14 kms then how can rene take over eva???
this statement is not sufficient.
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 [#permalink] New post 09 Oct 2006, 07:42
1) IF rene and eva move up at different speeds and their speeds are constant.so,if eva was ahead of rene for the first 14 kms then how can rene take over eva???
this statement is not sufficient.
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Re: DS: Eva and Renee [#permalink] New post 09 Oct 2006, 09:32
(2)this one is also not sufficient.i think u must know either of their speed to answer that.
i solved it mathematically.this is what i came up with.
let speed of rene = x km/min
speed of eva= y

time taken by eva=10/y
timetaken by rene=10/y+40=10/x
hence 4xy=x-y

now when rene has already reached the top,distance covered by eva=speed*time=y*10/x=10y/x

and now rene takes 10/2x time to reach at the bottom.
so eva has 10/2x to climb to top and then descent whatec\ver distance she can.

while climbing up,distace left frr eva=10-10y/x
speed=y
therefore time taken by eva to climb up = (10-10y/x)/y =(10x-10y)/xy

so now time remaining for eva while she comes down the slope= (10/2x) -(10x-10y)/xy=(15y-10x)/xy

speed =2y

hence distance travelled by eva in this much time =(15y-10x)/xy *2y=(30y-20x)/x

right now rene is at the bottom,so the distance between the two =

10-(30y-20x)/x=(30x-30y)/x=30(x-y)/x

we can substitute x-y=4xy as derived earlier on,
we get,

distance between the two=30(x-y)/x=30*4xy/x=120y
since y isstill unknown,we can not determine the distance between the two.


i hope this is not the solution or this type doesn't come in actual gmat!
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 [#permalink] New post 09 Oct 2006, 10:03
A

Statement 1:

(10/R) + (4/2R) = 6/E

6R = 12E.

When Renee has covered 10 miles at R and 10 miles at 2R, E would have covered 7.5 miles.
So, 7.5 miles apart. SUFF

Statement 2: Insufficient
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 [#permalink] New post 09 Oct 2006, 20:02
Is their a consenus on the correct answer? If so, please explain.
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 [#permalink] New post 09 Oct 2006, 20:29
tx anand,

i just missed the thought that rene could meet eva on the way back.
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 [#permalink] New post 09 Oct 2006, 22:15
Kevin, i will choose A.

The gap will be 7.5...........
_________________

Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)

  [#permalink] 09 Oct 2006, 22:15
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At 6 PM, Renee and Eva begin cycling up a 10 kilometre slope

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