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If x, y, and z are positive integers, is y > x?
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28 Jul 2015, 00:50
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81% (01:33) correct 19% (01:10) wrong based on 59 sessions
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If x, y, and z are positive integers, is y > x? (1) y^2 = xz (2) z  x > 0 Kudos for a corrector solution.
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Re: If x, y, and z are positive integers, is y > x?
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28 Jul 2015, 03:24
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 GPNow 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?
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28 Jul 2015, 04:35
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 SUFFICIENTStatement 2: z  x > 0 i.e. z > x but x can't be compared with z evn now hence NOT SUFFICIENTCombining the two statementssince 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 SUFFICIENTAnswer: option C
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If x, y, and z are positive integers, is y > x?
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28 Jul 2015, 05:05
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?
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28 Jul 2015, 05:59
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 Engr2012The 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?
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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 Engr2012The 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?
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28 Jul 2015, 10:18
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 yx..so not sufficient. S2: Zx>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?
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28 Jul 2015, 13:42
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: zx>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?
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28 Jul 2015, 19:51
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?
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23 Mar 2019, 05:27
What if you use 16 and 1. Then x =z so should be E right?




Re: If x, y, and z are positive integers, is y > x?
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23 Mar 2019, 05:27






