It is currently 15 Dec 2017, 15:43

Decision(s) Day!:

CHAT Rooms | Olin (St. Louis) R1 | Tuck R1 | Ross R1 | Fuqua R1


Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Is |x - y| > |x + y|?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Manager
Manager
avatar
Joined: 14 Apr 2010
Posts: 216

Kudos [?]: 244 [0], given: 1

Is |x - y| > |x + y|? [#permalink]

Show Tags

New post 02 Aug 2010, 23:16
6
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

60% (01:29) correct 40% (02:12) wrong based on 197 sessions

HideShow timer Statistics

Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2
[Reveal] Spoiler: OA

Kudos [?]: 244 [0], given: 1

Manager
Manager
avatar
Status: Waiting to hear from University of Texas at Austin
Joined: 24 May 2010
Posts: 76

Kudos [?]: 75 [0], given: 4

Location: Changchun, China
Schools: University of Texas at Austin, Michigan State
Re: approach to these kind of qs?? [#permalink]

Show Tags

New post 03 Aug 2010, 04:17
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 :-)


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!

Kudos [?]: 75 [0], given: 4

Director
Director
avatar
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
Posts: 678

Kudos [?]: 171 [0], given: 15

Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Re: approach to these kind of qs?? [#permalink]

Show Tags

New post 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!
_________________

Consider kudos, they are good for health

Kudos [?]: 171 [0], given: 15

Intern
Intern
avatar
Joined: 23 Nov 2009
Posts: 4

Kudos [?]: 1 [0], given: 0

Re: approach to these kind of qs?? [#permalink]

Show Tags

New post 03 Aug 2010, 10:24
1
This post was
BOOKMARKED
IMO C

Stmt 1: After simplifying we get (x+y)(x-y)=9. So insuff.

Stmt 2: x-y=2

now if x=5 and y=3 then,
|5-3| > |5+3| does not hold true.

but if x=1 and y=-1 then
|1-(-1)| > |1-1| holds true.

So stmt 2 is also insuff.

combining 1 & 2,

we have value of x-y=2 from stmt 2,
so subsituting x-y=2 in (x-y)(x+y)=9, we get x+y=4.5.

Now 2 < 4.5 ....so |x-y|<|x+y|....

please correct if I have mistaken anything here....

Kudos [?]: 1 [0], given: 0

Expert Post
2 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42618

Kudos [?]: 135758 [2], given: 12708

Is |x - y| > |x + y|? [#permalink]

Show Tags

New post 04 Aug 2010, 00:34
2
This post received
KUDOS
Expert's post
5
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 cannot say from this whether \(x\) and \(y\) have opposite signs. Not sufficient.

(2) x-y=2 --> we cannot 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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135758 [2], given: 12708

Director
Director
avatar
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 17 Jul 2010
Posts: 678

Kudos [?]: 171 [0], given: 15

Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Re: approach to these kind of qs?? [#permalink]

Show Tags

New post 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...
_________________

Consider kudos, they are good for health

Kudos [?]: 171 [0], given: 15

Manager
Manager
avatar
Joined: 24 Apr 2010
Posts: 60

Kudos [?]: 10 [0], given: 0

Re: approach to these kind of qs?? [#permalink]

Show Tags

New post 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

any help here?

Kudos [?]: 10 [0], given: 0

Intern
Intern
avatar
Joined: 01 Aug 2006
Posts: 34

Kudos [?]: 42 [0], given: 0

Re: Is |x - y| > |x + y|? [#permalink]

Show Tags

New post 24 Jan 2014, 14:24
Squaring and rearranging, is xy <0? (x>0, y<0; or x<0, y>0)
1.x^2-y^2 = 9 -> x and y can take same/diff signs
2. x - y = 2 -> cannot be sure

Combining, (x+y)(x-y) = 9; since x -y = 2, x +y =4.5

No need to solve because we get distinct values for x and y => we know the signs of x and y => we know the sign of xy.

C.

Kudos [?]: 42 [0], given: 0

Current Student
User avatar
Joined: 06 Sep 2013
Posts: 1965

Kudos [?]: 759 [0], given: 355

Concentration: Finance
GMAT ToolKit User
Re: approach to these kind of qs?? [#permalink]

Show Tags

New post 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

Kudos [?]: 759 [0], given: 355

Expert Post
Math Revolution GMAT Instructor
User avatar
P
Joined: 16 Aug 2015
Posts: 4481

Kudos [?]: 3158 [0], given: 0

GPA: 3.82
Premium Member
Re: Is |x - y| > |x + y|? [#permalink]

Show Tags

New post 10 Nov 2015, 10:46
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2

If we modify the question, |x - y| > |x + y|?, or (|x - y|)^2 > (|x + y|)^2?, or (x - y)^2 > (x + y)^2?, or x^2-2xy+y^2>x^2+2xy+y^2?, or -2xy>2xy?, or 0>4xy?, or xy<0?.
The conditions do not have xy<0, so there are 2 variables (X,y) and 2 equations are given from the 2 conditions, so there is high chance (C) will be our answer.
Looking at the conditions together, x-y=2, x+y=9/2. This is sufficient and the answer becomes (C).

For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself
See our Youtube demo

Kudos [?]: 3158 [0], given: 0

Intern
Intern
avatar
B
Joined: 20 Jan 2017
Posts: 8

Kudos [?]: 7 [0], given: 35

Location: India
Concentration: Entrepreneurship, General Management
Schools: ISB '19 (A)
GMAT 1: 720 Q50 V38
GPA: 3.88
WE: General Management (Transportation)
Re: Is |x - y| > |x + y|? [#permalink]

Show Tags

New post 22 Jul 2017, 02:54
bibha wrote:
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2



Let us try to solve it step by step.

1. It says (x+y)(x-y)=9.
It means either both of them are positive or negative.
In that case, we can not say anything about their magnitudes.
Hence Insufficient.

2. It says x-y=2

If x=5, y=3, then x-y<x+y

If x=1, y=-2, x-y>x+y

If x=-5, y=-7, x-y<x+y(absolute values)

Hence Insufficient.

1+2, we have to take x+y and x-y of same signs.
it discards the usage of X=1 and y=-2.

Hence sufficient

Kudos [?]: 7 [0], given: 35

Senior Manager
Senior Manager
User avatar
G
Joined: 06 Jul 2016
Posts: 437

Kudos [?]: 128 [0], given: 98

Location: Singapore
Concentration: Strategy, Finance
GMAT ToolKit User Premium Member CAT Tests
Re: Is |x - y| > |x + y|? [#permalink]

Show Tags

New post 26 Jul 2017, 11:10
bibha wrote:
Is |x - y| > |x + y|?

(1) x^2 - y^2 = 9
(2) x - y = 2


|x-y| > |x+y| ? This can only happen if they are opposite signs, but looking the statements, plugging in number seems to be the fastest approach.

1) \(x^2\)-\(y^2\) = 9
=> (x+y)(x-y)=9
x=5, y=4 ; x+y=9 and x-y=1 => No
x=5, y= -4 ; x=y=1 and x-y=9 => Yes

Insufficient.

2) x-y = 2
x+y = ?
Insufficient.

1+2)
(x+y)(x-y) = 9
x-y = 2
=> x+y = 4.5
Sufficient.

C is the answer.
_________________

Put in the work, and that dream score is yours!

Kudos [?]: 128 [0], given: 98

Re: Is |x - y| > |x + y|?   [#permalink] 26 Jul 2017, 11:10
Display posts from previous: Sort by

Is |x - y| > |x + y|?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.