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

Question

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

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

Kudos for a corrector  solution .

VenoMfTw · Answer

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

GMATinsight · Answer

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   and therefore x must be the smallest among x, y and z
i.e. y > x for all cases
 SUFFICIENT

Answer: option C

ENGRTOMBA2018 · Answer

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 xx (y= 3, z=9, x=1) C is the correct answer.

GMATinsight · 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

ENGRTOMBA2018 · Answer

Yes. Thanks. I meant to write z=9, y =3, x=1. It was a typo.

M3sSiaH · Answer

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.

ashokk138 · Answer

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

zmtalha · Answer

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.

sidgvpta · Answer

(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.

bumpbot · Answer

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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?

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