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 xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x + [#permalink]

Show Tags

29 Mar 2009, 18:57

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

90% (02:01) correct
10% (01:00) wrong based on 29 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If xyz ≠ 0, is x (y + z) ≥ 0? (1) │y + z│ = │y│ + │z│ (2) │x + y│ = │x│ + │y│ A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.
_________________

Statement 1, insufficient since we don't know x. Statement 1, insufficient since we don't know z.

Taking both the statements together, it is xyz not equal to 0. So we can have multiple scenarios. 1 x, y and z are +ve - sufficient 2 x, y and z are -ve - sufficient 2. x is -ve while y and z are positive - not sufficient.

1) y and z have the same sign. So we have two possibilities, y<0 and z<0 OR y>0 and z>0

2) x and y have the same sign, So we have two possibilites, x<0 and y<0 OR x>0 and y>0

Combining, if we take y<0 and z<0 from (1) then from (2) we will get x<0 ( since from 2, if y<0 then x WILL be < 0). Hence, x,y,z all have the same sign. Similarly for the other case. Hence C.

If xyz ≠ 0, is x (y + z) ≥ 0? (1) │y + z│ = │y│ + │z│ (2) │x + y│ = │x│ + │y│ A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.

x (y + z) ≥ 0? Deducing the signs of x and (y+z) will be sufficient to answer the question.

(1) │y + z│ = │y│ + │z│

is possible only if both Y and Z are positive or 0. threfore (y+z) is positive or 0. but we dont know the sign of x. so not sufficient.

(2) │x + y│ = │x│ + │y│

For the above reasons, x is also positive.

Combining both we can say that x (y + z) ≥ 0. Therefore C.