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DS Inequality

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Director
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DS Inequality [#permalink]

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New post 29 Sep 2008, 20:08
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A
B
C
D
E

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Re: DS Inequality [#permalink]

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New post 29 Sep 2008, 22:11
A should be the answer. I will use substitution method for such questions.

For stmt1, if I take y = 0 then it reduces to x^2 > z^2. This means
either x > z for all positive values of x and z or
x < z for all negative integer values of z and x.

And, for both these conditions, x^4 > z^4. Hence, sufficient.

For stmt2, if y = 0 then x > z for all values of x and z. However, for negative fraction values of x and z, x^4 < z^4 and for all other values, x^4 > z^4. Hence, insufficient.

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Director
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Re: DS Inequality [#permalink]

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New post 02 Oct 2008, 06:57
unfortunately that not the OA ....


ANY TAKERS ???

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Re: DS Inequality [#permalink]

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New post 02 Oct 2008, 22:18
rao_1857 wrote:
unfortunately that not the OA ....


ANY TAKERS ???


My mistake. I should not have used y = 0. Giving another try. It should be C.

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Manager
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Re: DS Inequality [#permalink]

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New post 03 Oct 2008, 01:05
rao_1857 wrote:
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(1) if square both sides, left x^4 + y^4 + 2x^2y^2, so we know there is a gap, should be insufficient

x^2 + y^2 > z^2 <=> t + u > v

Give it an example: 1 + 1.8 > 2 but 1^2 + 1.8^2 < 2^2

(2) try x = 4, y = 5, z = 6

insufficient

(1) & (2), try example same as (2)

insufficient

Hence E

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Re: DS Inequality [#permalink]

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New post 03 Oct 2008, 02:16
Agreed with Lyla.
It should be E.

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Re: DS Inequality [#permalink]

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New post 03 Oct 2008, 03:04
Even I agree with Lylya.
The answer is E.
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Varun


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Re: DS Inequality [#permalink]

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New post 03 Oct 2008, 07:07
x^4 + y^4 >z^4?

x^2 + y^2>z^2

(x^2+y^2)(x^2+y^2)=x^4+y^4+2x^2y^2 >z^4 but we have this (2x^2y^2) on the left side which..we dont know anything about..therefore insuff

2) is obviously insuff too

E is my take

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Re: DS Inequality [#permalink]

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New post 03 Oct 2008, 11:51
should be E. solved with substitution.

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Re: DS Inequality   [#permalink] 03 Oct 2008, 11:51
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