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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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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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Is r > s ? [#permalink]

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New post 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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