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Is x^4 + y^4 > z^4? 1) x^2 + y^2 > z^2 2) x + y > z

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Is x^4 + y^4 > z^4? 1) x^2 + y^2 > z^2 2) x + y > z  [#permalink]

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New post 24 Jul 2011, 10:25
2
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

49% (00:56) correct 51% (01:22) wrong based on 67 sessions

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Is x^4 + y^4 > z^4 ?

(1) x^2 + y^2 > z^2
(2) x+y > z

OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/is-x-4-y-4-z-101358.html
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Re: Is x^4 + y^4 > z^4? 1) x^2 + y^2 > z^2 2) x + y > z  [#permalink]

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New post 24 Jul 2011, 10:34
kannn wrote:
Is x^4 + y^4 > z^4?

1) x^2 + y^2 > z^2
2) x + y > z

OA later.
Source GMATPrep


Should be E.

Square first statment and you get x^4+y^4+2x^2y^2>z^4. As 2x^2y^2 is missing, th quality may or may not hold.

Statement 2 is similar, so wont answer either.


See this link, already disussed.

gmatprep-ds-is-x-4-y-4-z-66899.html
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Re: Is x^4 + y^4 > z^4? 1) x^2 + y^2 > z^2 2) x + y > z  [#permalink]

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New post 26 Jul 2011, 08:59
E both are not sufficient
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Re: Is x^4 + y^4 > z^4? 1) x^2 + y^2 > z^2 2) x + y > z  [#permalink]

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New post 27 Nov 2017, 04:09
Is \(x^4+y^4>z^4\)?

The best way to deal with this problem is plugging numbers. Remember on DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

(1) \(x^2+y^2>z^2\)
It's clear that we get YES answer very easily with big x and y (say 10 and 10), and small z (say 0).

For NO answer let's try numbers from Pythagorean triples:
\(x^2=3\), \(y^2=4\) and \(z^2=5\) (\(x^2+y^2=7>5=z^2\)) --> \(x^4+y^4=9+16=25=z^4\), so we have answer NO (\(x^4+y^4\) is NOT more than \(z^4\), it's equal to it).

Not sufficient.

(2) \(x+y>z\). This one is even easier: again we can get YES answer with big x and y, and small z.

As for NO try to make z some big enough negative number: so if \(x=y=1\) and \(z=-5\), then \(x^4+y^4=1+1=2<25=z^4\).

Not sufficient.

(1)+(2) As we concluded YES answer is easily achievable. For NO try the case of \(x^2=3\), \(y^2=4\) and \(z^2=5\) again: \(x+y=\sqrt{3}+\sqrt{4}>\sqrt{5}\) (\(\sqrt{3}+2\) is more than 3 and \(\sqrt{5}\) is less than 3), so statement (2) is satisfied, we know that statement (1) is also satisfied (\(x^2+y^2=7>5=z^2\)) and \(x^4+y^4=9+16=25=z^4\). Not sufficient.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/is-x-4-y-4-z-101358.html
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Re: Is x^4 + y^4 > z^4? 1) x^2 + y^2 > z^2 2) x + y > z &nbs [#permalink] 27 Nov 2017, 04:09
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