GMAT Changed on April 16th - Read about the latest changes here

It is currently 26 Apr 2018, 00:34

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 |xy| > xy?

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

Hide Tags

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 44674
Is |xy| > xy? [#permalink]

Show Tags

New post 13 Sep 2017, 21:48
Expert's post
6
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

56% (01:00) correct 44% (01:14) wrong based on 204 sessions

HideShow timer Statistics

4 KUDOS received
PS Forum Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1061
Location: India
GPA: 3.82
GMAT ToolKit User Premium Member Reviews Badge CAT Tests
Is |xy| > xy? [#permalink]

Show Tags

New post 13 Sep 2017, 22:17
4
This post received
KUDOS
2
This post was
BOOKMARKED
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


If both \(x\) and \(y\) are positive or negative then \(|xy|=xy\) and if either of the two is negative and other is positive then \(|xy|>xy\)

Statement 1: LHS \(= \sqrt{(xy)^2}\) \(= |xy|\)
so we have \(|xy|=xy\). Hence we have a NO for our question stem. Sufficient

Statement 2: this implies magnitude of both \(x\) & \(y\) equals summation of both \(x\) & \(y\). Hence both \(x\) & \(y\) are positive. Hence \(|xy|=xy\)
Hence we have a NO for our question stem. Sufficient

Option \(D\)
Intern
Intern
avatar
Joined: 26 Sep 2017
Posts: 3
Re: Is |xy| > xy? [#permalink]

Show Tags

New post 26 Sep 2017, 22:14
niks18 wrote:
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


If both \(x\) and \(y\) are positive or negative then \(|xy|=xy\) and if either of the two is negative and other is positive then \(|xy|>xy\)

Statement 1: LHS \(= \sqrt{(xy)^2}\) \(= |xy|\)
so we have \(|xy|=xy\). Hence we have a NO for our question stem. Sufficient

Statement 2: this implies magnitude of both \(x\) & \(y\) equals summation of both \(x\) & \(y\). Hence both \(x\) & \(y\) are positive. Hence \(|xy|=xy\)
Hence we have a NO for our question stem. Sufficient

Option \(D\)



if either of the two is negative and other is positive then \(|xy|>xy\) - Can you explain this further?
1 KUDOS received
PS Forum Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1061
Location: India
GPA: 3.82
GMAT ToolKit User Premium Member Reviews Badge CAT Tests
Re: Is |xy| > xy? [#permalink]

Show Tags

New post 26 Sep 2017, 22:18
1
This post received
KUDOS
Manveetha wrote:
niks18 wrote:
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


If both \(x\) and \(y\) are positive or negative then \(|xy|=xy\) and if either of the two is negative and other is positive then \(|xy|>xy\)

Statement 1: LHS \(= \sqrt{(xy)^2}\) \(= |xy|\)
so we have \(|xy|=xy\). Hence we have a NO for our question stem. Sufficient

Statement 2: this implies magnitude of both \(x\) & \(y\) equals summation of both \(x\) & \(y\). Hence both \(x\) & \(y\) are positive. Hence \(|xy|=xy\)
Hence we have a NO for our question stem. Sufficient

Option \(D\)



if either of the two is negative and other is positive then \(|xy|>xy\) - Can you explain this further?


Hi Manveetha

Take few examples, lets assume \(x=-2\) and \(y =2\), then \(xy=-2*2=-4\)

but \(|xy|=|-4|=|4|\)

hence \(|xy|>xy\) in this case
1 KUDOS received
Senior Manager
Senior Manager
avatar
G
Status: Countdown Begins...
Joined: 03 Jul 2016
Posts: 304
Location: India
Concentration: Technology, Strategy
Schools: IIMB
GMAT 1: 580 Q48 V22
GPA: 3.7
WE: Information Technology (Consulting)
GMAT ToolKit User Reviews Badge
Re: Is |xy| > xy? [#permalink]

Show Tags

New post 27 Sep 2017, 21:54
1
This post received
KUDOS
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


Bunuel, I think if we consider that x and y both are non zero then D would be the answer. But question stem does not mention that way. Can you please explain if my assessment is wrong?
_________________

Need Kudos to unlock GMAT Club tests

Director
Director
avatar
G
Joined: 02 Sep 2016
Posts: 751
Premium Member CAT Tests
Is |xy| > xy? [#permalink]

Show Tags

New post 27 Sep 2017, 22:26
We have to just figure out whether the signs of x and y are same or different.
If they have the same signs, then the two sides would be equal. If one of the two values or both are zero, then the two sides are equal and not less than or greater than.

Both the statements answer this question: The signs are same.
_________________

Help me make my explanation better by providing a logical feedback.

If you liked the post, HIT KUDOS !!

Don't quit.............Do it.

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 44674
Re: Is |xy| > xy? [#permalink]

Show Tags

New post 27 Sep 2017, 23:09
Expert's post
1
This post was
BOOKMARKED
RMD007 wrote:
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


Bunuel, I think if we consider that x and y both are non zero then D would be the answer. But question stem does not mention that way. Can you please explain if my assessment is wrong?


Is \(|xy| > xy\)?

Notice that the question basically asks whether xy is negative. Only when xy is negative \(|xy| > xy\).

(1) \(\sqrt{x^2y^2}=xy\). This implies that |xy| = xy. So, we have a NO answer to the question. Notice that |xy| = xy means that xy is positive or 0, so not negative. Sufficient.

Or: (1) says that xy equals to the square root of some number (x^2y^2). Since the square root function cannot give negative result, then it xy must be non-negative.

(2) \(|x|+|y|=x+y\). This also implies that both x and y must be non-negative. So, again the answer to the question is NO. Sufficient.

Answer: D.
_________________

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

Expert Post
Math Revolution GMAT Instructor
User avatar
D
Joined: 16 Aug 2015
Posts: 5282
GMAT 1: 800 Q59 V59
GPA: 3.82
Re: Is |xy| > xy? [#permalink]

Show Tags

New post 01 Oct 2017, 13:59
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)



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.

The first step of VA (Variable Approach) method is modifying the original condition and the question, and rechecking the number of variables and the number of equations.

The question \(|xy| > xy\) is equivalent to \(xy < 0\).

Condition 1)
\(|xy| = xy\) is equivalent ot \(xy \ge 0\).
Since the answer is no, this condition is sufficient by CMT(Common Mistake Type) 1.

Condition 2)
\(|x| + |y| = x + y\)
=> \((|x| + |y|)^2 = (x + y)^2\)
<=> \(|x|^2 + 2|x||y| + |y|^2 = x^2 + 2xy + y^2\)
<=> \(x^2 + 2|x||y| + y^2 = x^2 + 2xy + y^2\)
<=> \(2|x||y| = 2xy\)
<=> \(|xy| = xy\)
<=> \(xy \ge 0\)
Then the answer is no and "no" is also an answer.
By CMT 1, this condition is also sufficient.

Therefore, D is the answer.
_________________

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.
"Only $79 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Intern
Intern
avatar
B
Joined: 11 Feb 2018
Posts: 18
Is |xy| > xy? [#permalink]

Show Tags

New post 06 Apr 2018, 09:49
Bunuel wrote:
RMD007 wrote:
Bunuel wrote:
Is \(|xy| > xy\)?


(1) \(\sqrt{x^2y^2}=xy\)

(2) \(|x|+|y|=x+y\)


Bunuel, I think if we consider that x and y both are non zero then D would be the answer. But question stem does not mention that way. Can you please explain if my assessment is wrong?


Is \(|xy| > xy\)?

Notice that the question basically asks whether xy is negative. Only when xy is negative \(|xy| > xy\).

(1) \(\sqrt{x^2y^2}=xy\). This implies that |xy| = xy. So, we have a NO answer to the question. Notice that |xy| = xy means that xy is positive or 0, so not negative. Sufficient.

Or: (1) says that xy equals to the square root of some number (x^2y^2). Since the square root function cannot give negative result, then it xy must be non-negative.

(2) \(|x|+|y|=x+y\). This also implies that both x and y must be non-negative. So, again the answer to the question is NO. Sufficient.

Answer: D.


Hi Bunuel.

I understood Statement-2 but confused with statement-1, my logic is as below

From question we can figure out that if xy>0, then lXYl= XY ; so XY>XY so we have to neglect this option
if xy<0, then l XY l = -(xy), So question becomes Is -(xy)>xy, this is what we have to look out from the question set.

So, -(xy)>xy is possible when we have different signs of x and Y (one + and another -); then we can definitely say "Yes" to the question Is -(xy)>xy
If X and Y have same signs then -(xy)<xy ; then a definite "NO" (to our question which is -(xy)>xy)
now in statement-1

X and Y both can be positive/negative; or X negative and Y positive (or vice versa) as the square will negate the effect of positivity/negativity of X/Y
So we can't be sure from statement 1 whether X/Y have same or different signs. Hence insufficient

I hope you understood my thinking and plz help to correct where me going wrong with my understanding
Thanks
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 44674
Re: Is |xy| > xy? [#permalink]

Show Tags

New post 06 Apr 2018, 12:22
cruiseav wrote:
Bunuel wrote:
Is \(|xy| > xy\)?

Notice that the question basically asks whether xy is negative. Only when xy is negative \(|xy| > xy\).

(1) \(\sqrt{x^2y^2}=xy\). This implies that |xy| = xy. So, we have a NO answer to the question. Notice that |xy| = xy means that xy is positive or 0, so not negative. Sufficient.

Or: (1) says that xy equals to the square root of some number (x^2y^2). Since the square root function cannot give negative result, then it xy must be non-negative.

(2) \(|x|+|y|=x+y\). This also implies that both x and y must be non-negative. So, again the answer to the question is NO. Sufficient.

Answer: D.


Hi Bunuel.

I understood Statement-2 but confused with statement-1, my logic is as below

From question we can figure out that if xy>0, then lXYl= XY ; so XY>XY so we have to neglect this option
if xy<0, then l XY l = -(xy), So question becomes Is -(xy)>xy, this is what we have to look out from the question set.

So, -(xy)>xy is possible when we have different signs of x and Y (one + and another -); then we can definitely say "Yes" to the question Is -(xy)>xy
If X and Y have same signs then -(xy)<xy ; then a definite "NO" (to our question which is -(xy)>xy)
now in statement-1

X and Y both can be positive/negative; or X negative and Y positive (or vice versa) as the square will negate the effect of positivity/negativity of X/Y
So we can't be sure from statement 1 whether X/Y have same or different signs. Hence insufficient

I hope you understood my thinking and plz help to correct where me going wrong with my understanding
Thanks


For \(\sqrt{x^2y^2}=xy\) to be true x and y cannot have different signs because if they do, then the right hand side would become negative: \(xy=negative\) and it cannot equal to the square root of any number (\(\sqrt{x^2y^2}\)), which is always positive or 0.
_________________

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

Re: Is |xy| > xy?   [#permalink] 06 Apr 2018, 12:22
Display posts from previous: Sort by

Is |xy| > xy?

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


cron

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®.