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Is r > s ?

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Is r > s ? [#permalink]

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New post 01 Oct 2010, 08:58
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Is r > s ?

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

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New post 01 Oct 2010, 09:03
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agnok wrote:
Is r > s ?

(1) -r + s < 0
(2) r < | s |
Please explain:


Is \(r>s\)?

(1) \(-r+s<0\) --> 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.

Answer: A.
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Re: Is r>s? [#permalink]

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New post 01 Oct 2010, 09:25
Answer A is correct, try to solve such problems by number plug-in
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Re: Is r > s ? [#permalink]

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New post 13 Mar 2016, 06:57
HERE statement 1 is sufficient as we directly arrive at the result of r>s

statement 2 is insufficient as it tells us => s>r or s<-r => not sufficient
hence A is sufficient
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Re: Is r > s ? [#permalink]

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New post 18 Jul 2016, 01:41
agnok wrote:
Is r > s ?

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


This is a yes or no question. Any value as long as it is definite yes or definite no will be correct
Is r > s ?
(1) -r + s < 0
r>s
Is r > s = YES
SUFFICIENT


(2) r < | s |
r<s or r<-s

First case
r<s
Is r > s NO

Second case
r <- s sometimes yes, sometimes no

Is 4 < -5 no
Is 4 < -(-5) yes

INSUFFICIENT

ANSWER IS A
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Re: Is r > s ? [#permalink]

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New post 24 Jul 2016, 06:49
2) r < |s|
doesn't it mean that r lies between s and -s -s < r < s ... so r definitely is smaller than s ?
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Re: Is r > s ? [#permalink]

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New post 24 Jul 2016, 08:02
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Re: Is r > s ?   [#permalink] 24 Jul 2016, 08:02
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