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# x > y^2 > z^4

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VP
Joined: 08 Apr 2009
Posts: 1183
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
x > y^2 > z^4 [#permalink]

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16 Jul 2009, 20:02
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0% (00:00) correct 100% (00:50) wrong based on 6 sessions

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If $$x > y^2 > z^4$$, which of the folloing statements could be true?
I. x > y > z
II. z > y > x
III. x > z > y

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III
Manager
Joined: 16 Apr 2009
Posts: 233
Schools: Ross
Re: x > y^2 > z^4 [#permalink]

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16 Jul 2009, 21:01
x>y^2>z^4

Let x be 100, y be -9 and z be -2
100>(-9)^2>(-2)^4

In this case
x>z>y

holds true

Let x be 100, y be 9 and z be 2
100>(9)^2>(2)^4

holds true
In this case
x>y>z

Hence OA -C
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Senior Manager
Joined: 04 Jun 2008
Posts: 289
Re: x > y^2 > z^4 [#permalink]

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16 Jul 2009, 22:56
Ans would be E

stmt II can also be true.

Take all 3 to be positive fractions
Say z = 0.81
y = 0.80
x = 0.7

z>y>x

and x>y^2>z^4
VP
Joined: 08 Apr 2009
Posts: 1183
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
Re: x > y^2 > z^4 [#permalink]

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17 Jul 2009, 08:33
the OA is E.
However, is there a more methodical way to approach this type of problem besides plug-in #s?
Current Student
Joined: 18 Jun 2009
Posts: 356
Location: San Francisco
Re: x > y^2 > z^4 [#permalink]

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17 Jul 2009, 10:16
Nope there is no rule except for the fact that

1) -ve numbers when raised to power of multiples of two becomes +ve
2) Fractions when raised to any power decreases in its value.

These two rules should make it easier to plug in the right values
Manager
Joined: 16 Apr 2009
Posts: 233
Schools: Ross
Re: x > y^2 > z^4 [#permalink]

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17 Jul 2009, 17:03
Thanks rashiminet84 and gmanjesh..
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Manager
Joined: 25 May 2011
Posts: 152
Re: x > y^2 > z^4 [#permalink]

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23 Oct 2011, 03:42
it seems that just picking numbers works for this kind of question
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India
Re: x > y^2 > z^4 [#permalink]

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24 Oct 2011, 23:30
asimov wrote:
If $$x > y^2 > z^4$$, which of the folloing statements could be true?
I. x > y > z
II. z > y > x
III. x > z > y

A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II, and III

Yes, just plugging in numbers work best for such questions. The only thing to keep in mind is that you should plug in the right numbers. How do you know the right numbers?
When I see $$x > y^2 > z^4$$, I think that $$y^2$$ and $$z^4$$ are non negative. Since $$y^2 > z^4$$, $$y^2$$ cannot be 0. Only z can be 0. x has to be positive. Also, I have to take into account two ranges: 0 to 1 and 1 to infinity. The powers behave differently in these two ranges. I will consider negative numbers only if I have to since with powers, they get confusing to deal with.

The question says: "Which of the following could be true?"
We have to find examples where each relation holds.

I. x > y > z
This is the most intuitive of course.
z = 0, y = 1 and x = 2
$$2 > 1^2 > 0^4$$

II. z > y > x
Let me consider the 0 to 1 range here. Say z = 1/2, y = 1/3 and x = 1/4
$$1/4 > 1/9 > 1/16$$

III. x > z > y
Let's stick to 0 to 1 range. z > y as in case II above but x has to be greater than both of them. Say z = 1/2, y = 1/3 and x = 1
$$1>1/9 > 1/16$$

So all three statements could be true.
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Re: x > y^2 > z^4   [#permalink] 24 Oct 2011, 23:30
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