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Intern  Joined: 23 Dec 2011
Posts: 9
GMAT 1: 620 Q47 V28 Is r > s ? (1) -r + s < 0 (2) r < | s |  [#permalink]

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5
8 00:00

Difficulty:   15% (low)

Question Stats: 76% (01:08) correct 24% (01:22) wrong based on 549 sessions

### HideShow timer Statistics 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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Originally posted by DesecratoR on 29 Jan 2012, 18:54.
Last edited by MikeScarn on 05 Jul 2019, 09:08, edited 1 time in total.
Hid posters comment than contains potential spoiler.
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s |  [#permalink]

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5
6
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!

Welcome to GMAT Club. Below is a solution for this problem.

Is $$r>s$$?

(1) $$-r+s<0$$ --> rearrange (or 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.

Basically 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.
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Intern  Joined: 23 Dec 2011
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GMAT 1: 620 Q47 V28 Re: Is r > s ? (1) -r + s < 0 (2) r < | s |  [#permalink]

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Clear, thanks! See you around!
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Re: Is r > s ? (1) -r + s < 0 (2) r < | s |  [#permalink]

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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 |  [#permalink]

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

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

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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 |  [#permalink]

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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!

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

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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 |  [#permalink]

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

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3
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 |  [#permalink]

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1
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.

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

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2
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 |  [#permalink]

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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 Sufficient

r < | s |

means r < s or r <−s Insufficient

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

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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 $$r > -s$$ ? Re: Is r > s ? (1) -r + s < 0 (2) r < | s |   [#permalink] 30 Dec 2018, 17:22
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