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Re: For positive integer x, y, and z, is y > z? [#permalink]

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17 Mar 2011, 19:36

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

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]

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11 Aug 2013, 05:44

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)
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Re: For positive integer x, y, and z, is y > z? [#permalink]

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25 Feb 2016, 07:29

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