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Re: Tough DS Algebra [#permalink]
07 Jan 2011, 09:15

10

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

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.

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.

Answer: C _________________

"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

Is 1/(x-y) < y - x? [#permalink]
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 yare 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.

Please, your comments. _________________

"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."

Re: Is 1/(x-y) < y - x? [#permalink]
27 Aug 2012, 07:14

1

This post received KUDOS

Expert's post

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 yare 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.

Please, your comments.

Merging similar topics. You are right, answer should be C, not E. _________________

Re: Is 1 / (x-y) < y - x [#permalink]
03 Sep 2012, 04:16

5

This post received KUDOS

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.

Answer C. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Is 1 / (x-y) < y - x [#permalink]
26 Nov 2012, 01:28

1

This post received KUDOS

Expert's post

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.

Re: Is 1 / (x-y) < y - x [#permalink]
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.

Re: Tough DS Algebra [#permalink]
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.

Answer: C.

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.”

Re: Is 1/(x-y) < y - x [#permalink]
07 Aug 2014, 21:46

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 [#permalink]
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.

Answer: C.

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?

Re: Is 1/(x-y) < y - x [#permalink]
15 Sep 2014, 20:00

Expert's post

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.

Answer: C.

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. _________________

Hey everyone, today’s post focuses on the interview process. As I get ready for interviews at Kellogg and Tuck (and TheEngineerMBA ramps up for his HBS... ...

Hey everyone, today’s post focuses on the interview process. As I get ready for interviews at Kellogg and Tuck (and TheEngineerMBA ramps up for his HBS... ...