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If xyz ≠ 0, is x(y + z) >= 0? [#permalink]
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26 Jul 2008, 00:41
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If xyz ≠ 0, is x(y + z) >= 0? (1) y + z = y + z (2) x + y = x + y OPEN DISCUSSION OF THIS QUESTION IS HERE: ifxyz0isxyz107551.html
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Last edited by Bunuel on 08 Feb 2014, 02:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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Re: DS absolute value 2 (n3.7) [#permalink]
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26 Jul 2008, 07:22
judokan wrote: If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) x + y = x + y (2) z + y = y + z From 1) Either both X and Y are positive or both are negative. Then only 1) can hold true. If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative. From 2) Either both Z and Y are positive or both are negative. Then only 2) can hold true. If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative. Combining 1) and 2) When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0 Thus C



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Re: DS absolute value 2 (n3.7) [#permalink]
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26 Jul 2008, 08:19
daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C What's the OA? rahulgoyal1986 wrote: judokan wrote: If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) x + y = x + y (2) z + y = y + z From 1) Either both X and Y are positive or both are negative. Then only 1) can hold true. If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative. From 2) Either both Z and Y are positive or both are negative. Then only 2) can hold true. If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative. Combining 1) and 2) When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0 Thus C



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Re: DS absolute value 2 (n3.7) [#permalink]
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26 Jul 2008, 10:21
haidzz wrote: daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C What's the OA? rahulgoyal1986 wrote: judokan wrote: If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) x + y = x + y (2) z + y = y + z From 1) Either both X and Y are positive or both are negative. Then only 1) can hold true. If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative. From 2) Either both Z and Y are positive or both are negative. Then only 2) can hold true. If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative. Combining 1) and 2) When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0 Thus C You guys do well. OA is C



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Re: DS absolute value 2 (n3.7) [#permalink]
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06 Aug 2008, 13:05
judokan wrote: If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) x + y = x + y (2) z + y = y + z 1) x and y should have same sign don't know about z.. it can be +ve or ve leads mutliple answers.. insuffciient 2) y and z should have same sign don't know about x.. it can be +ve or ve leads mutliple answers.. combine. x,y,z must have same signs. (x)(y + z) > 0 always true.. combined sufficient. C good question.
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Re: DS absolute value 2 (n3.7) [#permalink]
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06 Aug 2008, 14:48
good one



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Re: DS absolute value 2 (n3.7) [#permalink]
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06 Aug 2008, 18:09
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judokan wrote: If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) x + y = x + y (2) z + y = y + z simplest way to solve this : whether we say a+b= a+b a,b are of same sign either a,b are +ve or both are ve hence now consider : (1) x,y should be same sign but (x)(y + z) > 0 depends on z also hence INSUFFI info (2)z,y are same sign but (x)(y + z) > 0 depends on x hence INSUFFI info now (1) and (2) say that x,y,z are of same sign hence (x)(y+z)>0 always SUFFI hence IMO C
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Re: If x, y, and z are nonzero numbers, is (x)(y + z) > 0? [#permalink]
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07 Feb 2014, 20:03
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) x + y = x + y > sq and simplifying, xy = xy => x and y have same sign. NOT SUFFICIENT
(2) z + y = y + z > Sq and simplifying, zy = zy => y and z have same sign. NOT SUFFICIENT.
Combining, x,y, and z have same sign. when all +ve, (x)(y + z) > 0. when all ve, (x)(y + z) > 0. SUFFICIENT.
C.



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Re: If xyz ≠ 0, is x(y + z) >= 0? [#permalink]
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08 Feb 2014, 02:11
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If xyz ≠ 0, is x (y + z) >= 0?xyz ≠ 0 means that neither of unknowns is 0. (1) y + z = y + z > either both \(y\) and \(z\) are positive or both are negative, because if they have opposite signs then \(y+z\) will be less than \(y+z\) (3+1<3+1). Not sufficient, as no info about \(x\). (2) x + y = x + y > the same here: either both \(x\) and \(y\) are positive or both are negative. Not sufficient, as no info about \(z\). (1)+(2) Either all three are positive or all three are negative > but in both cases the product will be positive: \(x(y+z)=positive*(positive+positive)=positive>0\) and \(x(y+z)=negative*(negative+negative)=negative*negative=positive>0\). Sufficient. Answer: C. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifxyz0isxyz107551.html
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Re: If xyz ≠ 0, is x(y + z) >= 0?
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