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If x, y, and z are positive integers, is y > x?

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If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 00:50
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C
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E

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If x, y, and z are positive integers, is y > x?

(1) y^2 = xz
(2) z - x > 0

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C
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Re: If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 03:24
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Bunuel wrote:
If x, y, and z are positive integers, is y > x?

(1) y^2 = xz
(2) z - x > 0

Kudos for a corrector solution.



IMO: C

St 1: y^2 = xz
we can infer from the above equation that x, y, z are in GP
Now it can be
increasing GP --> in this case x < y
or decreasing GP --> in this case x > y
or GP with r=1 --> x= y
So not suff

St 2: z > x
cannot infer from this statement. Not sufficient

Combined:
it will boil down to increasing GP. Thus x < y
suff
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Re: If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 04:35
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Bunuel wrote:
If x, y, and z are positive integers, is y > x?

(1) y^2 = xz
(2) z - x > 0

Kudos for a corrector solution.


Given : x, y, and z are positive integers,

Question : Is y > x?

Statement 1: y^2 = xz
@y = 4, x may be 2 and z may be 6 i.e. y > x
@y = 4, x may be 8 and z may be 2 i.e. y < x
NOT SUFFICIENT

Statement 2: z - x > 0
i.e. z > x but x can't be compared with z evn now hence
NOT SUFFICIENT

Combining the two statements
since y^2 = x*z and z > x
then z must be greater than y as well [e.g.4(y)^2 = 8(z)*2(x) ] and therefore x must be the smallest among x, y and z
i.e. y > x for all cases
SUFFICIENT

Answer: option C
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If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 05:05
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Bunuel wrote:
If x, y, and z are positive integers, is y > x?

(1) y^2 = xz
(2) z - x > 0

Kudos for a corrector solution.



Is y>x?

Statement 1, y^2=xz ---> if x=y=z =1 ,t hen "no" but if y=3, x=1, z=3, then "yes". This statement is not sufficient.

Statement 2, z>x, not sufficient.

Combining, y^2=xz and z>x *in other words, x,y,z form a GP with x<y<z. The only way this is possible is by y>x (y= 3, z=9, x=1)

C is the correct answer.
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Re: If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 05:59
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Engr2012 wrote:
Bunuel wrote:
If x, y, and z are positive integers, is y > x?

(1) y^2 = xz
(2) z - x > 0

Kudos for a corrector solution.



Is y>x?

Statement 1, y^2=xz ---> if x=y=z =1 ,t hen "no" but if y=3, x=1, z=3, then "yes". This statement is not sufficient.

Statement 2, z>x, not sufficient.

Combining, y^2=xz and z>x. The only way this is possible is by y>x (y=z=3, x=1)

C is the correct answer.


Hi Engr2012

The highlighted part seems some mistake. @y=z, y and z will also be equal to x which is not possible because it's given that z>x
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Re: If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 06:14
GMATinsight wrote:
Engr2012 wrote:
Bunuel wrote:
If x, y, and z are positive integers, is y > x?

(1) y^2 = xz
(2) z - x > 0

Kudos for a corrector solution.



Is y>x?

Statement 1, y^2=xz ---> if x=y=z =1 ,t hen "no" but if y=3, x=1, z=3, then "yes". This statement is not sufficient.

Statement 2, z>x, not sufficient.

Combining, y^2=xz and z>x. The only way this is possible is by y>x (y=z=3, x=1)

C is the correct answer.


Hi Engr2012

The highlighted part seems some mistake. @y=z, y and z will also be equal to x which is not possible because it's given that z>x


Yes. Thanks. I meant to write z=9, y =3, x=1. It was a typo.
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Re: If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 10:18
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Bunuel wrote:
If x, y, and z are positive integers, is y > x?

(1) y^2 = xz
(2) z - x > 0

Kudos for a corrector solution.


S1: y^2=xz so y is geometric mean of x and z... which means y lies between x and z now possible combinations are x -y- z or z -y-x..so not sufficient.
S2: Z-x>0 so z>x ?? what about y?? y could still be greater than or less than x.
s1+s2 - so only one combination left - x y z hence sufficient.

Ans C.
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Re: If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 13:42
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Bunuel wrote:
If x, y, and z are positive integers, is y > x?

(1) y^2 = xz
(2) z - x > 0

Kudos for a corrector solution.


St 1:

y^2 =xz

Let Y^2 = 9, then
x = 3, z = 3 => y = x
x = 9, z = 1 => y < x
x = 1, z = 9 => y > x

Not sufficient

St 2:
z-x>0 => z>x

But we do not know any thing about y. Hence Not Sufficient.

Combining 1 and 2,

Only one condition holds here,
y^2= 9 then x = 1, z = 9 => y > x

Hence sufficient. Option C
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If x, y, and z are positive integers, is y > x? [#permalink] New post   28 Jul 2015, 19:51
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St1: x=1; y=1; z=1 then, y=x

Or x=1; y=2; z=4 then y>x insufficient.

Stmt2: z>x. Nothing about y. Insufficient.

Stmt1&2:

y2 = xz
=> z = y2/x;

from stmt 2, z>x

so, y2/x > x
y2 > x2
y>x

Sufficient. Answer is C.
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Re: If x, y, and z are positive integers, is y > x? [#permalink] New post   04 Oct 2020, 12:06
(1) (y^2) = x*z is insufficient, as we cannot make any judgement about whether or not y > x, as there are 3 unknowns.

(2) z - x > 0 is insufficient, as no information is given about y.

Combining (1) and (2):

From (2), we can say z > x. Since (y^2) = x*z, we can replace z in (1) with x, and now say that (y^2) > (x^2), as z > x.

Then, since x and y are both positive integers, we can safely say that given (y^2) > (x^2), y must be greater than x.
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Re: If x, y, and z are positive integers, is y > x? [#permalink] New post   08 Jan 2022, 09:36
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Re: If x, y, and z are positive integers, is y > x? [#permalink]
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