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If xyz 0, is x (y + z) 0? (1) y + z = y + z (2) x + y = x + [#permalink]
 29 Mar 2009, 19:57
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25% (01:00) wrong based on 1 sessions
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
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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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Manager
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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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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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Director
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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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Senior Manager
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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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CEO
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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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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.
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