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Is r > s ? (1) r + s < 0 (2) r <  s 
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29 Jan 2012, 17:54
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76% (00:46) correct 24% (00:53) wrong based on 714 sessions
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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!
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Re: Is r > s ?
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29 Jan 2012, 18:02




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Re: Is r > s ? (1) r + s < 0 (2) r <  s 
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29 Jan 2012, 18:07
Clear, thanks! See you around!
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Re: Is r > s ? (1) r + s < 0 (2) r <  s 
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21 Feb 2012, 21:38
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 
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Re: Is r > s ? (1) r + s < 0 (2) r <  s 
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22 Feb 2012, 03:11
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 
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15 Nov 2012, 20:44
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 
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04 Dec 2012, 01:40
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! Answer: A
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Re: Is r > s ? (1) r + s < 0 (2) r <  s 
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04 Dec 2012, 19:36
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 
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08 Jul 2017, 10:12
Is r > s ?
(1) r + s < 0 = I re arranged the statement, so r < s so sufficient (2) r <  s  = here there were two cases given the modulus, so insufficient: case 1) r < s 2) r > s
But I combined 1) + 2) where the common answer was r < s. But A is OA. Could anyone explain the flaw in my reasoning?



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Is r > s ? (1) r + s < 0 (2) r <  s 
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08 Jul 2017, 10:24
Madhavi1990 wrote: Is r > s ?
(1) r + s < 0 = I re arranged the statement, so r < s so sufficient (2) r <  s  = here there were two cases given the modulus, so insufficient: case 1) r < s 2) r > s
But I combined 1) + 2) where the common answer was r < s. But A is OA. Could anyone explain the flaw in my reasoning? 1) r + s < 0 r < s Multiple both side by 1, when u do this reverse the inequality r>s So, it is sufficient. Once u have statement 1 as sufficient, you can eliminate option B,C,E. Only option A and D are left now. As statement 2 is not sufficient in itself , we eliminate option D and only possible answer is A.



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Re: Is r > s ? (1) r + s < 0 (2) r <  s 
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08 Jul 2017, 10:28
Madhavi1990 wrote: Is r > s ?
(1) r + s < 0 = I re arranged the statement, so r < s so sufficient (2) r <  s  = here there were two cases given the modulus, so insufficient: case 1) r < s 2) r > s
But I combined 1) + 2) where the common answer was r < s. But A is OA. Could anyone explain the flaw in my reasoning? According to statement 1 > r + s < 0 => r < s Multiply both by 1 <— we need to flip the sign R > S Sufficient. Statement 2 has 2 scenarios, when s <0 and when s>0 hence its insufficient. A. Hope this helps. Posted from my mobile device
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Re: Is r > s ? (1) r + s < 0 (2) r <  s 
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08 Jul 2017, 10:30
Madhavi1990 wrote: Is r > s ?
(1) r + s < 0 = I re arranged the statement, so r < s so sufficient (2) r <  s  = here there were two cases given the modulus, so insufficient: case 1) r < s 2) r > s
But I combined 1) + 2) where the common answer was r < s. But A is OA. Could anyone explain the flaw in my reasoning? Statement 1 which reads r + s < 0 Is nothing but r < s(when you subtract s from both sides) When we multiply 1 on both sides, the greater than becomes lesser than (or) lesser than becomes greater than Therefore, the inequality becomes r>s which alone is sufficient. These are the options in a GMAT DS question (A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. (B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. (C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. (D) EACH statement ALONE is sufficient. (E) Statements (1) and (2) TOGETHER are NOT sufficient. C comes into play only when either 1 or 2 is not enough to prove the statement. Hope that helps!
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Re: Is r > s ? (1) r + s < 0 (2) r <  s 
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10 Mar 2018, 22:31
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! Question : Is r > s ? or r  s > 0 ? or r + s < 0 ? St 1: r + s < 0 Sufficientr <  s  means r < s or r <−s Insufficient(A)
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Re: Is r > s ? (1) r + s < 0 (2) r <  s 
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30 Dec 2018, 16:22
Bunuel wrote: 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. Bunuel, In above solution 2nd part, isn't it be \(r < s\) or \(r > s\) ?




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