GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 26 Feb 2020, 09:30

### 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

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

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 61508

### Show Tags

09 Jul 2012, 03:49
7
61
00:00

Difficulty:

45% (medium)

Question Stats:

66% (01:49) correct 34% (01:41) wrong based on 1839 sessions

### HideShow timer Statistics

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

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

Diagnostic Test
Question: 41
Page: 26
Difficulty: 650

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 61508
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

09 Jul 2012, 03:49
12
23
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.

_________________
Intern
Joined: 05 Mar 2012
Posts: 40
Schools: Tepper '15 (WL)
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

10 Jul 2012, 07:52
8
1
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
##### General Discussion
Senior Manager
Joined: 28 Mar 2012
Posts: 292
Location: India
GMAT 1: 640 Q50 V26
GMAT 2: 660 Q50 V28
GMAT 3: 730 Q50 V38
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

09 Jul 2012, 04:38
1
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.

Regards,
Board of Directors
Joined: 01 Sep 2010
Posts: 3393
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

10 Jul 2012, 13:46
For me is E the solution.

I wait for an explanation by Bunuel to understand if something went wrong in my reasoning
_________________
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 892
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

10 Jul 2012, 15:28
4
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

x -y < 0
x+y <0
--------
2x <0
So,
$$x<0$$

_________________
"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
Intern
Joined: 11 Jul 2012
Posts: 47
GMAT 1: 650 Q49 V29
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

11 Jul 2012, 09:30
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
Senior Manager
Joined: 06 Aug 2011
Posts: 316
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

11 Jul 2012, 10:28
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 !
Current Student
Joined: 10 Mar 2013
Posts: 456
Location: Germany
Concentration: Finance, Entrepreneurship
Schools: WHU MBA"20 (A$) GMAT 1: 580 Q46 V24 GPA: 3.88 WE: Information Technology (Consulting) Re: If (x+y)/z>0, is x<0? [#permalink] ### Show Tags 23 Jun 2015, 12:14 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 Intern Joined: 10 Oct 2015 Posts: 1 Re: If (x+y)/z>0, is x<0? [#permalink] ### Show Tags 10 Oct 2015, 03:20 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! Math Expert Joined: 02 Sep 2009 Posts: 61508 Re: If (x+y)/z>0, is x<0? [#permalink] ### Show Tags 11 Oct 2015, 04:28 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. _________________ Retired Moderator Joined: 29 Oct 2013 Posts: 248 Concentration: Finance GPA: 3.7 WE: Corporate Finance (Retail Banking) Re: If (x+y)/z>0, is x<0? [#permalink] ### Show Tags 06 Dec 2015, 12:28 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8601 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If (x+y)/z>0, is x<0? [#permalink] ### Show Tags 09 Dec 2015, 06:14 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. "Only$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4355

### Show Tags

Updated on: 17 Apr 2018, 07:05
1
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

Cheers,
Brent
_________________
Test confidently with gmatprepnow.com

Originally posted by GMATPrepNow on 03 Oct 2017, 19:14.
Last edited by GMATPrepNow on 17 Apr 2018, 07:05, edited 1 time in total.
Manager
Joined: 03 May 2017
Posts: 82
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

04 Oct 2017, 21:44
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
Intern
Joined: 20 Jun 2017
Posts: 6
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

29 Oct 2017, 04:34
Sir in the 2nd statement i am having doubt. How we can rewrite Z<0 as X+Y <0. Please Explain
Math Expert
Joined: 02 Sep 2009
Posts: 61508
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

29 Oct 2017, 04:39
Raj94* wrote:
Sir in the 2nd statement i am having doubt. How we can rewrite Z<0 as X+Y <0. Please Explain

$$\frac{x+y}{z}>0$$, means that x + y and z have the same sign, either both are positive or both are negative. (2) says that z is positive, so x + y must also be positive.

Hope it's clear.
_________________
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 16153
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

09 Dec 2017, 16:20
1
Hi All,

We're told that (X+Y)/Z is greater than 0. We're asked if X is less than 0. This is a YES/NO question. We can answer it with a mix of Number Properties and/or TESTing VALUES.

To start, for (X+Y)/Z to be greater than 0, one of two options must occur:
(X+Y) > 0 and Z > 0
(X+Y) < 0 and Z < 0

1) X < Y
IF....
X = 1, Y = 2, Z = 1, then the answer to the question is NO.
X = -2, Y = -1, Z = -1, then the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) Z < 0
IF...
X = -2, Y = -1, Z = -1, then the answer to the question is YES.
X = 1, Y = -2, Z = -1, then the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, we know:
X < Y
Z < 0

Since Z is NEGATIVE, then we know that (X+Y) must ALSO be negative. Since X < Y, we know that either just X or both X and Y must be NEGATIVE. Either way, the answer to the question is ALWAYS YES.
Combined, SUFFICIENT.

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Intern
Joined: 28 Feb 2015
Posts: 25
GMAT 1: 720 Q50 V37
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

11 Nov 2018, 16:35
1) not suf
2) not suf
Together : when z < o , x+y must < 0 since (1) x< y ..x must <0 ..otherwise x+y > 0 ( x>=0,y>0)
So suf
C

Posted from my mobile device
Non-Human User
Joined: 09 Sep 2013
Posts: 14143
Re: If (x+y)/z>0, is x<0?  [#permalink]

### Show Tags

06 Dec 2019, 04:52
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.
_________________
Re: If (x+y)/z>0, is x<0?   [#permalink] 06 Dec 2019, 04:52
Display posts from previous: Sort by