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If 5x=3y=2z, is xyz>0? (1) (xyz)^2 is an integer (2)

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If 5x=3y=2z, is xyz>0? (1) (xyz)^2 is an integer (2) [#permalink] New post 01 Sep 2006, 08:02
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If 5x=3y=2z, is xyz>0?

(1) (xyz)^2 is an integer

(2) 2x+3y+z< 50
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 [#permalink] New post 01 Sep 2006, 08:29
is it E or am i missing somehting?
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 [#permalink] New post 01 Sep 2006, 08:32
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

Last edited by yezz on 01 Sep 2006, 08:45, edited 2 times in total.
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 [#permalink] New post 01 Sep 2006, 08:35
stem 2 jus says the sum is less 50..it cud be -50 for all we know

yezz wrote:
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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 [#permalink] New post 01 Sep 2006, 08:46
I guess i have to practise a lot more .. sorry folks i edited my answer
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 [#permalink] New post 01 Sep 2006, 09:48
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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 [#permalink] New post 01 Sep 2006, 09:58
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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 [#permalink] New post 01 Sep 2006, 12:15
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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 [#permalink] New post 01 Sep 2006, 12:51
yezz wrote:
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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 [#permalink] New post 01 Sep 2006, 23:56
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:

Last edited by yezz on 02 Sep 2006, 05:38, edited 1 time in total.
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Re: DS: Positive Product [#permalink] New post 02 Sep 2006, 05:17
kevincan wrote:
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.

Last edited by GMATT73 on 02 Sep 2006, 05:58, edited 1 time in total.
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Re: DS: Positive Product [#permalink] New post 02 Sep 2006, 05:51
kevincan wrote:
If 5x=3y=2z, is xyz>0?

(1) (xyz)^2 is an integer

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



Note that y=5x/3 and z=5x/2 and that (x,y,z)=(0,0,0) is a solution that satisfies both conditions. If x=y=z=0, xyz is not positive. So the question is, could xyz be positive?

(1) implies that (25x^3/6)^2= 5^4*x^6/3^2*2^2 is an integer


Thus it could be that |x|=6^(1/3)*k^(1/6) where k is an integer

If k=0, x=0 and xyx=0 , but if k=1, x=6^1/3>0. Thus y and z would be positive and xyz would be positive, too. NOT SUFF

(2) 2x+5x+5x/2=19x/2<50 => x<100/19 NOT SUFF

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


Answer: E
Re: DS: Positive Product   [#permalink] 02 Sep 2006, 05:51
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