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

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New post 29 Mar 2009, 19:57
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A
B
C
D
E

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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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Re: DS - Modulas  [#permalink]

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New post 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
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Re: DS - Modulas  [#permalink]

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New post 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?
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Re: DS - Modulas  [#permalink]

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New post 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.
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Re: DS - Modulas  [#permalink]

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New post 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.
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Re: DS - Modulas  [#permalink]

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New post 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.
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Re: DS - Modulas  [#permalink]

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New post 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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Re: DS - Modulas  [#permalink]

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New post 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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Re: If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x +  [#permalink]

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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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