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GMATPrep: Geometry DS

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GMATPrep: Geometry DS [#permalink]

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New post 28 May 2009, 23:02
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A
B
C
D
E

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Please explain your logic. Thanks
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Re: GMATPrep: Geometry DS [#permalink]

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New post 28 May 2009, 23:26
C.
Lets say we have a point between r and s and we depict it by O (and i assume that it is 0 as 0 had to be to the left of s).

Now combining 1 and 2 we get..
TR(dist. b/w t and r) = -ST (distance between -s and r)

RO+OS+ST = -SO + OS + ST
this shows R = -S (means r is nothing but -S) so 0 had to be exactly in between them.

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Re: GMATPrep: Geometry DS [#permalink]

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New post 28 May 2009, 23:31
Couldn't you have gotten that with B alone? Why do you need to know that S is to the right of 0?
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Re: GMATPrep: Geometry DS [#permalink]

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New post 29 May 2009, 04:03
Hades wrote:
Couldn't you have gotten that with B alone? Why do you need to know that S is to the right of 0?


The question asks is 0&r = 0&s?
From B : t&r = t&-s, is t = 0? may be / may be not...so B is insuff

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Re: GMATPrep: Geometry DS [#permalink]

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New post 29 May 2009, 07:22
pm4553 wrote:
Hades wrote:
Couldn't you have gotten that with B alone? Why do you need to know that S is to the right of 0?


The question asks is 0&r = 0&s?
From B : t&r = t&-s, is t = 0? may be / may be not...so B is insuff


I disagree

t=0 is not the question, in fact t can be any value

here is my logic

CLEARLY from the diagram r<s<t

say 0<r<s<t
not possible unless r=s=0

say r<0<s<t
Possible, if r=-s, and we have a YES answer

say r<s<0<t
not possible, unless t=0 & r=s

say r<s<t<0
not possible, unless t=0 & r=s


Hence B is sufficient
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Re: GMATPrep: Geometry DS [#permalink]

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New post 02 Jun 2009, 02:23
Bump-- anyone?

I'm assuming my inequality assumptions are too rigid...
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Re: GMATPrep: Geometry DS [#permalink]

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New post 04 Jun 2009, 00:53
from B,

|t-r| = |t+s|

as t>s>r,

t-r is always positive...

but not sure abt t+s..

t-r = t+s
s=-r
s+r = 0.. ( implies s and r are equidistant from zero)

t-r = -t-s

2t = r-s .. ( implies no sensible relationship between a and r )

Combining 1, s>0 implies t >0, implies t+s >0

so only the first scenario holds gud..

Hence the answer C

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Re: GMATPrep: Geometry DS [#permalink]

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New post 04 Jun 2009, 05:13
I don't see how -r = s is not sufficient.
can you please give and example when without A then B would be insufficient?
thanks

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Re: GMATPrep: Geometry DS [#permalink]

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New post 05 Jun 2009, 02:12
With B we can have 2 combinations:
1. __(r=-s)___(0)___(s)________(t)____ or
2. __(r)___(s)___(t)____(0)___(-s), for example: r=-4; s=-2; t=-1 and -s=2
Then, only C is sufficient.

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Re: GMATPrep: Geometry DS [#permalink]

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New post 05 Jun 2009, 04:59
Sunchaser20 wrote:
With B we can have 2 combinations:
1. __(r=-s)___(0)___(s)________(t)____ or
2. __(r)___(s)___(t)____(0)___(-s), for example: r=-4; s=-2; t=-1 and -s=2
Then, only C is sufficient.



Why is the -s to the right side of +s.
I understand that you mean s could mean a negative value already. for e.g. s= -4.

Then, it would be odd, or probably wrong to mention -s to mean +4 !!!

any other thoughts on this are appreciated.

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Re: GMATPrep: Geometry DS [#permalink]

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New post 05 Jun 2009, 13:27
osnat wrote:
I don't see how -r = s is not sufficient.
can you please give and example when without A then B would be insufficient?
thanks


true that -r = s is suffient to say..
but this is not the equation given to us .. but rather derived from the calculation.. i.e the statement B.

-r = s is only one possibility....

when we consider the other possibility.. i.e 2t = r-s i.e t = -1, r = -6, s = -4..
it also holds gud...but r and s are not equidistant form 0.

and hence we cannot say that it is B alone is sufficient..

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Re: GMATPrep: Geometry DS [#permalink]

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New post 06 Jun 2009, 05:21
Hades wrote:
Please explain your logic. Thanks

Can you explain it in a more plausible way

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Re: GMATPrep: Geometry DS [#permalink]

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New post 09 Jun 2009, 00:26
I agree with C
If S to the left of zero, from 2) 0 can be between or to the right of r and s

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Re: GMATPrep: Geometry DS   [#permalink] 09 Jun 2009, 00:26
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