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# Is r > s ?

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Manager
Joined: 17 Nov 2009
Posts: 239
Followers: 1

Kudos [?]: 24 [0], given: 17

Is r > s ? [#permalink]

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01 Oct 2010, 08:58
00:00

Difficulty:

15% (low)

Question Stats:

82% (01:42) correct 18% (00:34) wrong based on 68 sessions

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Is r > s ?

(1) -r + s < 0
(2) r < | s |
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 32623
Followers: 5654

Kudos [?]: 68653 [1] , given: 9816

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01 Oct 2010, 09:03
1
KUDOS
Expert's post
agnok wrote:
Is r > s ?

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

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.

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Manager
Joined: 06 Nov 2009
Posts: 177
Concentration: Finance, Strategy
Followers: 1

Kudos [?]: 8 [0], given: 3

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01 Oct 2010, 09:25
Answer A is correct, try to solve such problems by number plug-in
Director
Joined: 12 Aug 2015
Posts: 686
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Kudos [?]: 57 [0], given: 160

Re: Is r > s ? [#permalink]

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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
Re: Is r > s ?   [#permalink] 13 Mar 2016, 06:57
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