Solution:
It is given that x > (y^2) > (z^4).
Let x = 65, y = 8 and z = 2.
y^2 = 64 and z^4 = 16.
Here x > (y^2) > (z^4), and x > y > z.
So I can be true.
Next let x = 1/8, y = 1/3, z = ½.
y^2 = 1/9 and z^4 = 1/16.
Now again x > (y^2) > (z^4), and x < y < z.
So II can be true.
Next let x = 1, y = -1/2, z = -1/3.
y^2 = 1/4, z^4 = 1/81.
Here x > (y^2) > (z^4). and x > z > y.
So even III can be true.
Since all three can be true correct answer is (E).
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Rahul@gurome
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Rahul Lakhani
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Gurome, Inc.
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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+91-99201 32411 (India)
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lokesh r
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Rahul@gurome wrote:Solution:
It is given that x > (y^2) > (z^4).
Let x = 65, y = 8 and z = 2.
y^2 = 64 and z^4 = 16.
Here x > (y^2) > (z^4), and x > y > z.
So I can be true.
Next let x = 1/8, y = 1/3, z = ½.
y^2 = 1/9 and z^4 = 1/16.
Now again x > (y^2) > (z^4), and x < y < z.
So II can be true.
Next let x = 1, y = -1/2, z = -1/3.
y^2 = 1/4, z^4 = 1/81.
Here x > (y^2) > (z^4). and x > z > y.
So even III can be true.
Thanks a lot for solving this question for me..
Since all three can be true correct answer is (E).
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nasir
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i tried to plug in same values for the given options and got the answer incorrect. i couldn't figure out that i have to plug in different values. sometimes its difficult to understand the " Question "
