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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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29 Mar 2009, 19:57
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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.
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Re: DS  Modulas [#permalink]
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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



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Re: DS  Modulas [#permalink]
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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?



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Re: DS  Modulas [#permalink]
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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.



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Re: DS  Modulas [#permalink]
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02 Apr 2009, 20:45
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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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Re: DS  Modulas [#permalink]
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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.



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Re: DS  Modulas [#permalink]
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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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Re: DS  Modulas [#permalink]
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17 Apr 2009, 10:06
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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.










