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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]  30 May 2007, 16:01
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is x^4+y^4>z^4

1.x^2+y^2>z^ 2
2.x+y>z
Director
Joined: 03 Sep 2006
Posts: 885
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Kudos [?]: 259 [0], given: 33

Re: DS [#permalink]  30 May 2007, 18:01
ajisha wrote:
is x^4+y^4>z^4

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

I will say "E".

From ( i );

x^2 + y^2 > z^2

Squaring both sides of the inequality we will get;

x^4 + y^4 + 2*x^2*y^2 > z^4

Thus we can't say whether x^4+y^4>z^4 or not!

Nothing comes up from the (ii).

Why did you guys reply the answer as "A"?
VP
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[#permalink]  30 May 2007, 20:15
the answer is (A)

x^4 + y^4 > z^4 - is this true ?

statement 1

x^2 + y^2 > z

lets plug in:

(1) x,y = 1/2 z = 2

or alternatively

(2) x,y = 2 z = 2

I'm only useing negative numbers since n^2 or n^4 will give a positive outcome.

for (1):

1/4+1/4 > 4 (this is false)

for (2):

4+4 > 4 (this is true)

we will applay this to the stem (x^4 + y^4 > z^4)

to prove it's correctness !

sufficient

statement 1

lets plug in:

(1) x,y = 2/3 z = 1

or alternatively

(2) x,y = 2 z = 2

for (1):

2/3+2/3 > 1 (this is true)

for (2):

2+2 > 2 (this is true)

if we will applay (1) & (2) to the stem since they are both true (x^4 + y^4 > z^4)

we will once get true (2) and once false (1)

insufficient

hence the answer is (A)

[#permalink] 30 May 2007, 20:15
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