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Re: SC) For positive integer x, y, and z, is y > z? (1) When x [#permalink]
17 Mar 2011, 18:36
statement 1: x=4+y, z is unknown, insufficient statement 2: x=6+z, y is unknown, insufficient both statement 1+2: 4+y=6+z so y=2+z because all x, y, z are positive integer so y is always 2 bigger than z C
Re: For positive integer x, y, and z, is y > z? [#permalink]
21 Jun 2012, 11:25
Expert's post
Galiya wrote:
(1)+(2)
x=4k+y x=6m+z 4k+y=6m+z, hence E the answer i wonder why some of you skipped the quotients in front of 4 and 6.Is it possible to solve so?
You are right the answer is E, and yes you should put a quotient in front of divisor.
For positive integer x, y, and z, is y > z?
(1) When x is divided by 4, the remainder is y --> \(x=4q+y\) and \(0<y<4\). Not sufficient. (2) When x is divided by 6, the remainder is z --> \(x=6p+z\) and \(0<z<6\). Not sufficient.
(1)+(2) Still insufficient. Consider: \(x=13\), \(y=1\) and \(z=1\) for a NO answer; \(x=7\), \(y=3\) and \(z=1\) for an YES answer.
Re: For positive integer x, y, and z, is y > z? [#permalink]
11 Aug 2013, 04:44
Clearly (E) it is
Explanation:
Is y>z
(1).
x=4A + y No info about z hence Insufficient
(2).
x=6B + z No info about y hence Insufficient
Combining we get:
4A + y = 6B + z
=> (y-z) = 6B - 4A ( now from here y can be less than z or more than z) _________________
Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________
gmatclubot
Re: For positive integer x, y, and z, is y > z?
[#permalink]
11 Aug 2013, 04:44
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