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 350,000 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: approach to these kind of qs?? [#permalink]
03 Aug 2010, 04:17

bibha wrote:

Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2

Please take the trouble of explaining one step at a time

To me I try to address (ii) first because it looks easiest

Pick x and y that meet the conditions given (x=5 y=3) |5-3| > |5+3| FALSE Try different x and y thinking we are using absolutes I picked two negatives (x=-3 y=-5) |-3-(-5)| > |-3+-5| 2 > 8 False So we see that ii alone is sufficient. The answer is B or D

To investigate i I am lost at the algebra, so I just try to think about which squares are 9 apart

3^2=9 4^2=16 5^2=25 luckily on the GMAT you typically work with smaller squares

So I pick X=5 and Y=4 because \(5^2-4^2=9\) Since we are squaring keep in mind that X can be 5 or -5 While Y can be 4 or -4

The test in the given equation |x-y| > |x+y| (5,4) gives us 1 > 9 False (5,-4) gives us 9 > 1 True (-5,4) gives us 9 > 1 True (-5,-4) gives us 1 > 9 False

So A alone doesn't work

The correct answer is B.

In proofreading my work I found a major error in my calculation with the absolutes. This changed my answer on this question. On the real GMAT I would have guessed probably E after 3-4 minutes of frustration!

Re: approach to these kind of qs?? [#permalink]
03 Aug 2010, 09:28

TallJTinChina wrote:

bibha wrote:

Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2

Please take the trouble of explaining one step at a time

To me I try to address (ii) first because it looks easiest

Pick x and y that meet the conditions given (x=5 y=3) |5-3| > |5+3| FALSE Try different x and y thinking we are using absolutes I picked two negatives (x=-3 y=-5) |-3-(-5)| > |-3+-5| 2 > 8 False So we see that ii alone is sufficient. The answer is B or D

To investigate i I am lost at the algebra, so I just try to think about which squares are 9 apart

3^2=9 4^2=16 5^2=25 luckily on the GMAT you typically work with smaller squares

So I pick X=5 and Y=4 because \(5^2-4^2=9\) Since we are squaring keep in mind that X can be 5 or -5 While Y can be 4 or -4

The test in the given equation |x-y| > |x+y| (5,4) gives us 1 > 9 False (5,-4) gives us 9 > 1 True (-5,4) gives us 9 > 1 True (-5,-4) gives us 1 > 9 False

So A alone doesn't work

The correct answer is B.

In proofreading my work I found a major error in my calculation with the absolutes. This changed my answer on this question. On the real GMAT I would have guessed probably E after 3-4 minutes of frustration!

In plugging in numbers approach - like here, (ii) works for 2 examples selected, but how do we universally say it is true? Is there not a more general approach? I think for speed we just select a few examples, but how do we guarantee it? I am always confused on this point. So I jump to prove it for good and I made a mistake here

I saw x^2-y^2 = 9, took it to mean (x+y)(x-y)=9 and said x+y,x-y have to be both 3 or -3 and jumped at A. Wrong, does not say that they have to be integers...

Please explain or point me to the strategy in dealing with these please! _________________

Re: approach to these kind of qs?? [#permalink]
04 Aug 2010, 00:34

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

bibha wrote:

Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2

Please take the trouble of explaining one step at a time

I'd start solving by analyzing the stem.

First approach:

Is \(|x-y|>|x+y|\)? In which cases \(|x-y|\) will be more than \(|x+y|\)?

If \(x\) and \(y\) have the same sign (both positive or both negative) then absolute value of their sum (\(|x+y|\)) will be more than absolute value of their difference (\(|x-y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x+y|\), for example \(|5+3|\) or \(|-5-3|\) and in \(|x-y|\) they will not.

If \(x\) and \(y\) have opposite signs then absolute value of their difference (\(|x-y|\)) will be more than absolute value of their sum (\(|x+y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x-y|\), for example \(|5-(-3)|\) or \(|-5-3|\) and in \(|x+y|\) they will not.

So basically the question is: do \(x\) and \(y\) have opposite signs?

Second approach:

Is \(|x-y|>|x+y|\)? As both sides of inequality are non-negative we can safely square them: is \((x-y)^2>(x+y)^2\)? --> is \(x^2-2xy+y^2>x^2+2xy+y^2\)? --> is \(xy<0\)? Again the question becomes: do \(x\) and \(y\) have opposite signs?

Next:

(1) x^2-y^2 = 9 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.

(2) x-y=2 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.

(1)+(2) \(x+y=4.5\) and \(x-y=2\) --> directly tells us that \(|x-y|=2<|x+y|=4.5\). Sufficient. (If you solve system of equations you'll see that \(x\) and \(y\) have the same sign: they are both positive).

Answer: C.

PLUGGING NUMBERS:

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

(1) \(x^2-y^2=9\) --> try \(x=5\) and \(y=4\) (\(x^2-y^2=25-16=9\)) --> \(|x-y|=1<|x+y|=9\), so we have answer YES. Now let's try another set of numbers to get NO: \(x=5\) and \(y=-4\) --> \(|x-y|=9>|x+y|=1\), so now we have answer NO. Thus this statement is not sufficient.

We can do the same with statement (2) as well.

For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient.

Re: approach to these kind of qs?? [#permalink]
07 Aug 2010, 03:37

Bunuel wrote:

bibha wrote:

Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2

Please take the trouble of explaining one step at a time

I'd start solving by analyzing the stem.

First approach:

Is \(|x-y|>|x+y|\)? In which cases \(|x-y|\) will be more than \(|x+y|\)?

If \(x\) and \(y\) have the same sign (both positive or both negative) then absolute value of their sum (\(|x+y|\)) will be more than absolute value of their difference (\(|x-y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x+y|\), for example \(|5+3|\) or \(|-5-3|\) and in \(|x-y|\) they will not.

If \(x\) and \(y\) have opposite signs then absolute value of their difference (\(|x-y|\)) will be more than absolute value of their sum (\(|x+y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x-y|\), for example \(|5-(-3)|\) or \(|-5-3|\) and in \(|x+y|\) they will not.

So basically the question is: do \(x\) and \(y\) have opposite signs?

Second approach:

Is \(|x-y|>|x+y|\)? As both sides of inequality are non-negative we can safely square them: is \((x-y)^2>(x+y)^2\)? --> is \(x^2-2xy+y^2>x^2+2xy+y^2\)? --> is \(xy<0\)? Again the question becomes: do \(x\) and \(y\) have opposite signs?

Next:

(1) x^2-y^2 = 9 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.

(2) x-y=2 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.

(1)+(2) \(x+y=4.5\) and \(x-y=2\) --> directly tells us that \(|x-y|=2<|x+y|=4.5\). Sufficient. (If you solve system of equations you'll see that \(x\) and \(y\) have the same sign: they are both positive).

Answer: C.

PLUGGING NUMBERS:

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

(1) \(x^2-y^2=9\) --> try \(x=5\) and \(y=4\) (\(x^2-y^2=25-16=9\)) --> \(|x-y|=1<|x+y|=9\), so we have answer YES. Now let's try another set of numbers to get NO: \(x=5\) and \(y=-4\) --> \(|x-y|=9>|x+y|=1\), so now we have answer NO. Thus this statement is not sufficient.

We can do the same with statement (2) as well.

For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient.

Hope it helps.

Excellent Bunuel!! Simply Excellent. When it is said that x-y=2 and x^2-y^2=9 why does it not jump at me that x+y must be 4.5!!! Got your point about plugging in numbers.. Can disprove but proving universally is difficult... _________________

Re: approach to these kind of qs?? [#permalink]
07 Aug 2010, 08:09

well I eliminated A and B out as in both case x and y can be both and negative in either case.... which will obvoiusly give dual case... so D also out....

now here was big problem C and E.... I was not able to decide here i was trying to establish a case where x must be + or - same with y if we can do that we will have 1 another yes or no as answer

Re: approach to these kind of qs?? [#permalink]
30 Jan 2014, 16:20

Bunuel wrote:

bibha wrote:

Is |x-y| > |x+y| i. x^2-y^2 = 9 ii. x-y=2

Please take the trouble of explaining one step at a time

I'd start solving by analyzing the stem.

First approach:

Is \(|x-y|>|x+y|\)? In which cases \(|x-y|\) will be more than \(|x+y|\)?

If \(x\) and \(y\) have the same sign (both positive or both negative) then absolute value of their sum (\(|x+y|\)) will be more than absolute value of their difference (\(|x-y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x+y|\), for example \(|5+3|\) or \(|-5-3|\) and in \(|x-y|\) they will not.

If \(x\) and \(y\) have opposite signs then absolute value of their difference (\(|x-y|\)) will be more than absolute value of their sum (\(|x+y|\)), as in this case \(x\) and \(y\) will contribute to each other in \(|x-y|\), for example \(|5-(-3)|\) or \(|-5-3|\) and in \(|x+y|\) they will not.

So basically the question is: do \(x\) and \(y\) have opposite signs?

Second approach:

Is \(|x-y|>|x+y|\)? As both sides of inequality are non-negative we can safely square them: is \((x-y)^2>(x+y)^2\)? --> is \(x^2-2xy+y^2>x^2+2xy+y^2\)? --> is \(xy<0\)? Again the question becomes: do \(x\) and \(y\) have opposite signs?

Next:

(1) x^2-y^2 = 9 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.

(2) x-y=2 --> we can not say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.

(1)+(2) \(x+y=4.5\) and \(x-y=2\) --> directly tells us that \(|x-y|=2<|x+y|=4.5\). Sufficient. (If you solve system of equations you'll see that \(x\) and \(y\) have the same sign: they are both positive).

Answer: C.

PLUGGING NUMBERS:

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

(1) \(x^2-y^2=9\) --> try \(x=5\) and \(y=4\) (\(x^2-y^2=25-16=9\)) --> \(|x-y|=1<|x+y|=9\), so we have answer YES. Now let's try another set of numbers to get NO: \(x=5\) and \(y=-4\) --> \(|x-y|=9>|x+y|=1\), so now we have answer NO. Thus this statement is not sufficient.

We can do the same with statement (2) as well.

For this question I don't recommend to use number plugging for (1)+(2), as it's quite straightforward algebraically that taken together 2 statements are sufficient.

Hope it helps.

I like second approach best, squaring both sides is definetely something you can do here in order to manipulate the problem more easily getting rid of the absolute values in both sides and saving time instead of plugging numbers

Just my 2cents

Cheers J

gmatclubot

Re: approach to these kind of qs??
[#permalink]
30 Jan 2014, 16:20

Hello everyone! Researching, networking, and understanding the “feel” for a school are all part of the essential journey to a top MBA. Wouldn’t it be great... ...

As part of our focus on MBA applications next week, which includes a live QA for readers on Thursday with admissions expert Chioma Isiadinso, we asked our bloggers to...

Booth allows you flexibility to communicate in whatever way you see fit. That means you can write yet another boring admissions essay or get creative and submit a poem...