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# If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0

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If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]

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19 Jun 2010, 03:45
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If x#-y is (x-y)/(x+y)>1?

(1) x>0
(2) y<0
[Reveal] Spoiler: OA
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Re: gmat prep DS [#permalink]

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19 Jun 2010, 04:34
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gmatcracker2010 wrote:
Please explain the attached problem.

My take on it:

is (X+y)/ (x-y) > 1

ie. x + y > x -y
ie. 0 > y

which is B but OA is E.

If $$x\neq{-y}$$ is $$\frac{x-y}{x+y}>1$$?

Is $$\frac{x-y}{x+y}>1$$? --> Is $$0>1-\frac{x-y}{x+y}$$? --> Is $$0>\frac{x+y-x+y}{x+y}$$? --> Is $$0>\frac{2y}{x+y}$$?

(1) $$x>0$$ --> Not sufficient.

(2) $$y<0$$ --> Not sufficient.

(1)+(2) $$x>0$$ and $$y<0$$ --> numerator (y) is negative, but we cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.

The problem with your solution is that when you are then writing $$x-y>x+y$$, you are actually multiplying both sides of inequality by $$x+y$$: never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if $$x+y>0$$ you should write $$x-y>x+y$$ BUT if $$x+y<0$$, you should write $$x-y<x+y$$ (flip the sign when multiplying by negative expression).

So again: given inequality can be simplified as follows: $$\frac{x-y}{x+y}>1$$ --> $$0>1-\frac{x-y}{x+y}<0$$ --> $$0>\frac{x+y-x+y}{x+y}$$ --> $$0>\frac{2y}{x+y}$$ --> we can drop 2 and finally we'll get: $$0>\frac{y}{x+y}$$.

Now, numerator is negative ($$y<0$$), but we don't know about the denominator, as $$x>0$$ and $$y<0$$ can not help us to determine the sign of $$x+y$$. So the answer is E.

Hope it helps.
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Re: GMATPrep DS question [#permalink]

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01 Aug 2010, 15:03
Hi,

twi ways to approach:
1) Check it out by plugging in numbers:
a) x < y: makes the numerator negative, so doesn't work
b) x = y: creates a 0, so doesn't work
c) x > y: makes the denominator greater than the numerator, so doesn't work
d) y < 0: makes numerator greater than denominator, so WORKS

2) solve the equation:
(x-y)/(x+y)>1
=> x-y > x + y
=> x-x > y + y
=> 0 > 2 y
=> y must be negative

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Re: GMATPrep DS question [#permalink]

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01 Aug 2010, 15:28
The OA is E. I thought it would be B. I think the catch is that x cannot be cancelled from both sides...

If x - y > x + y
Cancel x

y < 0
Should be B. OA says E.
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Re: GMATPrep DS question [#permalink]

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01 Aug 2010, 15:37
$$\frac{x-y}{x+y}>1$$
Let us make cross multiplication with respect to the sign of x+y.
- Case 1:
x-y>x+y>0
or x+y>0 and y<0
- Case 2:
x-y<x+y<0
or x+y<0 and y>0

Both statement 1 and 2 together cannot say about the sign of x+y, thereofore correct answer is E.
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Re: GMATPrep DS question [#permalink]

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01 Aug 2010, 15:56
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But Bogos, what is wrong in just cancelling x from both sides of
x-y > x+y?

You mentioned about w.r.t the sign of x+y, why is that important?
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Re: GMATPrep DS question [#permalink]

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03 Aug 2010, 10:22
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bogos wrote:
You can refer to this review (3rd point):
inequalities-review-69016.html

Bogos
I read about inequalities and it says that one can subtract the same amout from both sides
so if x-y/x+y > 1 => x-y > x+y => y<0

where is the flaw?
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Re: GMATPrep - DS - Inequalities [#permalink]

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27 Oct 2010, 14:20
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thirst4edu wrote:
Rephrasing the question
$$\frac{(x-y)}{(x+y)} > 1$$
$$(x-y) > (x+y)$$ , Since x <> -y, we can multiply (x+y) both sides
Adding -x both sides
-y > y
Adding y both sides
0 > y , which is y < 0, given by statement 2, so why Answer is E(both together not sufficient)? Am I missing something?

What you did is only true is (x+y)>0.
If you multiply both sides of an inequality with a number, the sign remains same if the number or expression is positive, whereas it flips if it is negative.

A counter example to show answer is indeed (e) is taking x=5 and y=-20 ... you'll get 25/-15 = -5/3 which is less than 1 ... on the other hand x=5 & y=-1 would get you 6/4 which is greater than 1.
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Re: gmat prep DS [#permalink]

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22 Dec 2010, 07:59
Thanks Bunuel, that explains it perfectly
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Re: gmat prep DS [#permalink]

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13 Mar 2011, 08:17
(x-y/x+y)>1
What is the problem if i try numbers
(1) x>0 no information about y clearly insuf.
(2) y <0 no information about x. insuf.

For C
x=2, y=-1
2+1/2-1>1
again
x=2, y=-3
2+3/2-3<1

so ans E.

Help if i wrong.
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if x is different from -y, in (x-y)/(x+y) greater than 1 ? [#permalink]

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14 Apr 2011, 02:55
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if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0
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Re: Did Sal had this question wrong ? [#permalink]

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14 Apr 2011, 04:10
whichscore wrote:
if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0

$$\frac{x-y}{x+y}>1$$
$$\frac{x-y}{x+y}-1>0$$
$$\frac{x-y-x-y}{x+y}>0$$
$$\frac{-2y}{x+y}>0$$
$$-\frac{y}{x+y}>0$$
$$\frac{y}{x+y}<0$$

Case I:
If y<0 or -y>0 -----A
Then, x+y>0
x>-y-----B
Combining A and B:
x>-y>0

Case II:
If y>0 or -y<0 -----C
Then, x+y<0
x<-y-----D
Combining C and D:
x<-y<0

1. x > 0
CaseI:
x>0
Is -y between 0 and x

If
x=5
y=-3
-y=-(-3)=3 will be between 0 and x(5) and the expression will be true.

If
x=5
y=-7
-y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false.
Not Sufficient.

2. y < 0
CaseI:
x>0
Is -y between 0 and x

If
x=5
y=-3; Here y<0
-y=-(-3)=3 will be between 0 and x(5) and the expression will be true.

If
x=5
y=-7; Here y<0
-y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false.
Not Sufficient.

Combing both; we use the same sample set:
x=5
y=-3
AND
x=5
y=-7
To prove its insufficiency.

Ans: "E"
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Re: Did Sal had this question wrong ? [#permalink]

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14 Apr 2011, 05:11
I plugged numbers to get the answer :

A per (1)

x = 5, y = 5

then 0/10 < 1

x = 5, y = -1

6/4 > 1

So insufficient

As per (2)

x = 6, y = -2

8/4 > 1

x = 2 , y = -3

5/-1 < 1

So insufficient

As evident above, (1) and (2) together are also insufficient

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Re: Did Sal had this question wrong ? [#permalink]

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14 Apr 2011, 18:18
whichscore wrote:
if x is different from -y, in (x-y)/(x+y) greater than 1 ?
1. x > 0
2. y< 0

Even though fluke has already provided the detailed solution, there are a couple of points I would like to reinforce here.

People often get confused when they read 'if x is different from -y'. Why do they give this information? They do that because you have (x+y) in the denominator and hence it cannot be 0. That is, x + y = 0 or x = -y is not valid. Hence, they are just clarifying that the fraction is indeed defined.

Also, we are used to having 0 on the right of an inequality. What do we do when we have a 1? We take the 1 to the left hand side and get a 0 on the right.
Is $$\frac{(x-y)}{(x+y)} > 1$$? (Don't forget it is a question, not given information)
Is $$\frac{(x-y)}{(x+y)} - 1 > 0$$?
which simplifies to: Is y/(x+y) < 0 ?

We know how to deal with this inequality! Note here that we have no information about x and y as yet (except that x is not equal to -y which is more of a technical issue rather than actual information)

1. x > 0
No information about y so not sufficient.
2. y< 0
No information about x so not sufficient.

Both together, we know that y is negative. We need the sign of (x+y) now. But (x+y) may be positive or negative depending on whether x or y has greater absolute value. Hence not sufficient.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing. Affiliations: University of Chicago Booth School of Business Joined: 03 Feb 2011 Posts: 920 Followers: 13 Kudos [?]: 315 [0], given: 123 Re: Did Sal had this question wrong ? [#permalink] ### Show Tags 14 Apr 2011, 22:31 1) + 2) When x > 0 and y < 0, x- y will always be +ve but nothing can be concluded about (x + y). y can be very small -Infinity or y can be close to zero. Hence sign of x - y is unknown. Hence the division i.e x-y / (x + y) may or may not be greater than 1. I don't think any calculation is required in this problem. Manager Status: ==GMAT Ninja== Joined: 08 Jan 2011 Posts: 247 Schools: ISB, IIMA ,SP Jain , XLRI WE 1: Aditya Birla Group (sales) WE 2: Saint Gobain Group (sales) Followers: 5 Kudos [?]: 72 [0], given: 46 Re: Did Sal had this question wrong ? [#permalink] ### Show Tags 17 Apr 2011, 05:55 fluke wrote: whichscore wrote: if x is different from -y, in (x-y)/(x+y) greater than 1 ? 1. x > 0 2. y< 0 $$\frac{x-y}{x+y}>1$$ $$\frac{x-y}{x+y}-1>0$$ $$\frac{x-y-x-y}{x+y}>0$$ $$\frac{-2y}{x+y}>0$$ $$-\frac{y}{x+y}>0$$ $$\frac{y}{x+y}<0$$ Case I: If y<0 or -y>0 -----A Then, x+y>0 x>-y-----B Combining A and B: x>-y>0 Case II: If y>0 or -y<0 -----C Then, x+y<0 x<-y-----D Combining C and D: x<-y<0 1. x > 0 CaseI: x>0 Is -y between 0 and x If x=5 y=-3 -y=-(-3)=3 will be between 0 and x(5) and the expression will be true. If x=5 y=-7 -y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false. Not Sufficient. 2. y < 0 CaseI: x>0 Is -y between 0 and x If x=5 y=-3; Here y<0 -y=-(-3)=3 will be between 0 and x(5) and the expression will be true. If x=5 y=-7; Here y<0 -y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false. Not Sufficient. Combing both; we use the same sample set: x=5 y=-3 AND x=5 y=-7 To prove its insufficiency. Ans: "E" Dear Fluke i had below approach please correct me if i am wrong the term simplifies to $$-\frac{y}{x+y}>0$$ Case I Numerator and denominator both are positive -y > 0 y < 0 and x + y > 0 x > -y but since -y>0 so x also also be >0 Case 2 Numerator and denominator both are negative -y < 0 y > 0 and x + y < 0 x < -y but since -y<0 so x will also be < 0 now Clearly none of the statement alone is sufficient and taking them together gives us only Case I but not takes care about the case 2 so answer is E _________________ WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post. VP Status: There is always something new !! Affiliations: PMI,QAI Global,eXampleCG Joined: 08 May 2009 Posts: 1353 Followers: 17 Kudos [?]: 217 [0], given: 10 Re: Did Sal had this question wrong ? [#permalink] ### Show Tags 19 May 2011, 23:45 by resolving - x/ (x+y) > 1 a + b, x>0 y<0 x=2,y= -1 LHS > RHS x=1,y =-2 LHS < RHS E _________________ Visit -- http://www.sustainable-sphere.com/ Promote Green Business,Sustainable Living and Green Earth !! Manager Joined: 27 Apr 2012 Posts: 62 Location: United States GMAT Date: 06-11-2013 GPA: 3.5 WE: Marketing (Consumer Products) Followers: 1 Kudos [?]: 45 [0], given: 21 If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink] ### Show Tags 15 Nov 2012, 18:55 Hi, Request help with the following question.Thanks.. If x is not equal to -y,is (x -y)/(x+y) >1? 1.x > 0 2.y < 0 P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question. Director Status: Done with formalities.. and back.. Joined: 15 Sep 2012 Posts: 647 Location: India Concentration: Strategy, General Management Schools: Olin - Wash U - Class of 2015 WE: Information Technology (Computer Software) Followers: 43 Kudos [?]: 508 [0], given: 23 Re: If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink] ### Show Tags 15 Nov 2012, 19:14 shivanigs wrote: Hi, Request help with the following question.Thanks.. If x is not equal to -y,is (x -y)/(x+y) >1? 1.x > 0 2.y < 0 P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question. Here you go: First, lets rephrase the question- $$(x -y)/(x+y) >1$$ $$(x -y)/(x+y) -1 >0$$ $$-2y/(x+y) >0$$ or is $$y/(x+y) < 0$$ ? Statement 1: x >0 This doesnt help us understanding the sign of numerator or denominator. Not sufficient. Statement 2: y<0 now we know, that numerator is postive, but we still dont know what is the sign for x+y. Not sufficient. Combining 1 and 2. x>0, y<0 We want to know if y/(x+y) < 0 Numerator is negative here, and denominator can be positive or negative depending upon absolute values of x and y. Not sufficien Hence, Ans E it is! _________________ Lets Kudos!!! Black Friday Debrief Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6830 Location: Pune, India Followers: 1926 Kudos [?]: 11965 [0], given: 221 Re: If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink] ### Show Tags 15 Nov 2012, 19:21 shivanigs wrote: Hi, Request help with the following question.Thanks.. If x is not equal to -y,is (x -y)/(x+y) >1? 1.x > 0 2.y < 0 P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question. Responding to a pm: Question: Is (x -y)/(x+y) >1? What do you do when you are given >1 ? It is hard to find implications of '>1'. It is much easier to handle '>0' Is $$\frac{(x -y)}{(x+y)} -1 > 0$$ ? Is $$\frac{-2y}{(x+y)} > 0$$ ? When will this be positive? In 2 cases: 1. When y is negative and (x+y) is positive i.e. x is positive and x has greater absolute value than y's absolute value. 2. When y is positive and (x+y) is negative i.e. x is negative and x has greater absolute value than y's absolute value. Both statements together tell us that y is negative and x is positive but we don't know whether 'x's absolute value is greater than y's absolute value'. Hence (E) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If x is not equal to -y,is (x -y)/(x+y) >1?   [#permalink] 15 Nov 2012, 19:21

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