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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

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

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

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 !

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.

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.
_________________

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

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.
_________________

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

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).

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...