It is currently 23 Oct 2017, 10:59

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

If (x+y)/z>0, is x<0?

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

Hide Tags

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129477 [1], given: 12201

If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 09 Jul 2012, 04:49
1
This post received
KUDOS
Expert's post
34
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

63% (01:26) correct 37% (01:17) wrong based on 1041 sessions

HideShow timer Statistics

Kudos [?]: 129477 [1], given: 12201

Expert Post
2 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129477 [2], given: 12201

Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 09 Jul 2012, 04:49
2
This post received
KUDOS
Expert's post
17
This post was
BOOKMARKED
SOLUTION

If \(\frac{x+y}{z}>0\), is x<0?

Noticee that: \(\frac{x+y}{z}>0\) means that \(x+y\) and \(z\) have the same sign: either both are positive or both are negative.

(1) x < y. No info about \(z\). Not sufficient.
(2) z < 0. This statement implies that \(x+y\) must also be negative: \(x+y<0\). But we cannot say whether \(x<0\). Not sufficient.

(1)+(2) From (1) we have that \(x < y\) and from (2) we have that \(x+y<0\). Sum these two inequalities (remember we can add inequalities with the sign in the same direction): \(x+y+x<y\) --> \(2x<0\) --> \(x<0\). Sufficient.

Answer: C.
_________________

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 [?]: 129477 [2], given: 12201

Kellogg MMM ThreadMaster
User avatar
B
Joined: 29 Mar 2012
Posts: 321

Kudos [?]: 515 [0], given: 23

Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
GMAT ToolKit User
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 09 Jul 2012, 05:38
Bunuel wrote:
If \(\frac{x+y}{z}>0\), is x<0?

(1) x < y
(2) z < 0

Hi,

Difficulty: 650

\(\frac{x+y}{z}>0\)

Using (1),
x < y,
Possible values, x= -2, y = -1 & z = -1
or x= 1, y = 2 & z = 1
Both satisfy the given condition, thus Insufficient.

Using (2),
z < 0
or x + y < 0,
Possible values, x = -1, y = -1 or x = 1, y = -2. Insufficient.

Using both the statements,
z < 0 & x < y, we have x + y < 0
if y=2, x < -2 for x+y < 0 to hold true.
or if y=-2, x<-2. Sufficient.

Answer (C)

Regards,

Kudos [?]: 515 [0], given: 23

4 KUDOS received
Manager
Manager
avatar
Joined: 05 Mar 2012
Posts: 62

Kudos [?]: 11 [4], given: 7

Schools: Tepper '15 (WL)
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 10 Jul 2012, 08:52
4
This post received
KUDOS
If (x+y)/z is greater than 0, then it is basically saying that either both z and (x+y) are negative or they are both positive. Question asks if x is negative.

Statement 1 says that x is less than y but does not say anything about z. 3 unknown variables, 2 equations, insufficient data.

Statement 2 says that z is negative. The result is that (x+y) is negative then, but there are a lot of possibilities for x+y to be negative.

If we were to combine the two, then we know that x+y is negative with x being less than y. No matter what combination of numbers we put together, x can never be positive. If y is negative and x < y, x is negative. If y is positive then x must be negative to the point that x + y is still negative.

C.

EDIT: people might not like my explanation since I was taught to simplify or reword questions. Might not be for everyone :(

Kudos [?]: 11 [4], given: 7

Moderator
Moderator
User avatar
G
Joined: 01 Sep 2010
Posts: 3355

Kudos [?]: 9093 [0], given: 1155

Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 10 Jul 2012, 14:46

Kudos [?]: 9093 [0], given: 1155

4 KUDOS received
Retired Moderator
User avatar
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1633

Kudos [?]: 1107 [4], given: 109

Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 10 Jul 2012, 16:28
4
This post received
KUDOS
Bunuel wrote:


If \(\frac{x+y}{z}>0\), then dividend and divisor must be or positive or negative at the same time.
Let's analyze the clues:
(1) x < y
We don't know whether z is positive or negative. We cannot know whether x is positive or negarive.

(2) z<0
Then x + y <0. However, with this info we cannot claim that x<0.

Combining (1) and (2):
(1) x<y ----> x-y<0
Also, based on (2), we know that x+y<0

Adding the inequalities:
x -y < 0
x+y <0
--------
2x <0
So,
\(x<0\)

C is the answer.
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 1107 [4], given: 109

Manager
Manager
avatar
B
Joined: 11 Jul 2012
Posts: 60

Kudos [?]: 49 [0], given: 25

GMAT 1: 650 Q49 V29
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 11 Jul 2012, 10:30
if x+y/z > 0, is x<0?

(1) x < y
(2) z < 0

Statement 1: if x<y, then necessarily x-y < 0---(1), substitute, x = +ve then always y will have to be a +ve value exceeding x to satisfy the equ.
substitute x = -ve then either y will be a -ve value OR
y will be a +ve value so that |x| > |y|
substitute x = 0, y will always be a +ve value
Statement 2: if z < 0, then necessarily x+y < 0--(2) substitute x = + ve then y will be -ve so that |x| < |y|
substitute x = -ve then y can be +ve if |x| > |y| OR
y wll be -ve
substitute x = 0, y will always be a -ve value

Individually nothing can be concluded but evaluating (1) and (2) together we can see that when x < 0 only then can the given equ. be satisfied,
Hence C.

I have a question can we go ahead and solve two inequalities like
x - y < 0
x + y < 0
------------
2x < 0
x < 0

I tried to substitute some values in equations but it didn't fit..
like if we have 2x + 4y < 0 and 2x - 4y < 0, then we cannot conclusively say anything about the value of x or y

Please comment and provide an explanation

Kudos [?]: 49 [0], given: 25

Senior Manager
Senior Manager
avatar
Joined: 06 Aug 2011
Posts: 390

Kudos [?]: 235 [0], given: 82

Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 11 Jul 2012, 11:28
Its C...

statement 1.. if x+y/z>0 that mean numerator or demonitor both have same sign...

x<y..that means x cud be +ve or negative and no any information abt z..so insufficient..

statement 2: .. z<0 .. no any information abt x and y..

state1 and stat 2: .. if x is less than y that means x will be negative and y will be positive or both will be negative.. so both are sufficient ..


ans c..
_________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

Kudos [?]: 235 [0], given: 82

Director
Director
User avatar
Joined: 10 Mar 2013
Posts: 592

Kudos [?]: 463 [0], given: 200

Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
GMAT ToolKit User
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 23 Jun 2015, 13:14
Here's an approach with algebra
Numerator and denominator of this expression must have same signs to be positive

(x + y) z
+ +
- -
St1
x<y Not sufficient - no info about z

St2
z<0 , ok, this means x+y<0 Not sufficient (let's say x=2, y=-5 NO or x=-, y=2 YES)

St1+St2
SO we have overall this information: z<0, x+y<0, x<y
In order x+y<0 we have 3 Scenarios: a) x and y both negative b) x +,y - and c)y +, x -
We have here scenario c) x<y --> x is definitely negative
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Kudos [?]: 463 [0], given: 200

Intern
Intern
avatar
Joined: 10 Oct 2015
Posts: 1

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

Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 10 Oct 2015, 04:20
Hi guys! While doing this question I was wondering, if the answer we'd get combining the 2 options gave us x>0, would C still be the correct answer? I'm just trying to understand whether data sufficiency in the gmat is confirming the information in the question or also can be used as a negative answer, yet sufficient to give it.

Hope you understand what I mean and can help me out with this.

Many thanks and good luck with your preparation!

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

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 41913

Kudos [?]: 129477 [0], given: 12201

Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 11 Oct 2015, 05:28
Nuktrue wrote:
Hi guys! While doing this question I was wondering, if the answer we'd get combining the 2 options gave us x>0, would C still be the correct answer? I'm just trying to understand whether data sufficiency in the gmat is confirming the information in the question or also can be used as a negative answer, yet sufficient to give it.

Hope you understand what I mean and can help me out with this.

Many thanks and good luck with your preparation!


A definite NO answer to the question is still considered to be sufficient.
_________________

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 [?]: 129477 [0], given: 12201

Senior Manager
Senior Manager
avatar
Joined: 29 Oct 2013
Posts: 294

Kudos [?]: 479 [0], given: 197

Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)
GMAT ToolKit User
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 06 Dec 2015, 13:28
Or another way to look at 1+2:
St2-->when we know x+y<0 both x and y cant be positive. Possible scenarios:
1) Both are negative: -->Yes
2) x<0 & y>0 ;|x|>|y|-->No
3) x>0 & y<0; |x|<|y|--> Yes
Since st1 tells us that x<y scenario3 is not possible

Ans: C
_________________

Please contact me for super inexpensive quality private tutoring

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

Kudos [?]: 479 [0], given: 197

Expert Post
Math Revolution GMAT Instructor
User avatar
P
Joined: 16 Aug 2015
Posts: 4155

Kudos [?]: 2923 [0], given: 0

GPA: 3.82
Premium Member CAT Tests
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 09 Dec 2015, 07:14
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.

If x+yz >0 , is x<0?

(1) x < y
(2) z < 0

There are 3 variables (x,y,z) and 1 equation ((x+y)/z>0) in the original condition, and there are 2 more equations given from the 2 conditions, so there is high chance of (C) becoming the answer.
Looking at the conditions together,
as z<0, x+y<0, x<-y. But as x<y, if both sides are added together,
x+x<y-y=0, 2x<0, x<0 this answers the question 'yes' and the conditions become sufficient, making the answer become (C).

For cases where we need 2 more equations, 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 [?]: 2923 [0], given: 0

Expert Post
Top Contributor
SVP
SVP
User avatar
G
Joined: 12 Sep 2015
Posts: 1798

Kudos [?]: 2476 [0], given: 357

Location: Canada
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 03 Oct 2017, 20:14
Expert's post
Top Contributor
Bunuel wrote:
If \(\frac{x+y}{z}>0\), is x<0?

(1) x < y
(2) z < 0


Target question: Is x NEGATIVE?

Given: (x+y)/z > 0
What does tell us?
Not much.
It tells us that (x+y)/z is POSITIVE, which means EITHER (x+y) and z are both positive OR (x+y) and z are both negative

Statement 1: x < y
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of x, y and z that satisfy statement 1 AND the given information. Here are two:
Case a: x = -2, y = -1, and z = -1. In this case x is NEGATIVE
Case b: x = 1, y = 2, and z = 1. In this case x is POSITIVE
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: z < 0
There are several values of x, y and z that satisfy statement 2 AND the given information. Here are two:
Case a: x = -2, y = -1, and z = -1. In this case x is NEGATIVE
Case b: x = 1, y = -2, and z = -1. In this case x is POSITIVE
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 tells us that z is NEGATIVE
If z is NEGATIVE and (x+y)/z > 0, then it must be the case that (x+y) is also NEGATIVE
In other words, x + y < 0

Statement 1 tells us that x < y
If we subtract y from both sides of the inequality, we get: x - y < 0

So, we now have the following two inequalities:
x + y < 0
x - y < 0

When we ADD the two inequalities, we get: 2x < 0
Divide both sides by 2 to get: x < 0
In other words, x is NEGATIVE
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer:
[Reveal] Spoiler:
C


RELATED VIDEOS FROM OUR COURSE



_________________

Brent Hanneson – Founder of gmatprepnow.com

Image

Kudos [?]: 2476 [0], given: 357

Manager
Manager
User avatar
B
Joined: 03 May 2017
Posts: 95

Kudos [?]: 15 [0], given: 13

CAT Tests
Re: If (x+y)/z>0, is x<0? [#permalink]

Show Tags

New post 04 Oct 2017, 22:44
Bunuel wrote:
If \(\frac{x+y}{z}>0\), is x<0?

(1) x < y
(2) z < 0

Diagnostic Test
Question: 41
Page: 26
Difficulty: 650



both statements are insufficient alone as the sign of x depends on those of y and z.
Combine both and they are sufficient.
i.e in (2), z<0, z is negative, then x+y must be negative for (x+y)/z to be positive and this is true from (1) x<y (x must be negative for x+y to be negative).

Hence C

Kudos [?]: 15 [0], given: 13

Re: If (x+y)/z>0, is x<0?   [#permalink] 04 Oct 2017, 22:44
Display posts from previous: Sort by

If (x+y)/z>0, is x<0?

  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®.