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) 4x = 3y --> \(x=\frac{3}{4}y\), well this one is relatively easy. This statement only tells that x and y have the same sign. When they are both positive then \(x>y\), BUT when they are both negative \(y>x\). Not sufficient.

(2) \(|y - x| = x - y\) --> \(|y - x| = -(y-x)\). This means that \(y - x\leq{0}\) --> \(x\geq{y}\). Thus, this statement says that x can be more than or equal to y. Not sufficient.

(1)+(2) From (1) (\(x=\frac{3}{4}y\)) it follows that \(x\) is not equal to \(y\) (bearing in mind that xy ≠ 0), hence from (2): \(x>y\). Sufficient.

okay so my answer is wrong..can't figure out what i did wrong on statement 2

What you missed is that (2) is true not only when x>y but also when x=y. Hence we can not say for sure that x>y, as x>=y. Not sufficient. _________________

\(|y - x| = x - y\). Look at the LHS it's absolute value, it's never negative, hence RHS or \(x-y\) is never negative, \(x-y>=0\) or \(y-x<=0\). If so then only one possibility for \(|y - x|\), it must be \(-y+x\).

So we'll have: Condition: \(y-x<=0\), which is the same as \(x>=y\)

The above means that \(|y - x| = x - y\) is always true for \(x>=y\). But as we concluded earlier it's not enough. We need to be sure that \(x>y\). And \(x>=y\) leaves the possibility that they are equal, which is removed with the statement (1) when considering together. Hence C. _________________

st 1) y=1, x=3/4 - false y=-1, x=-3/4 - true Not sufficient st 2) |y-x| =x-y--> x>y Sufficient coz |-x| = -(-x) = x -- basically a negative numbers should be negated again to get the result for absolute value all the cases as cay that x>y but fails when x=y

combining we can answer the question

C

<didnt consider x=y option, so answered as B >

Last edited by chix475ntu on 11 Feb 2010, 10:06, edited 1 time in total.

st 1) y=1, x=3/4 - false y=-1, x=-3/4 - true Not sufficient st 2) |y-x| =x-y--> x>y Sufficient coz |-x| = -(-x) = x -- basically a negative numbers should be negated again to get the result for absolute value

B

B alone is not sufficient as we dont know whether x=y?

So A gives information that x is not = y.

So Final ans is C This was nice question...even I would have done the same mistake in the exam..Its really important to think of all the possibilities on the G day. _________________

1) Now x=(3/4)y so y > x means statement 1 alone is sufficient to answer the question.

2) Using some values for x & y we can find if x>y then statement |y - x| = x - y is true and if y>x then statement |y - x| = x - y is not true. Hence statement 2 alone is sufficient to answer question.

Either statement is sufficient to answer question so answer is D.

(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).

(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.

(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.

Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]

Show Tags

26 Sep 2013, 09:28

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. _________________

(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).

(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.

(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.

Answer: C.

Hope it helps.

Dear Bunuel:

I am a bit confused with this problem. I don't have a problem to understand the first statement, but I do have a problem trying to understand the second one..

I saw a user on this thread was solving the second statement as this:

2. |y - x| = x - y

y - x = x - y 2x = 2y x=y

-y + x = x-y 0=0

Isn't his way of solving the correct way to solve this kind of problems? I have seen some other exercises and they always negate the LHS that has the absolute value on it regarding of the inequality symbol or equality. I would like to learn why we are not doing this on this problem.

If I am wrong, would you be so kind to revamp the way of solving statement 2, I already read the way you solved it, but it was not 100% clear for me.

(1) \(4x = 3y\) --> if \(x\) and \(y\) are both positive, then \(x<y\) BUT if they are both negative then \(x>y\). Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as \(xy\neq0\) then from statement (1) \(x\neq{y}\).

(2) \(|y-x|=x-y\). Now as LHS is absolute value, which is never negative, RHS must also be \(\geq0\) --> so \(x-y\geq0\) --> then \(|y-x|=-(y-x)\), and we'll get \(-(y-x)=x-y\) --> \(0=0\), which is true. This means that equation \(|y-x|=x-y\) holds true when \(x-y\geq0\) or, which is the same, when \(x\geq{y}\). But this not enough as \(x=y\) is still possible. Not sufficient.

(1)+(2) From (1) we got that \(x\neq{y}\) and from (2) \(x\geq{y}\), hence \(x>y\). Sufficient.

Answer: C.

Hope it helps.

Dear Bunuel:

I am a bit confused with this problem. I don't have a problem to understand the first statement, but I do have a problem trying to understand the second one..

I saw a user on this thread was solving the second statement as this:

2. |y - x| = x - y

y - x = x - y 2x = 2y x=y

-y + x = x-y 0=0

Isn't his way of solving the correct way to solve this kind of problems? I have seen some other exercises and they always negate the LHS that has the absolute value on it regarding of the inequality symbol or equality. I would like to learn why we are not doing this on this problem.

If I am wrong, would you be so kind to revamp the way of solving statement 2, I already read the way you solved it, but it was not 100% clear for me.

Thanks in advance.

There can be many correct ways to solve a questions.

Try this one: (2) says that \(|y-x|=-(y-x)\). We know that \(|x|=-x\), when \(x\leq{0}\), thus \(y-x\leq{0}\). _________________

Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]

Show Tags

27 Nov 2013, 06:13

Economist wrote:

If xy ≠ 0, is x > y?

(1) 4x = 3y

(2) |y - x| = x - y

Question is: Is x>y?

Statement 1 4x = 3y x/y=3/4 Not sufficient, we need to know their signs

Statement 2 |y - x| = x - y We can rearrange LHS as |-(-y + x|) = |(x-y|) So we end up with |(x-y|=x-y, meaning that |x-y|>=0 or x>=y Could be equal or could be greater. Insuff

Both together, they can't be equal to it is Suff

Answer is C Hope it helps Cheers! J

gmatclubot

Re: If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y
[#permalink]
27 Nov 2013, 06:13

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

Time is a weird concept. It can stretch for seemingly forever (like when you are watching the “Time to destination” clock mid-flight) and it can compress and...