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:

1) x and y both are positive or negative, so x could be -ve, or +ve insuff. 2) x and z both are positive or negative, so x could be -ve, or +ve insuff.

For C as xyz > 0, so if y and z both -ve, then x has to be +ve, and if y and z become +ve then x has to be +ve to become xyz > 0. _________________

Following chart lists down the possible scenarios in which XYZ > 0

X Y Z 1) - - + 2) - + - 3) + - - 4) + + +

As per statement I --> 1 & 4 have to be true for XY > 0. So I alone is insufficient as X can be +ve or -ve As per statement II --> 2& 4 have to be true for XZ > 0 So II alone is insufficient as X can be +ve or -ve If we combine both statements then only 4 is true and X has to be +ve

Obviously the answer is Yes or No; Get your kudos for reasoning out why the answer is C and keep an eye on time too

Triple kudos for someone suggesting quicker technique for DS stems with triple variables...!

Certainly each statement is not sufficient alone (the logic is identical for each: for S1, x and y can both be negative or both positive, and for S2, x and z can both be negative or both positive). You might notice that S1 tells us z is positive, and S2 tells us y is positive, but that's not what we're looking for.

Now, together, the inequality xy > 0 means 'x and y have the same sign'. So from the two statements we know x and y have the same sign, and x and z have the same sign, so x, y and z all have the same sign. If xyz > 0, they must all be positive, and the answer is C. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

C. For XYZ > 0 either all are positive or two of them negative. From St1 and St2 it is clear that both xy and xz has to be positive or both xz and xy has to be negative. in the latter case XYZ > 0 will be false.

Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]

Show Tags

18 Nov 2013, 08:29

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 xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]

Show Tags

26 Dec 2014, 10:21

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 xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]

Show Tags

09 Jan 2016, 21:53

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 xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]

Show Tags

11 Feb 2016, 10:38

This one is pure fun!

xyz>0 this means that we need two number with same sign and the Notre number must be positive. otherwise we would get a negative number.

1-xy>0 Both x and y are positive or negative. NS

2-xz>0 Both x and y are positive or negative. NS

1 and 2 - xz>0 and xy>0 - SUFFICIENT Lets assume that x is negative: if x is negative, then y must be negative and z must be negative. This would result in a negative number. This is not possible. So we are sure that X is positive.

Re: If xyz > 0, is x > 0? (1) xy > 0 (2) xz > 0 [#permalink]

Show Tags

12 Feb 2016, 05:57

Expert's post

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 xyz > 0, is x > 0?

(1) xy > 0 (2) xz > 0

When you modify the original condition and the question, they become xyz>0 -> x>0? -> yz>0?. There are 3 variables(x,y,z) and 1 equation(xyz>0), which should match with the number of equations. So, you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. When 1) & 2), they become x(y^2)z>0. When dividing x(y^2)z>0 with y^2, it becomes yz>0, which is yes and sufficient. Therefore, the answer is C.

l 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. _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...