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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)

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Re: For positive integer x, y, and z, is y > z?
[#permalink]
11 Aug 2013, 04:44