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:

Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink]

Show Tags

09 Sep 2009, 11:35

Expert's post

there can be 4 instances:- i)x +,y -....|x| - |y| = |x+y| => x-y=x-y ii) both +... x-y=x+y.. not possible as xy not equal zero iii) both -....x-y=-(x+y) .. not possible as xy not equal zero iv) x-,y+...x-y=y-x......x=y.. therefore x and y are of different signs, when multiplied ,it should be -ive _________________

if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?

A. x-y > 0 B. x-y < 0 C. x+y >0 D. xy>0 E. xy<0

\(|x|-|y|=|x+y|\) --> square both sides --> \((|x|-|y|)^2=(|x+y|)^2\) --> note that \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\).

Answer: E.

Another way:

Right hand side, \(|x+y|\), is an absolute value, which is always non-negative, but as \(xy\neq{0}\), then in this case it's positive --> \(RHS=|x+y|>0\). So LHS must also be more than zero \(|x|-|y|>0\), or \(|x|>|y|\).

So we can have following 4 scenarios: 1. ------0--y----x--: \(0<y<x\) --> \(|x|-|y|=x-y\) and \(|x+y|=x+y\) --> \(x-y\neq{x+y}\). Not correct. 2. ----y--0------x--: \(y<0<x\) --> \(|x|-|y|=x+y\) and \(|x+y|=x+y\) --> \(x+y={x+y}\). Correct. 3. --x------0--y----: \(x<0<y\) --> \(|x|-|y|=-x-y\) and \(|x+y|=-x-y\) --> \(-x-y={-x-y}\). Correct. 4. --x----y--0------: \(x<y<0\) --> \(|x|-|y|=-x+y\) and \(|x+y|=-x-y\) --> \(-x+y\neq{-x-y}\). Not correct.

So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) --> \(x\) and \(y\) have opposite signs --> \(xy<0\).

if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?

A. x-y > 0 B. x-y < 0 C. x+y >0 D. xy>0 E. xy<0

Also, if you're short on time, and since you know that xy is not 0, then you know it MUST be greater than or less than 0, so you can narrow your choices down to D or E.

if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?

A. x-y > 0 B. x-y < 0 C. x+y >0 D. xy>0 E. xy<0

Also, if you're short on time, and since you know that xy is not 0, then you know it MUST be greater than or less than 0, so you can narrow your choices down to D or E.

the only way lxl - lyl can equal lx+yl is when one number is positive and one number is negative or both are zero. So then xy must be negative. I think I read if somewhere in Bunuel's post

I want to know wether there is a rule involved that i am missing, or an effective strategy to tackle that kind of questions

Thank you

mnpqxyzt has given a great solution above. I would like to add here that it is possible that it doesn't occur to you that you should square both sides. If you do get stuck with such a question, notice that it says 'which of the following MUST be true'. So as a back up, you can rely on plugging in numbers. If you get even one set of values for which the condition does not hold, the condition is not your answer.

|x| - |y| = |x+y| First set of non-zero values that come to mind is x = 1, y = -1 This set satisfies only options (A) and (E). Now, the set x = -1, y = 1 will also satisfy the given equation. But this set will not satisfy option (A). Hence answer (E). _________________

Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink]

Show Tags

18 Jun 2011, 21:49

Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive. _________________

Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink]

Show Tags

19 Jun 2011, 04:57

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

AnkitK wrote:

Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.

Ok. Given: |x| - |y| = |x+y| There are infinite set of values for x and y that satisfy this equation. Let us try one of them. Say x = 1, y = -1 a) x-y> 0 .......... 1 - (-1) > 0; 2 > 0; True b) x-y< 0........... 1 - (-1) < 0; 2 < 0; Eliminate c) x+y> 0........... 1 + (-1) > 0; 0 > 0; Eliminate d) xy>0.............. 1 *(-1) > 0; -1 > 0; Eliminate e) xy<0.............. 1 *(-1) < 0; -1 < 0; True

So I have two options that satisfy the assumed values of x and y. We need to eliminate one of them. They are a) x-y> 0 e) xy<0

We see that x = 1, y = -1 satisfies both these inequalities. But option (a) is not symmetric i.e. if you interchange the values of x and y, it will not hold. That is, if x = -1 and y = 1, our original equation |x| - |y| = |x+y| is still satisfied but a) x-y> 0 .............. (-1) - (1) > 0; -2>0; False. Eliminate e) xy<0................. (-1)(1) < 0; True Since option (e) still holds, it is the answer.

xy<0 is certainly true when one of them is negative and the other is positive.

Takeaways:

|x| + |y| = |x+y| when x and y have the same signs - either both are positive or both are negative (or one or both of them are 0)

|x| - |y| = |x+y| when x and y have opposite signs - one is positive, the other negative (or y is 0 or both are 0) _________________

Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink]

Show Tags

21 Jun 2011, 10:25

VeritasPrepKarishma wrote:

AnkitK wrote:

Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.

Ok. Given: |x| - |y| = |x+y| There are infinite set of values for x and y that satisfy this equation. Let us try one of them. Say x = 1, y = -1 a) x-y> 0 .......... 1 - (-1) > 0; 2 > 0; True b) x-y< 0........... 1 - (-1) < 0; 2 < 0; Eliminate c) x+y> 0........... 1 + (-1) > 0; 0 > 0; Eliminate d) xy>0.............. 1 *(-1) > 0; -1 > 0; Eliminate e) xy<0.............. 1 *(-1) < 0; -1 < 0; True

So I have two options that satisfy the assumed values of x and y. We need to eliminate one of them. They are a) x-y> 0 e) xy<0

We see that x = 1, y = -1 satisfies both these inequalities. But option (a) is not symmetric i.e. if you interchange the values of x and y, it will not hold. That is, if x = -1 and y = 1, our original equation |x| - |y| = |x+y| is still satisfied but a) x-y> 0 .............. (-1) - (1) > 0; -2>0; False. Eliminate e) xy<0................. (-1)(1) < 0; True Since option (e) still holds, it is the answer.

xy<0 is certainly true when one of them is negative and the other is positive.

Takeaways:

|x| + |y| = |x+y| when x and y have the same signs - either both are positive or both are negative (or one or both of them are 0)

|x| - |y| = |x+y| when x and y have opposite signs - one is positive, the other negative (or y is 0 or both are 0)

Karishma, I think the 2nd takeaway has some extra condition missed out :

|x|-|y|=|x+y| if and only if 1) Both have opposite signs and 2) |x| >= |y|

because x, y, |x|-|y| , |x+y| 5 ,-6, -1,1 5 ,-5 , 0 , 0 5, -4, 1 , 1

so, |5| cannot be greater than |-6| for the condition to occur _________________

Re: If |x| - |y| = |x+y| and xy not equal zero , which of the following [#permalink]

Show Tags

21 Jun 2011, 11:42

Expert's post

krishp84 wrote:

Karishma, I think the 2nd takeaway has some extra condition missed out :

|x|-|y|=|x+y| if and only if 1) Both have opposite signs and 2) |x| >= |y|

because x, y, |x|-|y| , |x+y| 5 ,-6, -1,1 5 ,-5 , 0 , 0 5, -4, 1 , 1

so, |5| cannot be greater than |-6| for the condition to occur

The takeaway is that if x and y satisfy the condition |x|-|y|=|x+y|, then they must have opposite signs (or y is 0 or both are 0).

But, x and y having opposite signs is not sufficient to satisfy the condition |x|-|y|=|x+y|. As you said, in that case we will also need to check for their absolute values. (Good thinking, btw) _________________

If |x| - |y| = |x+y| and xy does not equal to 0, which of the fo [#permalink]

Show Tags

03 Dec 2012, 20:27

2

This post received KUDOS

When you see |x| or absolute values being tested in the GMAT, this means testing (1) positive and negative signs (2) zeroes (3) nature of |x| and |y| against each other

Looking at the equation we know |x| - |y| = |x + y| where xy is not 0. (1) We figure that |x| and |y| are non-zeroes (2) We figure that |x| - |y| > 0. This means |x| > |y|

Now, we figure the signs of the two variables by lining up possibilities (1) both negative ==> x=-5,y=-4 ==> |-5| - |-4| = |-9| FALSE (2) both positive ==> x=5,y=4 ==> |5| - |4| = |5 + 4| FALSE (3) x is positive, y is negative ==> |5| - |-4| = |5-4| TRUE (4) y is positive, x is negative ==> |-5| - |4| = |-5+4| TRUE

Now, let's find the answer A) x-y > 0 ==> -5-(-4) = -1 FALSE E) xy <0 ==> This is exactly what we need. x and y have both different signs.

if lxl - lyl = lx+yl anf xy does , not equal to o, which of the following must be true?

A. x-y > 0 B. x-y < 0 C. x+y >0 D. xy>0 E. xy<0

\(|x|-|y|=|x+y|\) --> square both sides --> \((|x|-|y|)^2=(|x+y|)^2\) --> note that \((|x+y|)^2=(x+y)^2\) --> \((|x|-|y|)^2=(x+y)^2\) --> \(x^2-2|xy|+y^2=x^2+2xy+y^2\) --> \(|xy|=-xy\) --> \(xy\leq{0}\), but as given that \(xy\neq{0}\), then \(xy<0\).

Answer: E.

Another way:

Right hand side, \(|x+y|\), is an absolute value, which is always non-negative, but as \(xy\neq{0}\), then in this case it's positive --> \(RHS=|x+y|>0\). So LHS must also be more than zero \(|x|-|y|>0\), or \(|x|>|y|\).

So we can have following 4 scenarios: 1. ------0--y----x--: \(0<y<x\) --> \(|x|-|y|=x-y\) and \(|x+y|=x+y\) --> \(x-y\neq{x+y}\). Not correct. 2. ----y--0------x--: \(y<0<x\) --> \(|x|-|y|=x+y\) and \(|x+y|=x+y\) --> \(x+y={x+y}\). Correct. 3. --x------0--y----: \(x<0<y\) --> \(|x|-|y|=-x-y\) and \(|x+y|=-x-y\) --> \(-x-y={-x-y}\). Correct. 4. --x----y--0------: \(x<y<0\) --> \(|x|-|y|=-x+y\) and \(|x+y|=-x-y\) --> \(-x+y\neq{-x-y}\). Not correct.

So we have that either \(y<0<x\) (case 2) or \(x<0<y\) (case 3) --> \(x\) and \(y\) have opposite signs --> \(xy<0\).

Answer: E.

Hope it helps.

Hi Bunuel, I thought an absolute value of a product can never be a negative. Could you please explain in your equation how did you progress after getting this negative?

Re: If |x| - |y| = |x+y| and xy does not equal to 0, which of [#permalink]

Show Tags

10 Feb 2013, 18:03

Basically, remember in these exercises to do square root in both sides to simplify: (|x+y|)^2 = (x+y)^2 or (|x|)^2 = x^2 And also remember that: |x||y|=|xy|

Therefore, with: -|x||y| = xy I would say: xy is always equal to something negative. Solution: E.

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...