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# Is r > s ? (1) -r + s < 0 (2) r < | s |

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Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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29 Jan 2012, 18:54
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Is r > s ?

(1) -r + s < 0
(2) r < | s |

Well, 1st one is clear but I have some difficulties with second statement. I guessed it's not sufficient but need clarification. Thank you!
[Reveal] Spoiler: OA

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Re: Is r > s ? [#permalink]

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29 Jan 2012, 19:02
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DesecratoR wrote:
Is r > s ?

(1) -r + s < 0
(2) r < | s |

Well, 1st one is clear but I have some difficulties with second statement. I guessed it's not sufficient but need clarification. Thank you!

Welcome to GMAT Club. Below is a solution for this problem.

Is $$r>s$$?

(1) $$-r+s<0$$ --> rearrange (or add $$r$$ to both parts): $$s<r$$, directly answers the question. Sufficient.

(2) $$r<|s|$$ --> either $$r<s$$ or $$r<-s$$ (for example $$r=1$$ and $$s=2$$ OR $$r=1$$ and $$s=-2$$). Not sufficient.

Basically this statement tells that absolute value of s is more than r, but s itself may be more, as well as less than r.

Hope it's clear.
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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29 Jan 2012, 19:07
Clear, thanks! See you around!
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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21 Feb 2012, 22:38
Thanks but I am struggling with statement II

The statement tells us that
a) r<-s or :
s -s r r<-s? r>s?
5 -5 6 No
-5 5 -6 Yes No

b) r<s : r cannot be greater than S

both the substatements say thar r is not greater than s therefor should be sufficient
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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21 Feb 2012, 23:41
devinawilliam83 wrote:
Thanks but I am struggling with statement II

The statement tells us that
a) r<-s or :
s -s r r<-s? r>s?
5 -5 6 No
-5 5 -6 Yes No

b) r<s : r cannot be greater than S

both the substatements say thar r is not greater than s therefor should be sufficient

(2) $$r<|s|$$ --> either $$r<s$$ OR $$r<-s$$. Now, try some number to see that this statement is not sufficient: if $$r=1$$ and $$s=2$$ then $$r<s$$ BUT if $$r=1$$ and $$s=-2$$ then $$r>s$$.

Again: this statement tells that absolute value of $$s$$ is more than $$r$$, but $$s$$ itself may be more, as well as less than $$r$$.

Hope it's clear.
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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22 Feb 2012, 04:11
A

(I) rearrange to s<r --> sufficient
(II) r < | s | --> r can be smaller s (5<6) or greater s (5>-6), absolute value hide the positive or negative sign --> not sufficient
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Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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15 Nov 2012, 21:44
Lolaergasheva wrote:
Is r > s ?

(1) -r + s < 0
(2) r < | s |

Is r>s?

(1) -r+s<0
so s<r
so for question answer will be "yes"
So (1) is sufficient

(2) r<|s|
If s is positive sufficient
If s is negative not sufficient.
So (2) is also not sufficient

A
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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04 Dec 2012, 02:40
Is r > s ?

(1) -r + s < 0
Manipulate -r + s > 0 ==> s > r ==> NO! ==> SUFFICIENT!

(2) r < | s |
Let r = 5, s = -6 ==> r > s ? YES!
Let r = 5, s = 6 ==> r > s ? NO!
INSUFFICIENT!

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Re: Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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04 Dec 2012, 20:36
Another way to look at the second statement is this. this will save u a lot of time. won't even have to plug in for this.

Remember that |x| >y (or any such relation, no matter >,< or =) can be written as
-y>x>y.
(that is, just remember to keep the relation the same and just put negative on the left hand side and positive on the right hand side.)

so... statement 2 says r<|S|
meaning |s|>r ==> -r>s>r

so on the left hand side, S lies further left of -r, such that if r is 2 and thus -r is -2, s is even less and so s is -3 thus s is less than r, and so the main queston r>s is answered yes. but on the right hand side, s>r so the main question is answered no. hence stmt 2 is not sufficient.

DesecratoR wrote:
Is r > s ?

(1) -r + s < 0
(2) r < | s |

Well, 1st one is clear but I have some difficulties with second statement. I guessed it's not sufficient but need clarification. Thank you!
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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15 Sep 2014, 22:50
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s | [#permalink]

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26 Dec 2015, 01:19
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Is r > s ? (1) -r + s < 0 (2) r < | s |   [#permalink] 26 Dec 2015, 01:19
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