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kevincan
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stem 2 jus says the sum is less 50..it cud be -50 for all we know

yezz
Well i will try to do my best on this one ... am still learning from you guys.specially both of you guys kevin and dahiya.

From the question stem x,y,z are all either +ve or -ve

and the least possible absolute values for x,y,z respectively is 6,10,15

from one... the square covers the signe of xyz and x,y,z are intigers ... insuff

from 2 still we have enough information about the signes of x,y,z except that sure that they are not +ve and may be all -ve or some +ve and some -ve

thus my answer is b
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I guess i have to practise a lot more .. sorry folks i edited my answer
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Statement 1: x,y,z could be -6, -5, -15 or 6, 5, 15 - INSUFF

Statement 2: x,y,z could be -6,-5,-15 or 2/5,2/3,1 - INSUFF

Together: x,y,z could be -6, -5, -15 or 6, 5, 15 - INSUFF

Answer E
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i believe that as long as you have a common set of answers from one and two ie(-6,-5,-15) thus c is the answer
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Kevin am i at least close to the truth or i should sleep on the idea of GMAT for a while :lol:
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yezz
Well i will try to do my best on this one ... am still learning from you guys.specially both of you guys kevin and dahiya.

From the question stem x,y,z are all either +ve or -ve ( intigers or fractions)


from one... the square covers the signe of xyz and x,y,z are intigers and
... insuff

from 2 we do not have enough information about the signes of x,y,z
( specially that they could be fractions

Taking one and two ( from one) x,y,z are intigers and from question stem ..the least possible absolute values for x,y,z respectively is 6,10,15

thus sure that they are not all +ve Thus and putting into consideration the question stem they are all -ve... suff
thus my answer is c


Don't see why x ,y,and z must be integers!
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Well I think that because (xyz)^2 is intiger therfore x,y,z are either intigers or a mix between intigers and complex numbers ( 7/2 and fractions ie : or their product might be 1 or a square root of an intiger ).... to keep the proportion of 5:3:2........

:roll:
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kevincan
If 5x=3y=2z, is xyz>0?

(1) (xyz)^2 is an integer

(2) 2x+3y+z< 50


Kevin, are there no fractional values that meet the conditions above and yield an integer when squared. Intuitively, I think (C) might be possible.
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[quote="kevincan"]If 5x=3y=2z, is xyz>0?

(1) (xyz)^2 is an integer

(2) 2x+3y+z0. Thus y and z would be positive and xyz would be positive, too. NOT SUFF

(2) 2x+5x+5x/2=19x/2 x<100/19 NOT SUFF

(1) and (2) Both possibilities for x cited in (1) satisfy (2) NOT SUFF


Answer: E



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