GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Jan 2019, 19:10

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### The winners of the GMAT game show

January 22, 2019

January 22, 2019

10:00 PM PST

11:00 PM PST

In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.
• ### Key Strategies to Master GMAT SC

January 26, 2019

January 26, 2019

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 13 May 2010
Posts: 99
If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

19 Jun 2010, 02:45
7
11
00:00

Difficulty:

75% (hard)

Question Stats:

52% (01:55) correct 48% (01:44) wrong based on 505 sessions

### HideShow timer Statistics

If x ≠ -y is $$\frac{x-y}{x+y}>1$$?

(1) x > 0
(2) y < 0
##### Most Helpful Expert Reply
Math Expert
Joined: 02 Sep 2009
Posts: 52390
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

19 Jun 2010, 03:34
11
2
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.

Answer: E.

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.
_________________
##### General Discussion
Intern
Joined: 03 Mar 2010
Posts: 37
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

01 Aug 2010, 14:37
$$\frac{x-y}{x+y}>1$$
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.
_________________

Hardworkingly, you like my post, so kudos me.

Intern
Joined: 18 Aug 2010
Posts: 5
If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

01 Aug 2010, 14:56
1
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?
_________________

"Learning never exhausts the mind."
--Leonardo da Vinci

Retired Moderator
Joined: 02 Sep 2010
Posts: 765
Location: London
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

27 Oct 2010, 13:20
2
1
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.
_________________
Senior Manager
Joined: 25 Jul 2010
Posts: 279
Location: United States
Concentration: Finance
Schools: Cornell (Johnson) - Class of 2014
WE: Engineering (Telecommunications)
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

22 Dec 2010, 06:59
Thanks Bunuel, that explains it perfectly
Director
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 516
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

13 Mar 2011, 07:17
(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.
_________________

Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Retired Moderator
Joined: 20 Dec 2010
Posts: 1810
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

14 Apr 2011, 03:10
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"
_________________
Retired Moderator
Joined: 16 Nov 2010
Posts: 1419
Location: United States (IN)
Concentration: Strategy, Technology
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

14 Apr 2011, 04:11
1
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

Answer - E
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8805
Location: Pune, India
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

14 Apr 2011, 17:18
2
2
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.
Answer (E)
_________________

Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

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: 728
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

14 Apr 2011, 21:31
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: 194
Schools: ISB, IIMA ,SP Jain , XLRI
WE 1: Aditya Birla Group (sales)
WE 2: Saint Gobain Group (sales)
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

17 Apr 2011, 04:55
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: 1013
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

19 May 2011, 22:45
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
Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 587
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE: Information Technology (Computer Software)
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

15 Nov 2012, 18:14
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: 8805
Location: Pune, India
If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

15 Nov 2012, 18:21
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

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >

Manager
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 121
Location: India
Concentration: General Management, Technology
GMAT 1: 640 Q42 V37
GPA: 3.5
WE: Web Development (Computer Software)
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

16 May 2013, 12:48
gmacforjyoab wrote:
if x ≠ -y, is (x-y)/(x+y) > 1?

1) x> 0
2) y<0

The answer is [E]. I marked it [B], but as I began to write the solution, the obviousness of the solution hit me!

The expression, $$(x-y)/(x+y) > 1$$ when simplified gives us y < 0. Clearly the above will drive us directly to [B]. But when we consider this, in the case y < 0 and x > 0, and no relation is given between x and y, what happens if |x| < |y|. In such a case x+ y < 0. But clearly, x-y is greater than 0! Hence the solution doesn't stick! This is the reason why [B] can not be the answer! Similarly if x < 0 and y < 0, we cannot for sure determine the sign of the fraction.

Also, clearly the individual choice x > 0 does not help! so, neither of the choices lead to our answer. To actually solve the above we need a relation determined between x and y so as to determine the sign of (x-y)/(x+y). Wonderful question gmacforjyoab!

Hope my procedure is what you expected as an answer!

Regards,
Arpan
_________________

Feed me some KUDOS! *always hungry*

My Thread : Recommendation Letters

Retired Moderator
Joined: 05 Jul 2006
Posts: 1722
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

16 May 2013, 12:58
if x ≠ -y, is (x-y)/(x+y) > 1?

1) x> 0
2) y<0

is [(x-y)-(x+y)]/(x+y) > 0 , i.e. is -2y / x+y > 0 i.e. is 2y / x+y < 0? true if 1) y <0 , x>0 and /y/ > /x/ 2) Y>0 AND X<0 AND /X/>/Y/

from 1

x is +ve .... insuff

from 2

y -ve insuff

both

this looks like the 1st case described above but we dont know whether /y/ > /x/ ..... insuff

E
Intern
Joined: 07 Sep 2014
Posts: 18
Location: United States (MA)
Concentration: Finance, Economics
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

03 Oct 2014, 14:13
in your explanation i just dont see the progression how the "1-" goes away and we change the numerator to "x+y-x+y" to arrive at 2y
Math Expert
Joined: 02 Sep 2009
Posts: 52390
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

03 Oct 2014, 14:17
bsmith37 wrote:
in your explanation i just dont see the progression how the "1-" goes away and we change the numerator to "x+y-x+y" to arrive at 2y

$$1-\frac{x-y}{x+y}$$;

$$\frac{x+y}{x+y}-\frac{x-y}{x+y}$$;

$$\frac{(x+y)-(x-y)}{x+y}$$;

$$\frac{2y}{x+y}$$.
_________________
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6827
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0  [#permalink]

### Show Tags

19 Oct 2015, 22:50
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) x>0
(2) y<0

If we modify the original condition, as squared numbers are always positive, the inequality sign does not change even if we multiply certain positive integers.
So if we multiply (x+y)^2, (x-y)(x+y)>(x+y)^2, --> (x-y)(x+y)-(x+y)^2>0, (x+y)(x-y-x-y)>0, (x+y)(-2y)>0, and if we divide both sides by -2,
the inequality sign changes, and ultimately we want to know whether (x+y)y<0.
There are 2 variables, so we need 2 equations, which are provided by the 2 conditions. If we look at the conditions together, the answer to what we want to know becomes 'yes' for x=2, y=-1, but 'no' for x=2,y=-3. Hence, the conditions are insufficient and the answer becomes (E).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only \$149 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Re: If x ≠ -y is (x - y)/(x + y) > 1? (1) x > 0 (2) y < 0 &nbs [#permalink] 19 Oct 2015, 22:50

Go to page    1   2    Next  [ 23 posts ]

Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.