Author 
Message 
Senior Manager
Joined: 08 Nov 2008
Posts: 278

If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x + [#permalink]
Show Tags
29 Mar 2009, 19:57
Question Stats:
91% (01:03) correct 9% (01:00) wrong based on 47 sessions
HideShow timer Statistics
This topic is locked. If you want to discuss this question please repost it in the respective forum. This topic is locked. If you want to discuss this question please repost 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.
_________________
"CEO in making"



Director
Joined: 14 Aug 2007
Posts: 704

Re: DS  Modulas [#permalink]
Show Tags
29 Mar 2009, 21:11
tenaman10 wrote: (1) │y + z│ = │y│ + │z│
=> either y & z are both +ve or both ve. Insuff as we don't know X. tenaman10 wrote: (2) │x + y│ = │x│ + │y│
=> either x&y are both +ve or both ve. Insuff as we don't know Z. Together, We have 2 cases 1) x,y,z are all positive: x (y + z) ≥ 0 2) x,y,z are all ve. x (y + z) ≥ 0 Suff. C



Manager
Joined: 19 May 2008
Posts: 160
Location: Mumbai

Re: DS  Modulas [#permalink]
Show Tags
29 Mar 2009, 21:28
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.
So I think the answer is E. What is the OA?



VP
Joined: 05 Jul 2008
Posts: 1377

Re: DS  Modulas [#permalink]
Show Tags
02 Apr 2009, 20:25
alpha_plus_gamma wrote: tenaman10 wrote: (1) │y + z│ = │y│ + │z│
=> either y & z are both +ve or both ve. Insuff as we don't know X. tenaman10 wrote: (2) │x + y│ = │x│ + │y│
=> either x&y are both +ve or both ve. Insuff as we don't know Z. Together, We have 2 cases 1) x,y,z are all positive: x (y + z) ≥ 0 2) x,y,z are all ve. x (y + z) ≥ 0 Suff. C How can it be C? x y z are all +ve answer for the Q x(y+z) >= 0 Yes x y z are all ve answer for the Q x(y+z) >=0 No One yes and one No makes it E.



Director
Joined: 01 Apr 2008
Posts: 847
Name: Ronak Amin
Schools: IIM Lucknow (IPMX)  Class of 2014

Re: DS  Modulas [#permalink]
Show Tags
02 Apr 2009, 20:45
1
This post received KUDOS
I go with C. What is OA?
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.



Manager
Joined: 19 Aug 2006
Posts: 233

Re: DS  Modulas [#permalink]
Show Tags
04 Apr 2009, 10:26
I go with C.
stmnt 1  indicates y and z are of the same sign, insuff. stmnt 2  indicates x and y are of the same sign, insuff.
combining, it's suffic.



SVP
Joined: 29 Aug 2007
Posts: 2457

Re: DS  Modulas [#permalink]
Show Tags
16 Apr 2009, 22:40
icandy wrote: alpha_plus_gamma wrote: tenaman10 wrote: (1) │y + z│ = │y│ + │z│
=> either y & z are both +ve or both ve. Insuff as we don't know X. tenaman10 wrote: (2) │x + y│ = │x│ + │y│
=> either x&y are both +ve or both ve. Insuff as we don't know Z. Together, We have 2 cases 1) x,y,z are all positive: x (y + z) ≥ 0 2) x,y,z are all ve. x (y + z) ≥ 0 Suff. C How can it be C? x y z are all +ve answer for the Q x(y+z) >= 0 Yes x y z are all ve answer for the Q x(y+z) >=0 No One yes and one No makes it E. If x, y, and z are all +ve, x(y+z) > 0. Suff. If x, y, and z are all ve, x(y+z) > 0. Suff.
_________________
Verbal: http://gmatclub.com/forum/newtotheverbalforumpleasereadthisfirst77546.html Math: http://gmatclub.com/forum/newtothemathforumpleasereadthisfirst77764.html Gmat: http://gmatclub.com/forum/everythingyouneedtoprepareforthegmatrevised77983.html
GT



Manager
Joined: 22 Feb 2009
Posts: 135
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)

Re: DS  Modulas [#permalink]
Show Tags
17 Apr 2009, 10:06
1
This post received KUDOS
tenaman10 wrote: 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.










