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# Is r > s ?

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Manager
Joined: 17 Nov 2009
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Is r > s ? [#permalink]

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01 Oct 2010, 08:58
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5% (low)

Question Stats:

79% (00:36) correct 21% (00:51) wrong based on 211 sessions

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Is r > s ?

(1) -r + s < 0
(2) r < | s |
[Reveal] Spoiler: OA

Kudos [?]: 59 [0], given: 17

Math Expert
Joined: 02 Sep 2009
Posts: 41886

Kudos [?]: 128774 [1], given: 12182

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01 Oct 2010, 09:03
1
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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
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Concentration: Finance, Strategy

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01 Oct 2010, 09:25
Answer A is correct, try to solve such problems by number plug-in

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BSchool Forum Moderator
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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
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Kudos [?]: 836 [0], given: 595

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Re: Is r > s ? [#permalink]

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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

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Re: Is r > s ? [#permalink]

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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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24 Jul 2016, 08:02
Expert's post
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thapliya wrote:
2) r < |s|
doesn't it mean that r lies between s and -s -s < r < s ... so r definitely is smaller than s ?

That's not correct.

(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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Re: Is r > s ? [#permalink]

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08 Jul 2017, 01:14
Bunuel wrote:
thapliya wrote:
2) r < |s|
doesn't it mean that r lies between s and -s -s < r < s ... so r definitely is smaller than s ?

That's not correct.

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

Bunuel, please correct my below understanding, if i am wrong:

Example: If we have an equation, |x| >1 it means simply x>1 or x<-1;

Coming back to the question, it states that s>r or s<-r
Case 1. s>r => r<s
Case 2. s<-r => -s>r => r<-s
combining the 2 cases we have r<-s;

Now in order for r<-s to be true we have further 2 cases:
s=+ve; then, r must be negative => s=1 then r<-1 clearly r is not greater than s
s=-ve; then, r can be wither + or -; s=-2 then r=1 or -3 in such case we don't know r>s or not. Hence it is insufficient to answer.

Is my reasoning correct?
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Re: Is r > s ?   [#permalink] 08 Jul 2017, 01:14
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