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# For positive integer x, y, and z, is y > z?

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For positive integer x, y, and z, is y > z? [#permalink]  17 Mar 2011, 13:42
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Question Stats:

82% (02:06) correct 18% (01:40) wrong based on 38 sessions
For positive integer x, y, and z, is y > z?

(1) When x is divided by 4, the remainder is y.
(2) When x is divided by 6, the remainder is z.
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Re: SC) For positive integer x, y, and z, is y > z? (1) When x [#permalink]  17 Mar 2011, 14:12
banksy wrote:
SC) For positive integer x, y, and z, is y > z?
(1) When x is divided by 4, the remainder is y.
(2) When x is divided by 6, the remainder is z.

if x = +ve, then x/4> x/6
if x= -ve, then x/6>x/4

therefore ans = e
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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
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Re: SC) For positive integer x, y, and z, is y > z? (1) When x [#permalink]  17 Mar 2011, 18:53
Argh.. missed the fact that the question says x,y,z are +ve.

Ans should be c
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Re: SC) For positive integer x, y, and z, is y > z? (1) When x [#permalink]  19 Mar 2011, 00:46
Hi

I think the answer is E, because :

x = 4k + y, no information about z, so not sufficient

so y = 1,2,3 (can't be 4 as then it can divide evenly)

x = 6m + z, no information about y, so not sufficient

so z = 1,2,3,4,5 (can't be 5 as then it can divide evenly)

Even from (1) and (2) also, y > z or z > y.

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Re: SC) For positive integer x, y, and z, is y > z? (1) When x [#permalink]  19 Mar 2011, 00:48
Btw, this is a DS question, not sure why it's in PS section.
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Re: SC) For positive integer x, y, and z, is y > z? (1) When x [#permalink]  19 Mar 2011, 02:46
Post in right forum.

statement 1: x=4+y, no information about Z, insufficient
statement 2: x=6+z, No information about y , insufficient

Considering C
4+y=6+z so y=z+2
so y>z
ans C
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Re: SC) For positive integer x, y, and z, is y > z? (1) When x [#permalink]  19 Mar 2011, 10:30
Since the individual statements only tell about the X,Y and X,Z resp.....both the not sufficient individually.

So the answer is either C or E.

Now take both the conditions:

=> X = 4a + Y and X = 6b + Z

=> 4a+y = 6b+ z

case 1: a=6 and b = 4..=> y=z...the answer is NO

case 2: a=5 and b = 5 => y>z..the answer is Yes.

so E.
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Re: For positive integer x, y, and z, is y > z? [#permalink]  21 Jun 2012, 11:11
(1)+(2)

x=4k+y
x=6m+z
i wonder why some of you skipped the quotients in front of 4 and 6.Is it possible to solve so?
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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
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.

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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
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