x > y^2 > z^4 : GMAT Problem Solving (PS)
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x > y^2 > z^4

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VP
Joined: 07 Apr 2009
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x > y^2 > z^4 [#permalink]

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16 Jul 2009, 19:02
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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
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Re: x > y^2 > z^4 [#permalink]

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16 Jul 2009, 20:01
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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Re: x > y^2 > z^4 [#permalink]

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16 Jul 2009, 21:56
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: 07 Apr 2009
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Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
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Re: x > y^2 > z^4 [#permalink]

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17 Jul 2009, 07:33
the OA is E.
However, is there a more methodical way to approach this type of problem besides plug-in #s?
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Re: x > y^2 > z^4 [#permalink]

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17 Jul 2009, 09:16
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
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Re: x > y^2 > z^4 [#permalink]

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17 Jul 2009, 16:03
Thanks rashiminet84 and gmanjesh..
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Re: x > y^2 > z^4 [#permalink]

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23 Oct 2011, 02:42
it seems that just picking numbers works for this kind of question
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Re: x > y^2 > z^4 [#permalink]

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24 Oct 2011, 22:30
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, 22:30
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