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# If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x +

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Manager
Joined: 08 Nov 2008
Posts: 232
If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x +  [#permalink]

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29 Mar 2009, 19:57
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Question Stats:

91% (01:03) correct 9% (01:00) wrong based on 53 sessions

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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.

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Director
Joined: 14 Aug 2007
Posts: 652

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29 Mar 2009, 21:11
1
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: 143
Location: Mumbai

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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: 1203

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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: 745
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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02 Apr 2009, 20:45
1
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: 206

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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: 2313

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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.
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Manager
Joined: 22 Feb 2009
Posts: 124
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)

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17 Apr 2009, 10:06
1
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.

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
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Joined: 09 Sep 2013
Posts: 10949
Re: If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x +  [#permalink]

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14 Apr 2019, 14:06
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Re: If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x +   [#permalink] 14 Apr 2019, 14:06
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