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

 It is currently 22 Nov 2018, 02:49

# GMAT Club Tests are Free and Open until midnight Nov 22, Pacific Time

### 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

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 November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### Key Strategies to Master GMAT SC

November 24, 2018

November 24, 2018

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.
• ### GMATbuster's Weekly GMAT Quant Quiz

November 24, 2018

November 24, 2018

09:00 AM PST

11:00 AM PST

We will start promptly at 09 AM Pacific Time. Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage.

# Is 1/(x-y) < y - x

Author Message
TAGS:

### Hide Tags

Director
Joined: 07 Jun 2004
Posts: 600
Location: PA
Is 1/(x-y) < y - x  [#permalink]

### Show Tags

Updated on: 10 Jul 2013, 01:31
14
21
00:00

Difficulty:

95% (hard)

Question Stats:

25% (02:44) correct 75% (02:38) wrong based on 877 sessions

### HideShow timer Statistics

Is 1/(x - y) < y - x

(1) 1/x < 1/y
(2) 2x = 3y

_________________

If the Q jogged your mind do Kudos me : )

Originally posted by rxs0005 on 07 Jan 2011, 08:09.
Last edited by Bunuel on 10 Jul 2013, 01:31, edited 2 times in total.
Edited the OA.
Math Expert
Joined: 02 Sep 2009
Posts: 50770
Is 1/(x-y) < y - x  [#permalink]

### Show Tags

07 Jan 2011, 09:15
18
15
rxs0005 wrote:
Is 1 / (x-y) < y - x

1 / x < 1 / y

2x = 3y

First of all if it were realistic GMAT question it would most likely state that $$xy\neq{0}$$ and $$x\neq{y}$$ (as x, y and x-y are in denominators).

Is $$\frac{1}{x-y}<y-x$$? --> is $$\frac{1}{x-y}+x-y<0$$ --> is $$\frac{1+(x-y)^2}{x-y}<0$$? as the nominator ($$1+(x-y)^2$$) is always positive then the question basically becomes whether denominator ($$x-y$$) is negative --> is $$x-y<0$$? or is $$x<y$$?

(1) $$\frac{1}{x}<\frac{1}{y}$$ --> if both unknowns are positive or both unknowns are negative then $$y<x$$ (if both are positive cross multiply to get $$y<x$$ and if both are negative cross multiply and flip the sigh twice to get $$y<x$$ again) and the answer will be NO but if $$x<0<y$$ given inequality also holds true and in this case the answer will be YES (if $$x$$ is any negative number and $$y$$ is any positive number then $$\frac{1}{x}=negative<positive=\frac{1}{y}$$). Not sufficient.

(2) $$2x=3y$$ --> $$x$$ and $$y$$ have the same sign, next: $$\frac{x}{y}=\frac{3}{2}$$: if both $$x$$ and $$y$$ are positive (for example 3 and 2 respectively) then $$0<y<x$$ and the answer will be NO but if both $$x$$ and $$y$$ are negative (for example -3 and -2 respectively) then $$x<y<0$$ and the answer will be YES. Not sufficient.

(1)+(2) As from (2) $$x$$ and $$y$$ have the same sign then from (1) $$y<x$$ and the answer to the question is NO. Sufficient.

_________________
Director
Joined: 22 Mar 2011
Posts: 601
WE: Science (Education)
Re: Is 1 / (x-y) < y - x  [#permalink]

### Show Tags

03 Sep 2012, 04:16
5
1
rxs0005 wrote:
Is 1 / (x-y) < y - x

(1) 1 / x < 1 / y
(2) 2x = 3y

If we denote by $$A=x-y$$, the question is "Is $$\frac{1}{A}<-A?$$"
If $$A>0$$, the above inequality cannot hold (a positive number is not smaller than a negative number).
So, the question can be reworded as is $$A<0$$, or is $$x-y<0$$ which is the same as is $$x<y?$$

(1) The given inequality is equivalent to $$\frac{x-y}{xy}>0.$$
If $$xy>0$$, or in other words if $$x$$ and $$y$$ have the same sign, then necessarily $$x$$ must be greater than $$y.$$
If $$xy<0$$, or in other words if $$x$$ and $$y$$ have opposite signs, then necessarily $$x$$ must be smaller than $$y.$$
Not sufficient.

(2) $$x=\frac{3}{2}y.$$ If $$y<0$$, then $$x<y.$$ But if $$y>0,$$ then $$x>y.$$
Not sufficient.

(1) and (2) together:
Since from (2) we have that $$x$$ and $$y$$ have the same sign, using (1) we deduce that necessarily $$x-y>0.$$
So, the answer to the question "Is $$x<y$$" is a definite NO.
Sufficient.

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

##### General Discussion
Manager
Joined: 07 Jun 2010
Posts: 81

### Show Tags

07 Jan 2011, 08:37
Proposition 1:

1/X < 1/Y

This is equal to X > Y | x, y both > 0 or both < 0

Case 1: Both > 0. X > Y, therefore X-Y >0 , Y-X <0, so 1/(X-Y) > 0 and Y-X<0 so FALSE

Case 2: X < 0, Y>0. Cross multiple and switch the signs, you get Y>X. There for Y-X>0, X-Y<0.

1/(X-Y)<0, Y-X>0, so case 2, 1/(X-Y) < (Y-X) is TRUE

Therefore statement 1 is insufficient by contradiction.

Proposition 2:

2X=3Y

Case 1: X=3, Y=2 - therefore X-Y=1, Y-X = -1

1/(X-Y) < Y-X ----> 1/1 < -1 = FALSE

Case 2: X=-3, Y=-2 - therefore X-Y=-1, Y-X = 1

1/(X-Y) < Y-X ----> 1/-1 < 1 = TRUE

Both Statments together - Intutitively, neither statement removes the ambiguity of the sign of X or Y. I can't think of a proof.

Manager
Joined: 20 Jul 2011
Posts: 111
GMAT Date: 10-21-2011

### Show Tags

07 Sep 2011, 03:24
**
Quote:
Is 1 / (x-y) < y - x?

1. 1 / x < 1 / y
2. 2x = 3y

equation: $$\frac{1}{(x-y)} < y-x$$

Statement 1
Given $$\frac{1}{x}<\frac{1}{y}$$, we have 2 possibilities:
One: if both x and y share the same signs (i.e. both are positive or both are negative) --> y<x
left side of equation will be a positive fraction; right side will be negative integer --> equation = true
Two: if x is negative and y is positive (e.g. x=-2 and y=3), left side of equation will be negative fraction; right side will be positive integer --> equation = false
Insufficient.

Statement 2
Given 2x=3y --> x/y = 3/2 -->i.e. x and y share the same sign (i.e. both positive or both negative). 2 possibilities:
One: if both x and y are positive (i.e. x=3, y=2), results in 1/1 < -1 --> equation = false
Two: if both x and y are negative (i.e. x=-3, y=-2), results in 1/-1 < 1 --> equation = true
Insufficient.

Statement 1 + 2
Applying statement 2 and 1 together, we know that x and y:
i. share the same sign, and
ii. that y<x
--> which means, equation = false i.e. No
Sufficient.

_________________

"The best day of your life is the one on which you decide your life is your own. No apologies or excuses. No one to lean on, rely on, or blame. The gift is yours - it is an amazing journey - and you alone are responsible for the quality of it. This is the day your life really begins." - Bob Moawab

Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1193
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Is 1/(x-y) < y - x?  [#permalink]

### Show Tags

27 Aug 2012, 06:45
Is $$\frac{1}{(x-y)} < y - x$$ ?

(1) $$\frac{1}{x} < \frac{1}{y}$$
(2) $$2x = 3y$$

I don't agree with the OA. It must be C.
If $$2x = 3y$$, then $$x$$ and $$y$$are both positive or both negative.
So, if we know that:
$$\frac{1}{x} < \frac{1}{y}$$, then $$x>y$$ --> $$x -y > 0$$

With that information we can conclude that the answer is No. C is correct.

_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Math Expert
Joined: 02 Sep 2009
Posts: 50770
Re: Is 1/(x-y) < y - x?  [#permalink]

### Show Tags

27 Aug 2012, 07:14
1
metallicafan wrote:
Is $$\frac{1}{(x-y)} < y - x$$ ?

(1) $$\frac{1}{x} < \frac{1}{y}$$
(2) $$2x = 3y$$

I don't agree with the OA. It must be C.
If $$2x = 3y$$, then $$x$$ and $$y$$are both positive or both negative.
So, if we know that:
$$\frac{1}{x} < \frac{1}{y}$$, then $$x>y$$ --> $$x -y > 0$$

With that information we can conclude that the answer is No. C is correct.

Merging similar topics. You are right, answer should be C, not E.
_________________
Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 88
Concentration: Strategy, General Management
GMAT 1: 460 Q35 V20
GPA: 3.6
WE: Consulting (Computer Software)
Re: Is 1 / (x-y) < y - x  [#permalink]

### Show Tags

25 Nov 2012, 18:54
Hi Buneul,

Can you please explain this part?
if both are negative cross multiply and flip the sigh twice to get y<x again)
Math Expert
Joined: 02 Sep 2009
Posts: 50770
Re: Is 1 / (x-y) < y - x  [#permalink]

### Show Tags

26 Nov 2012, 01:28
1
shankar245 wrote:
Hi Buneul,

Can you please explain this part?
if both are negative cross multiply and flip the sigh twice to get y<x again)

Given: 1/x<1/y. Now, if both x and y are negative, then when we multiply both parts by negative x we should flip the sign and write 1>x/y. Now, multiply both sides by negative y and flip the sign again to get y<x.

Hope it's clear.
_________________
Manager
Joined: 28 Dec 2012
Posts: 102
Location: India
Concentration: Strategy, Finance
WE: Engineering (Energy and Utilities)
Re: Is 1 / (x-y) < y - x  [#permalink]

### Show Tags

14 Jan 2013, 05:16
stem reduces to Is x<y ? [-1/(y-x) = (y-x) => (y-x) >0 ]

A. Depends on sign of x,y. (both positive or both negative for eg. Vs positive-negative give both YES and NO)
B. x= 3k, y= 2k. K>=0, Answer NO. K<0 Answer is YES.

C.

Using 2. x =3k when y =2k.
using the above in 1. 1/x < 1/ y or 1/3k < 1/2k holds only for k>0. For k>0, Using 2, we definitely get the single answer as NO.

KUDOS, if YOU LIKE
_________________

Impossibility is a relative concept!!

Director
Joined: 25 Apr 2012
Posts: 688
Location: India
GPA: 3.21

### Show Tags

14 Jan 2013, 22:43
Bunuel wrote:
rxs0005 wrote:
Is 1 / (x-y) < y - x

1 / x < 1 / y

2x = 3y

First of all if it were realistic GMAT question it would most likely state that $$xy\neq{0}$$ and $$x\neq{y}$$ (as x, y and x-y are in denominators).

Is $$\frac{1}{x-y}<y-x$$? --> is $$\frac{1}{x-y}+x-y<0$$ --> is $$\frac{1+(x-y)^2}{x-y}<0$$? as the nominator ($$1+(x-y)^2$$) is always positive then the question basically becomes whether denominator ($$x-y$$) is negative --> is $$x-y<0$$? or is $$x<y$$?

(1) $$\frac{1}{x}<\frac{1}{y}$$ --> if both unknowns are positive or both unknowns are negative then $$y<x$$ (if both are positive cross multiply to get $$y<x$$ and if both are negative cross multiply and flip the sigh twice to get $$y<x$$ again) and the answer will be NO but if $$x<0<y$$ given inequality also holds true and in this case the answer will be YES (if $$x$$ is any negative number and $$y$$ is any positive number then $$\frac{1}{x}=negative<positive=\frac{1}{y}$$). Not sufficient.

(2) $$2x=3y$$ --> $$x$$ and $$y$$ have the same sign, next: $$\frac{x}{y}=\frac{3}{2}$$: if both $$x$$ and $$y$$ are positive (for example 3 and 2 respectively) then $$0<y<x$$ and the answer will be NO but if both $$x$$ and $$y$$ are negative (for example -3 and -2 respectively) then $$x<y<0$$ and the answer will be NO. Not sufficient.

(1)+(2) As from (2) $$x$$ and $$y$$ have the same sign then from (1) $$y<x$$ and the answer to the question is NO. Sufficient.

Hello Bunuel,

Agree with your approach but I wanted to plug in nos and check so here it goes.

we get the expression ( 1+ (x-y)^2 )/(x-y) <0----> Q becomes Is x<y ?

From St 1, we get 1/x<1/y ----> 1/x-1/y <0 and therefore expression becomes

(y-x)/xy < 0 which means is y<x

Now lets take values

y=2 , x=3, the expression is true i.e <0
y=-2 and x=-3, the expression is false ie >0

So not sufficient

from St 2, we get x= 3/2y

Now y=4, x=6, Expression is False ie >0
y=-4, x=-6, expression is true i.e <0
So alone not sufficient

Now Combining we get y<x and x=3/2y ----> y< 3/2y which means Y > 0 and x>y. So we have x>y>0. For this condition, the expression is always false ie. ( 1+ (x-y)^2 )/(x-y) >0

Can't thank you enough in solving inequalities the way you just did.

Thanks
Mridul
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Senior Manager
Joined: 07 Apr 2012
Posts: 370
Re: Is 1/(x-y) < y - x  [#permalink]

### Show Tags

15 Sep 2014, 12:54
Bunuel wrote:
rxs0005 wrote:
Is 1 / (x-y) < y - x

1 / x < 1 / y

2x = 3y

First of all if it were realistic GMAT question it would most likely state that $$xy\neq{0}$$ and $$x\neq{y}$$ (as x, y and x-y are in denominators).

Is $$\frac{1}{x-y}<y-x$$? --> is $$\frac{1}{x-y}+x-y<0$$ --> is $$\frac{1+(x-y)^2}{x-y}<0$$? as the nominator ($$1+(x-y)^2$$) is always positive then the question basically becomes whether denominator ($$x-y$$) is negative --> is $$x-y<0$$? or is $$x<y$$?

(1) $$\frac{1}{x}<\frac{1}{y}$$ --> if both unknowns are positive or both unknowns are negative then $$y<x$$ (if both are positive cross multiply to get $$y<x$$ and if both are negative cross multiply and flip the sigh twice to get $$y<x$$ again) and the answer will be NO but if $$x<0<y$$ given inequality also holds true and in this case the answer will be YES (if $$x$$ is any negative number and $$y$$ is any positive number then $$\frac{1}{x}=negative<positive=\frac{1}{y}$$). Not sufficient.

(2) $$2x=3y$$ --> $$x$$ and $$y$$ have the same sign, next: $$\frac{x}{y}=\frac{3}{2}$$: if both $$x$$ and $$y$$ are positive (for example 3 and 2 respectively) then $$0<y<x$$ and the answer will be NO but if both $$x$$ and $$y$$ are negative (for example -3 and -2 respectively) then $$x<y<0$$ and the answer will be NO. Not sufficient.

(1)+(2) As from (2) $$x$$ and $$y$$ have the same sign then from (1) $$y<x$$ and the answer to the question is NO. Sufficient.

Hi Bunuel.
When taking both statements together, I substituted x=3/2 and y=2/3, or x=-3/2 and y=-2/3
When taking into account statement 1, I noticed that we can't have the negative value for the numbers, as then statement 1 won't hold true.
Is this true? Am I missing something?
Math Expert
Joined: 02 Sep 2009
Posts: 50770
Re: Is 1/(x-y) < y - x  [#permalink]

### Show Tags

15 Sep 2014, 20:00
ronr34 wrote:
Bunuel wrote:
rxs0005 wrote:
Is 1 / (x-y) < y - x

1 / x < 1 / y

2x = 3y

First of all if it were realistic GMAT question it would most likely state that $$xy\neq{0}$$ and $$x\neq{y}$$ (as x, y and x-y are in denominators).

Is $$\frac{1}{x-y}<y-x$$? --> is $$\frac{1}{x-y}+x-y<0$$ --> is $$\frac{1+(x-y)^2}{x-y}<0$$? as the nominator ($$1+(x-y)^2$$) is always positive then the question basically becomes whether denominator ($$x-y$$) is negative --> is $$x-y<0$$? or is $$x<y$$?

(1) $$\frac{1}{x}<\frac{1}{y}$$ --> if both unknowns are positive or both unknowns are negative then $$y<x$$ (if both are positive cross multiply to get $$y<x$$ and if both are negative cross multiply and flip the sigh twice to get $$y<x$$ again) and the answer will be NO but if $$x<0<y$$ given inequality also holds true and in this case the answer will be YES (if $$x$$ is any negative number and $$y$$ is any positive number then $$\frac{1}{x}=negative<positive=\frac{1}{y}$$). Not sufficient.

(2) $$2x=3y$$ --> $$x$$ and $$y$$ have the same sign, next: $$\frac{x}{y}=\frac{3}{2}$$: if both $$x$$ and $$y$$ are positive (for example 3 and 2 respectively) then $$0<y<x$$ and the answer will be NO but if both $$x$$ and $$y$$ are negative (for example -3 and -2 respectively) then $$x<y<0$$ and the answer will be NO. Not sufficient.

(1)+(2) As from (2) $$x$$ and $$y$$ have the same sign then from (1) $$y<x$$ and the answer to the question is NO. Sufficient.

Hi Bunuel.
When taking both statements together, I substituted x=3/2 and y=2/3, or x=-3/2 and y=-2/3
When taking into account statement 1, I noticed that we can't have the negative value for the numbers, as then statement 1 won't hold true.
Is this true? Am I missing something?

Yes, when we combine the statements we get that x > y and 2x = 3y, so x > y > 0.
_________________
Manager
Status: tough ... ? Naaahhh !!!!
Joined: 07 Sep 2015
Posts: 63
Location: India
Concentration: Marketing, Strategy
WE: Marketing (Computer Hardware)
Re: Is 1/(x-y) < y - x  [#permalink]

### Show Tags

07 Oct 2015, 04:33
Hi All,
I attended this question quite late from the posting date but i am getting the different answer D.

1) 1/x<1/y --- which means x>y..so for the positive values the given expression holds NO, and for the negaive values X=-1 and Y=-2 (X>Y) the expression says NO.

2) 2x=3y --- which means x>y..so the same solution like above

Can you help.
Math Expert
Joined: 02 Sep 2009
Posts: 50770
Re: Is 1/(x-y) < y - x  [#permalink]

### Show Tags

07 Oct 2015, 04:44
asethi wrote:
Hi All,
I attended this question quite late from the posting date but i am getting the different answer D.

1) 1/x<1/y --- which means x>y..so for the positive values the given expression holds NO, and for the negaive values X=-1 and Y=-2 (X>Y) the expression says NO.

2) 2x=3y --- which means x>y..so the same solution like above

Can you help.

_________________
CEO
Joined: 20 Mar 2014
Posts: 2635
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: Is 1/(x-y) < y - x  [#permalink]

### Show Tags

07 Oct 2015, 04:46
asethi wrote:
Hi All,
I attended this question quite late from the posting date but i am getting the different answer D.

1) 1/x<1/y --- which means x>y..so for the positive values the given expression holds NO, and for the negaive values X=-1 and Y=-2 (X>Y) the expression says NO.

2) 2x=3y --- which means x>y..so the same solution like above

Can you help.

Be very careful with multiplying or diving inequalities with variables for which you do not know the signs of.

In this case as well, you do not know whether x and y are both positive or both negative or one negative one positive etc.

Without this information you can not say x>y when you are given 1/x<1/y

Case in point, consider the cases (3,2) and (-2,-3) you get different answers and because of this the first statement is not sufficient.

You are making the same mistake when you are analyzing statement 2 alone.

Make sure to be extra careful when dealing with unknown variables in inequalities.

Hope this helps.
Non-Human User
Joined: 09 Sep 2013
Posts: 8871
Re: Is 1/(x-y) < y - x  [#permalink]

### Show Tags

29 Jul 2018, 02:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Is 1/(x-y) < y - x &nbs [#permalink] 29 Jul 2018, 02:55
Display posts from previous: Sort by