GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Jul 2018, 08:33

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

If y=|x+1|/x and x!=0, is xy>0?

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

Hide Tags

8 KUDOS received
VP
VP
User avatar
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1098
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
GMAT ToolKit User
If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post Updated on: 12 Jul 2014, 12:43
8
11
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

40% (01:13) correct 60% (01:35) wrong based on 262 sessions

HideShow timer Statistics

If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

(1) \(x^2+2x+1>0\)

(2) \(y\neq{0}\)

My own question, as always any feedback is appreciated
Click here for the OE

_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]


Originally posted by Zarrolou on 16 May 2013, 11:24.
Last edited by Bunuel on 12 Jul 2014, 12:43, edited 3 times in total.
Added OA and OE
Most Helpful Community Reply
8 KUDOS received
Director
Director
User avatar
G
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 613
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 16 May 2013, 11:47
8
2
Zarrolou wrote:
If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

A)\(x^2+2x+1>0\)

B)\(y\neq{0}\)

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!


I think Answer is D
question can be written as |x+1| = xy
so, to prove that xy> 0 we essentially need to prove that |x|1| > 0

STAT1
x^2 + 2x + 1 > 0
means,
(x+1) ^ 2 >0
taking square root on both the sides does not change the inequality
so, we have
|x+1| > 0
which means that xy > 0
So, SUFFICIENT

STAT2
y != 0
we know that
|x+1| = xy => xy is non negative. so, it is either 0 or positive
since we know that both x and y ar enot equal to zero so, xy > 0
so, SUFFICIENT

Hence, Answer will be D
_________________

Ankit

Check my Tutoring Site -> Brush My Quant

GMAT Quant Tutor
How to start GMAT preparations?
How to Improve Quant Score?
Gmatclub Topic Tags
Check out my GMAT debrief

How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities

General Discussion
1 KUDOS received
Intern
Intern
avatar
Joined: 15 Oct 2012
Posts: 23
GMAT ToolKit User
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post Updated on: 16 May 2013, 11:59
1
E it is.....................
For getting xy>0 we need to prove that both x and y are either greater than 0 I.e positive or both of them are negative......

Xy = |x+1|

Once negative:
Xy= -x-1
Xy+x=-1
X=-1/y+1

Once positive: similarly we will get x= 1/y-1

We are getting x once positive and negative ......so can't say xy >0 or not....

II
X^2+2x+1>0
Gives us (x+1)^2>0
Gives us x>-1 and from question stem we know that x is not equal to 0 - hence x will be positive.
But we still don't know their value/ sign of y. Hence this statement is also not sufficient.

Hope I'm correct.....

Originally posted by nikhilsehgal on 16 May 2013, 11:41.
Last edited by nikhilsehgal on 16 May 2013, 11:59, edited 1 time in total.
2 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 430
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 16 May 2013, 12:23
2
Agreed

I made the mistake of multiplying y by x and having to test four different cases:

yx = x+1
-yx = x+1

Then I had to test cases for X>0 and X<0 (i.e. four total cases) and each one turned out to hold true. Your way makes more sense and is quicker to boot.

for 2.) xy = |x+1| so xy could be 0 or greater than zero. We know that X isn't zero from the stem and we know that Y isn't zero from #2. Therefore x/y must be greater or less than zero yielding a positive or negative result. However, because xy = the absolute value of x+1 xy can only be greater than or equil to zero and because both x and y are not zero, product zy MUST be greater than zero.

Good work!

nktdotgupta wrote:
Zarrolou wrote:
If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

A)\(x^2+2x+1>0\)

B)\(y\neq{0}\)

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!


I think Answer is D
question can be written as |x+1| = xy
so, to prove that xy> 0 we essentially need to prove that |x|1| > 0

STAT1
x^2 + 2x + 1 > 0
means,
(x+1) ^ 2 >0
taking square root on both the sides does not change the inequality
so, we have
|x+1| > 0
which means that xy > 0
So, SUFFICIENT

STAT2
y != 0
we know that
|x+1| = xy => xy is non negative. so, it is either 0 or positive
since we know that both x and y ar enot equal to zero so, xy > 0
so, SUFFICIENT

Hence, Answer will be D
3 KUDOS received
Manager
Manager
User avatar
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 122
Location: India
Concentration: General Management, Technology
GMAT 1: 640 Q42 V37
GPA: 3.5
WE: Web Development (Computer Software)
GMAT ToolKit User
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 16 May 2013, 13:23
3
Zarrolou wrote:
If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

A)\(x^2+2x+1>0\)

B)\(y\neq{0}\)

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!


The answer sure does appear to be [D] :)

Statement 1: \(x^2+2x+1>0\)
Hence \((x+1)^2\) > 0

Now since the above is a squared term, the value will always be positive except for x = -1. Since the statement doesnt include 0 under the range we can assume that x is not equal to -1. Putting the same values into our main equation i.e. xy = |x+1| we, can be sure that x+1 is not equal to zero at any point. Hence the statement would be sufficient, since if we prove |x+1| is not equal to zero, then its always greater than 0! Hence xy = |x+1| > 0

Statement 2. Now if y is not equal to zero, we can assume |x+1| is never equal to zero! Hence we prove the fact by the same above method, that xy = |x+1| > 0

The idea in the above question, is to establish that at no point x = -1. At only that value y = 0 and xy = 0. Otherwise for all other possible values |x+1| > 0.

As always wonderful question Zarrolou! :) Hope my procedure is correct! :)

Regards,
Arpan
_________________

Feed me some KUDOS! :) :) *always hungry*

My Thread : Recommendation Letters

2 KUDOS received
VP
VP
User avatar
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1098
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
GMAT ToolKit User
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 16 May 2013, 13:46
2
1
Good job WholeLottaLove, nktdotgupta, arpanpatnaik !

Official explanation

If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

Rewrite the question as (multiply by x) \(xy=|x+1|\)

The \(|abs|\) is \(\geq{0}\), so basically we have to check if \(xy=|x+1|\neq{0}\)

A)\(x^2+2x+1>0\)
\((x+1)^2>0\)
So \(x\neq{-1}\) and \(|x+1|>0\)

This is what we are looking for. Sufficient

B)\(y\neq{0}\)

\(y=\frac{|x+1|}{x}\neq{0}\) so \(|x+1|\neq{0}\)

Sufficient

The correct answer is D

Hope everything is clear
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Expert Post
1 KUDOS received
MBA Section Director
User avatar
D
Status: Back to work...
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 5182
Location: India
City: Pune
GMAT 1: 680 Q49 V34
GPA: 3.4
WE: Business Development (Manufacturing)
GMAT ToolKit User Premium Member
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 16 May 2013, 15:41
1
1
@Zarrolou,
Regret for not responding within time.


If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

A)\(x^2+2x+1>0\)

B)\(y\neq{0}\)



First lets simplify the inequality

\(\frac{|x+1|}{x}\) –y = 0 ----------> \(\frac{|x+1|-xy}{x}\) = 0 -----> we know \(x\neq{}0\), then |x+1|-xy must be zero. Hence |x+1| - xy = 0 --------> |x+1| = xy


We are asked whether xy>0 --------> Whether |x+1| > 0 ? --------> We know the expression within modules can either be zero or greater than zero. For xy to be greater than zero |x+1| has to be greater than zero.
|x+1| will be zero only when x=-1 and for any other value of x, |x+1| will always be greater than zero
So the question can be rephrased as whether \(x\neq{-1}\)


Statement 1) \(x^2\) + 2x + 1 > 0

This is an quadratic inequality.
Rule :- For any quadratic inequation \(ax^2\) + bx + c > 0, if \(b^2\) – 4ac = 0 and a > 0 then the inequality holds true outside the interval of roots
In our case \(b^2\) – 4ac = 4 – 4 = 0 and a > 0 so \(x^2\) + 2X + 1 > 0 will hold true for all values beyond the Root(s) of equation (Towards any direction - Positive or Negative)
\(x^2\) + 2x + 1 = 0 --------> x(x+1) +1(x + 1) = 0 ---------> (x+1)(x+1) = 0 ----------> x=Root = -1
So \(x^2\) + 2x + 1 > 0 will hold true for any of x except for -1
This reveals that \(x\neq{-1}\) and xy>0 ----------------> Sufficient


Statement 2) \(y\neq{0}\)

From the question stem we know \(x\neq{}0\)
As per Statement 2, \(y\neq{0}\)-------------->That means both X and Y are nonzero.

|x+1| = xy
xy can be either Positive or Negative
|x+1| can be Zero or Positive
Combining both these inferences we can conclude that XY must be Positive. Sufficient

Answer = D

Regards,

Narenn

_________________

Chances of Getting Admitted After an Interview [Data Crunch]


Must Read Forum Topics Before You Kick Off Your MBA Application

New GMAT Club Decision Tracker - Real Time Decision Updates

SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1869
Concentration: Finance
GMAT ToolKit User
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 24 May 2014, 11:22
Lets make things simpler here.

If y=|x+1|/x, Is xy>0?

Now since |x+1| is always non negative then, xy>0 always except when x+1=0---> x=-1.

Thus, we need to prove that x is not equal to -1.

Statement 1.

We have that (x+1)^2>0

Therefore, we have that |x+1|>0

x+1>0---> x>-1 or x+1<0, x<-1. Therefore, in both cases x is different from -1.

Sufficient

Statement 2

If y is different from zero it means that x is also different from -1, since only the numerator can give 0. Remember x can't be zero in this case.

Sufficient

Answer: D

Hope this clarifies
Cheers!
J :)
Senior Manager
Senior Manager
User avatar
G
Joined: 18 Jun 2016
Posts: 270
Location: India
GMAT 1: 720 Q50 V38
GMAT 2: 750 Q49 V42
GPA: 4
WE: General Management (Other)
GMAT ToolKit User Reviews Badge
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 08 Sep 2016, 03:56
BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdr

Are you guys sure that Answer is D? because your own Solutions are Contradicting your answers.

Question: Is xy > 0
Given: \(x\neq{}0\)
\(y=\frac{|x+1|}{x}\)

|x+1| will always be Non-Negative but it can STILL BE 0.
Similarly, y CAN ALSO BE 0.


Statement 1: \(x^2+2x+1>0\)
=> \((x+1)^2 > 0\)
=> x > -1

There is no constraint on y.
So,
Case 1: If y = 1, x > 0;xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 2: y = -1, x = -0.5; xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 3: y = 0, x > -1; xy = 0 i.e. xy !> 0

Statement 1: Insufficient

Statement 2: \(y\neq{0}\)

Case 1: If y = 1, x > 0;xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 2: y = -1, x = -0.5; xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 3: y = -100, x = -1; xy = |x+1| = 0 i.e. y = 0 (because if x = -1 and xy = 0, y = 0). But this case violates the condition given in Statement 2 that \(y\neq{0}\).
Therefore, \(x\neq{-1}\) and as given \(y\neq{0}\)

Hence, xy > 0 (Always)

Statement 2: Sufficient
Answer: B
_________________

I'd appreciate learning about the grammatical errors in my posts

Please hit Kudos If my Solution helps

My Debrief for 750 - https://gmatclub.com/forum/from-720-to-750-one-of-the-most-difficult-pleatues-to-overcome-246420.html

My CR notes - https://gmatclub.com/forum/patterns-in-cr-questions-243450.html

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47037
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 08 Sep 2016, 04:11
umg wrote:
BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdr

Are you guys sure that Answer is D? because your own Solutions are Contradicting your answers.

Question: Is xy > 0
Given: \(x\neq{}0\)
\(y=\frac{|x+1|}{x}\)

|x+1| will always be Non-Negative but it can STILL BE 0.
Similarly, y CAN ALSO BE 0.


Statement 1: \(x^2+2x+1>0\)
=> \((x+1)^2 > 0\)
=> x > -1

There is no constraint on y.
So,
Case 1: If y = 1, x > 0;xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 2: y = -1, x = -0.5; xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 3: y = 0, x > -1; xy = 0 i.e. xy !> 0

Statement 1: Insufficient

Statement 2: \(y\neq{0}\)

Case 1: If y = 1, x > 0;xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 2: y = -1, x = -0.5; xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 3: y = -100, x = -1; xy = |x+1| = 0 i.e. y = 0 (because if x = -1 and xy = 0, y = 0). But this case violates the condition given in Statement 2 that \(y\neq{0}\).
Therefore, \(x\neq{-1}\) and as given \(y\neq{0}\)

Hence, xy > 0 (Always)

Statement 2: Sufficient
Answer: B


Notice that x^2+2x+1>0 is true for all value of x except when x = -1. So, for (1) \(x\neq -1\) and if \(x\neq -1\), then \(y\neq 0\). Your third case is not possible.
_________________

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

Senior Manager
Senior Manager
User avatar
G
Joined: 18 Jun 2016
Posts: 270
Location: India
GMAT 1: 720 Q50 V38
GMAT 2: 750 Q49 V42
GPA: 4
WE: General Management (Other)
GMAT ToolKit User Reviews Badge
If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 08 Sep 2016, 04:51
Bunuel wrote:
umg wrote:
BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdr

Are you guys sure that Answer is D? because your own Solutions are Contradicting your answers.

Question: Is xy > 0
Given: \(x\neq{}0\)
\(y=\frac{|x+1|}{x}\)

|x+1| will always be Non-Negative but it can STILL BE 0.
Similarly, y CAN ALSO BE 0.


Statement 1: \(x^2+2x+1>0\)
=> \((x+1)^2 > 0\)
=> x > -1

There is no constraint on y.
So,
Case 1: If y = 1, x > 0;xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 2: y = -1, x = -0.5; xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 3: y = 0, x > -1; xy = 0 i.e. xy !> 0

Statement 1: Insufficient

Statement 2: \(y\neq{0}\)

Case 1: If y = 1, x > 0;xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 2: y = -1, x = -0.5; xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 3: y = -100, x = -1; xy = |x+1| = 0 i.e. y = 0 (because if x = -1 and xy = 0, y = 0). But this case violates the condition given in Statement 2 that \(y\neq{0}\).
Therefore, \(x\neq{-1}\) and as given \(y\neq{0}\)

Hence, xy > 0 (Always)

Statement 2: Sufficient
Answer: B


Notice that x^2+2x+1>0 is true for all value of x except when x = -1. So, for (1) \(x\neq -1\) and if \(x\neq -1\), then \(y\neq 0\). Your third case is not possible.

:wow Wow! This question is crazier than I anticipated.

I see it now. I used that logic in Statement 2 but not in Statement 1. :wall

Thanks for pointing it out.
_________________

I'd appreciate learning about the grammatical errors in my posts

Please hit Kudos If my Solution helps

My Debrief for 750 - https://gmatclub.com/forum/from-720-to-750-one-of-the-most-difficult-pleatues-to-overcome-246420.html

My CR notes - https://gmatclub.com/forum/patterns-in-cr-questions-243450.html

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 7273
Premium Member
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

Show Tags

New post 17 Sep 2017, 20:58
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: If y=|x+1|/x and x!=0, is xy>0?   [#permalink] 17 Sep 2017, 20:58
Display posts from previous: Sort by

If y=|x+1|/x and x!=0, is xy>0?

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

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| 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®.