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:

If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0

daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C

What's the OA?

rahulgoyal1986 wrote:

judokan wrote:

If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0

daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C

What's the OA?

rahulgoyal1986 wrote:

judokan wrote:

If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0

If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

(1) |x + y| = |x| + |y|

(2) |z + y| = |y| + |z|

1) x and y should have same sign don't know about z.. it can be +ve or -ve leads mutliple answers.. insuffciient 2) y and z should have same sign don't know about x.. it can be +ve or -ve leads mutliple answers..

combine. x,y,z must have same signs.

(x)(y + z) > 0 always true..

combined sufficient. C

good question. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

(1) |y + z| = |y| + |z| --> either both \(y\) and \(z\) are positive or both are negative, because if they have opposite signs then \(|y+z|\) will be less than \(|y|+|z|\) (|-3+1|<|-3|+1). Not sufficient, as no info about \(x\).

(2) |x + y| = |x| + |y| --> the same here: either both \(x\) and \(y\) are positive or both are negative. Not sufficient, as no info about \(z\).

(1)+(2) Either all three are positive or all three are negative --> but in both cases the product will be positive: \(x(y+z)=positive*(positive+positive)=positive>0\) and \(x(y+z)=negative*(negative+negative)=negative*negative=positive>0\). Sufficient.

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...