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lokesh r: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sat Sep 04, 2010 2:27 am; Location: Leeds,UK; Thanked: 1 times

by lokesh r » Sat Sep 25, 2010 12:48 pm

Please help me in analyzing statement 3

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Source: Beat The GMAT — Problem Solving | Sat Sep 25, 2010 12:48 pm

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Sat Sep 25, 2010 8:01 pm

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 = Â½.
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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lokesh r: Master | Next Rank: 500 Posts; Posts: 156; Joined: Sat Sep 04, 2010 2:27 am; Location: Leeds,UK; Thanked: 1 times

by lokesh r » Sun Sep 26, 2010 11:09 am

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 = Â½.
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: Senior | Next Rank: 100 Posts; Posts: 55; Joined: Tue Oct 13, 2009 6:08 pm; Location: NY; Thanked: 3 times

by nasir » Sun Sep 26, 2010 11:49 am

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 "

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