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:

(1) |x-y| = |y-x| .... this is a fact i.e. |3| = |-3| \(|y-x| > 1\) insufficient because the value of "y-x" can be 3 or -3, so we can't answer the question.

(2) \(y > x\) so "y-x" will always be +ve. this is also a fact because the difference between a bigger term and a smaller term is always +ve. e.g. 3-2=1 (+ve) ... likewise ... -2-(-3)=1 (+ve)

\(y-x > \frac{1}{-(y-x) ?}\)

the above expression will always be true because LHS is always +ve and RHS is always -ve. sufficient.

This wont remain a 700 level question, if you just observe the question and answer choices and manipulate. Entire question and answere choices revloves around x-y and y-x. Lets assume y-x = z

question becomes: is z >1/-z or is z+1/z >0 ? So if we know z is positive then z+1/z is positive and we can ans, if we know z is negative then z+1/z is negative and we can ans. Only if we cant ans about the sign of z, we cant ans the qustion.

Statement 1: |z| >1 => z <-1 , z >1 Now we dont if z is positive or negative, therefore not sufficient.

Statement 2: y-x>0 => z>0 Now we know z is positive, and hence z+1/z is positive. Sufficient.

(1) |x-y|>1 --> well this one is clearly insufficient, try x-y=2 to get a NO answer and x-y=-2 to get an YES answer (notice that both examples satisfy┃x-y|>1). Not sufficient.

(2) y>x --> this can be rewritten as y-x>0 or 0>x-y, which means that LHS=y-x=positive and RHS=1/(x-y)=negative, thus y-x>1/(x-y). Sufficient.

Re: Is y - x > 1/(x-y)? (1) x - y > 1 (2) y > x [#permalink]

Show Tags

18 Jan 2012, 00:39

Let us paraphrase!!: Suppose: y-x=t => x-y= -t,so the question simply asks us: Is t>1/-t ? 1)|-t|>1 Since |-t|=|t| => |t|>1 =>t>1 or t< -1 => we have 2 signs for t ,NOT sufficient! 2) y>x=> y-x>0 =>t>0 => t has only one sign! So, SUFFICIENT!! The answer is:B To answer this question we need to know only the sign of t! Because for all real numbers (including integers,fractions...) the positive number is always bigger than negative numbers and vice versa...

This wont remain a 700 level question, if you just observe the question and answer choices and manipulate. Entire question and answere choices revloves around x-y and y-x. Lets assume y-x = z

question becomes: is z >1/-z or is z+1/z >0 ? So if we know z is positive then z+1/z is positive and we can ans, if we know z is negative then z+1/z is negative and we can ans. Only if we cant ans about the sign of z, we cant ans the qustion.

Statement 1: |z| >1 => z <-1 , z >1 Now we dont if z is positive or negative, therefore not sufficient.

Statement 2: y-x>0 => z>0 Now we know z is positive, and hence z+1/z is positive. Sufficient.

Ans B it is!

Hi Vipps can you just explain how did you get Z+1/Z >0

This wont remain a 700 level question, if you just observe the question and answer choices and manipulate. Entire question and answere choices revloves around x-y and y-x. Lets assume y-x = z

question becomes: is z >1/-z or is z+1/z >0 ? So if we know z is positive then z+1/z is positive and we can ans, if we know z is negative then z+1/z is negative and we can ans. Only if we cant ans about the sign of z, we cant ans the qustion.

Statement 1: |z| >1 => z <-1 , z >1 Now we dont if z is positive or negative, therefore not sufficient.

Statement 2: y-x>0 => z>0 Now we know z is positive, and hence z+1/z is positive. Sufficient.

Ans B it is!

Hi Vipps can you just explain how did you get Z+1/Z >0

question is (y-x) > 1/(x-y) ?

or is (y-x) -1/(x-y) >0 ? or is (y-x) + 1/(y-x) >0 ?

Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]

Show Tags

21 May 2013, 23:02

so My question is statement 1

X-y is positive so y - x is negative so the answer is a NO

and then second case is Y - X is positive so X - Y is negative so its a yes

statement 2 has Y - X is positive so its a definite yes

so this is the best approach to solve these problems or a numerical approach is better. How would solve such questions if its complex?
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Re: Is y - x > 1 / x-y ? (1) ┃x - y┃ > 1 (2) y > x [#permalink]

Show Tags

24 Jun 2013, 08:45

Hi!

I have a question.

For #1, we have (-) > 1/(+) Negative number > positive, the answer is NO

It seems to me that this simply isn't wrong, but not possible.

In another problem, we ended up with (x-1)^2<0. That's not possible, naturally, as anything squared cannot be less than zero. In that problem, which can be found here: is-x-134652.html

(x-1)^2<0 but obviously this is not possible. So even though it is incorrect, it doesn't make I.) insufficient, it's simply not counted. What am I missing here?

Thanks!

Zarrolou wrote:

Is \(y - x > \frac{1}{x-y}\) ?

(1) \(|x - y| > 1\) From this we get two cases: I)\(x-y>1\) In this one we would get

\((-) > \frac{1}{(+)}\) Negative number > positive, the answer is NO

II)\(y-x<-1\)

\((+) > \frac{1}{(-)}\) Positive number > negative, the answer is YES

Not sufficient

(2) \(y > x\) so \(y-x>0\)

\((+) > \frac{1}{(-)}\) Positive number > negative , the answer is YES

Sufficient B

This is the best approach for these questions (IMO)

A simpler solution without going too much into signs interpretation.

(y-x) > 1/(x-y) => (x-y) < 1/(x-y) (multiplying by minus both sides;basically ensuring both the variable expressions to be same) => The statement above is only possible when (x-y) is a fraction and is less than 1 or negative.

1 stmt gives both the possible cases. i.e. (x-y) is both + and -ve. 2 ensures that x-y is definitely -ve and less than 1.

In other words, (x-y) and (y-x) could be two OR -2 meaning we can't determine a single value for the inequality. INSUFFICIENT

(2) y > x

If y > x then y - x > 1/(x - y) will always have a positive # on the LHS and a fraction on the RHS. Is is true whether x and y are positive or negative. SUFFICIENT

(1) |x - y┃ > 1 (take note of x-y in the stem) x - y > 1 OR x - y < -1

We have two cases here, where x-y is either greater than 1 or less than negative 1. This means that the fraction could be either positive or negative meaning we can't be sure if the inequality y - x > 1/(x - y) holds true or not. INSUFFICIENT

(2) y > x If y>x then regardless of the signs of y and x (y, x) (y, -x) (-y, -x) y-x will be positive and 1/(x-y) will be a fraction or a negative number meaning that the inequality will always hold true. SUFFICIENT

Notice that the entire question uses (x - y) or (y - x) and not the two independently so we can easily proceed by taking y - x = a to reduce complications

Is \(a > \frac{-1}{a}\) Is \(a + \frac{1}{a} > 0\) Is \(\frac{(a^2 + 1)}{a}\) > 0 Obviously a^2 + 1 is positive so we just need to worry about the sign of 'a'. If we know whether a is positive or negative, we can answer the question.

(1) |x - y┃ > 1

|a┃> 1 implies a < -1 or a > 1. Not sufficient

(2) y > x y - x > 0 a > 0 Sufficient since we get that a > 0.

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.
_________________

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.
_________________

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...