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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]

Show Tags

04 Feb 2011, 17:29

2

This post received KUDOS

13

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

33% (02:35) correct
67% (02:20) wrong based on 501 sessions

HideShow timer Statistics

If x and y are nonzero integers, is \((x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}\)?

(1) x = 2y

(2) x + y > 0

I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.

If x and y are nonzero integers, is (x-1 + y-1)-1 > [(x-1)(y-1)]-1 ?

(1) x = 2y

(2) x + y > 0

I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.

Apologies again for giving out too much.

First of all the question should be:

If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > (x^(-1)*y^(-1))^(-1) ?

Is \((x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}\)? --> is \((\frac{1}{x}+\frac{1}{y})^{-1}>(\frac{1}{xy})^{-1}\)? --> is \((\frac{x+y}{xy})^{-1}>xy\) --> is \(\frac{xy}{x+y}>xy\)?

Now, from this point you cannot divide both parts of the inequality by \(xy\) and write \(\frac{1}{x+y}>1\) (as you did), because you don't know whether \(xy\) is positive or negative: if \(xy>0\) then you should write \(\frac{1}{x+y}>1\) BUT if \(xy<0\) then you should flip the sign and write \(\frac{1}{x+y}<1\). But even if you knew that \(xy>0\) then the next step of writing \(x+y<1\) from \(\frac{1}{x+y}>1\) would still be incorrect for the same exact reason: you don't k now whether \(x+y\) is positive or negative, hence you can not muliply both sides of the inequality by \(x+y\).

Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Thus the question is boiled down to: is \(\frac{xy}{x+y}>xy\)? Actually we can manipulate further but there is no need.

(1) x = 2y --> question becomes: is \(\frac{2y^2}{3y}>2y^2\)? Now, as we know that \(y\) is nonzero then \(2y^2>0\) and we can divide both parts by it --> is \(\frac{1}{3y}>1\)? As \(y\) is an integer (no matter positive or negative) then the answer to this question is always NO (if it's a positive integer then \(\frac{1}{3y}<1\) and if it's a negative integer then again: \(\frac{1}{3y}<0<1\)). Sufficient.

(2) x + y > 0 --> if \(x=y=1\) then the answer will be NO but if \(x=3\) and \(y=-1\) then the answer will be YES. Not sufficient.

Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]

Show Tags

28 Jul 2013, 21:25

1

This post received KUDOS

abilash10 wrote:

If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ? (1) x = 2y (2) x + y > 0 I'm not quite satisfied with the official answer for this question from MGMAT

The original question can be reduced to Is \(\frac{xy}{(x+y)}\) > \(xy\) ?

If Y is positive, the answer will be + -/+ --->Negative (-) If Y is negative, the answer will be + +/- --->Negative (-) Sufficient

Statement 2 x+y>0 If x = + & y = + , then \(\frac{xy}{(x+y)}\) > \(xy\) will be false If x = - & y = + , then \(\frac{xy}{(x+y)}\) > \(xy\) will be True Thus Insufficient
_________________

If you like my Question/Explanation or the contribution, Kindly appreciate by pressing KUDOS. Kudos always maximizes GMATCLUB worth-Game Theory

If you have any question regarding my post, kindly pm me or else I won't be able to reply

Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]

Show Tags

21 Oct 2014, 07:12

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.
_________________

If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]

Show Tags

16 Mar 2015, 03:35

I solved it this way:

from the original inequality --> is xy/y+x > xy --> is xy((1/x+y) -1) > 0? to satisfy the inequality either xy > 0 and (1/x+y)-1 > 0 or xy < 0 and (1/x+y)-1 < 0.

Notice that 1/x+y is a proper fraction. So unless x+y = 1 this expression (1/x+y) is going to be negative. If that expression is negative we want xy to be negative as well.

statement 1. x=2y --> x/y=2/1 x and y have the same sign. If x and y are negative ((1/x+y) -1) is surely negative and xy is positive. Thus the overall expression is not > 0. If x and y are positive ((1/x+y) -1) is also negative and xy is positive. Thus the overall expression is not > 0.

sufficient

statement 2. x+y >0 we don't know the exact values of x and y. Assume that x is negative and y is positive and x<y then x+y > 0 still ((1/x+y) -1) is negative and xy is negative too, making the overall expression > 0. Assume that both x and y are positive and you end up with the same scenario as statement 1. Assume that x+y=1 and the overall expression becomes 0.

not sufficient.

answer A.
_________________

learn the rules of the game, then play better than anyone else.

Re: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]

Show Tags

17 Mar 2016, 21:47

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: If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > [#permalink]

Show Tags

28 Jan 2017, 07:01

Bunuel wrote:

ajit257 wrote:

If x and y are nonzero integers, is (x-1 + y-1)-1 > [(x-1)(y-1)]-1 ?

(1) x = 2y

(2) x + y > 0

I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.

Apologies again for giving out too much.

First of all the question should be:

If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > (x^(-1)*y^(-1))^(-1) ?

Is \((x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}\)? --> is \((\frac{1}{x}+\frac{1}{y})^{-1}>(\frac{1}{xy})^{-1}\)? --> is \((\frac{x+y}{xy})^{-1}>xy\) --> is \(\frac{xy}{x+y}>xy\)?

Now, from this point you cannot divide both parts of the inequality by \(xy\) and write \(\frac{1}{x+y}>1\) (as you did), because you don't know whether \(xy\) is positive or negative: if \(xy>0\) then you should write \(\frac{1}{x+y}>1\) BUT if \(xy<0\) then you should flip the sign and write \(\frac{1}{x+y}<1\). But even if you knew that \(xy>0\) then the next step of writing \(x+y<1\) from \(\frac{1}{x+y}>1\) would still be incorrect for the same exact reason: you don't k now whether \(x+y\) is positive or negative, hence you can not muliply both sides of the inequality by \(x+y\).

Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Thus the question is boiled down to: is \(\frac{xy}{x+y}>xy\)? Actually we can manipulate further but there is no need.

(1) x = 2y --> question becomes: is \(\frac{2y^2}{3y}>2y^2\)? Now, as we know that \(y\) is nonzero then \(2y^2>0\) and we can divide both parts by it --> is \(\frac{1}{3y}>1\)? As \(y\) is an integer (no matter positive or negative) then the answer to this question is always NO ( if it's a positive integer then \(\frac{1}{3y}<1\) and if it's a negative integer then again: \(\frac{1}{3y}<0<1\) ). Sufficient.

(2) x + y > 0 --> if \(x=y=1\) then the answer will be NO but if \(x=3\) and \(y=-1\) then the answer will be YES. Not sufficient.

Answer: A.

Dear Bunuel, Could you elaborate on the highlighted sentence as I am confused with the sign
_________________

Be challenged at EVERY MOMENT.

Each stage of the journey is crucial to attaining new heights of knowledge.

If x and y are nonzero integers, is (x-1 + y-1)-1 > [(x-1)(y-1)]-1 ?

(1) x = 2y

(2) x + y > 0

I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.

Apologies again for giving out too much.

First of all the question should be:

If x and y are nonzero integers, is (x^(-1) + y^(-1))^(-1) > (x^(-1)*y^(-1))^(-1) ?

Is \((x^{-1}+y^{-1})^{-1}> (x^{-1}*y^{-1})^{-1}\)? --> is \((\frac{1}{x}+\frac{1}{y})^{-1}>(\frac{1}{xy})^{-1}\)? --> is \((\frac{x+y}{xy})^{-1}>xy\) --> is \(\frac{xy}{x+y}>xy\)?

Now, from this point you cannot divide both parts of the inequality by \(xy\) and write \(\frac{1}{x+y}>1\) (as you did), because you don't know whether \(xy\) is positive or negative: if \(xy>0\) then you should write \(\frac{1}{x+y}>1\) BUT if \(xy<0\) then you should flip the sign and write \(\frac{1}{x+y}<1\). But even if you knew that \(xy>0\) then the next step of writing \(x+y<1\) from \(\frac{1}{x+y}>1\) would still be incorrect for the same exact reason: you don't k now whether \(x+y\) is positive or negative, hence you can not muliply both sides of the inequality by \(x+y\).

Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.

Thus the question is boiled down to: is \(\frac{xy}{x+y}>xy\)? Actually we can manipulate further but there is no need.

(1) x = 2y --> question becomes: is \(\frac{2y^2}{3y}>2y^2\)? Now, as we know that \(y\) is nonzero then \(2y^2>0\) and we can divide both parts by it --> is \(\frac{1}{3y}>1\)? As \(y\) is an integer (no matter positive or negative) then the answer to this question is always NO ( if it's a positive integer then \(\frac{1}{3y}<1\) and if it's a negative integer then again: \(\frac{1}{3y}<0<1\) ). Sufficient.

(2) x + y > 0 --> if \(x=y=1\) then the answer will be NO but if \(x=3\) and \(y=-1\) then the answer will be YES. Not sufficient.

Answer: A.

Dear Bunuel, Could you elaborate on the highlighted sentence as I am confused with the sign

We know that y is an integer. Now, for integer y, the answer to the question whether \(\frac{1}{3y}>1\) will always be NO (no matter whether y is negative or positive). If y > 0, then \(\frac{1}{3y}<1\) and if y < 0, then again \(\frac{1}{3y}<1\).

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...