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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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Manager
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Is x^4 + y^4 > z^4? 1) x^2 + y^2 >z^2 2) x +y >z [#permalink]  02 Sep 2006, 17:46
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Is x^4 + y^4 > z^4?

1) x^2 + y^2 >z^2
2) x +y >z
Manager
Joined: 30 Jan 2006
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It is A

S1) x^2+y^2>z^2

if x =1 and y =2 and z =1

or any fraction also x = 0.5 y = 0.2 z = 0.4

also which satisifies it.

For S2)if x = 1 and y =2 and z = 2 which satisfies x^4+y^4>z^4

if x = 1 y = 1 and z = -4

which doesn't satisfy the equation above.

So A.
Manager
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smily_buddy,

For statement1, what if x=1/2; y=1/2; z=2/3 --> does not satify the original inequality.

jjhko
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Re: DS: x^4+y^4 > z^4? [#permalink]  02 Sep 2006, 21:53
It is E

jjhko wrote:
Is x^4 + y^4 > z^4?

1) x^2 + y^2 >z^2
If x^2=2, y^2=3 and z=2
Then x^4 + y^4 > z^4 is FALSE
If x,y,z = 1
Then x^4 + y^4 > z^4 is TRUE
INSUFF

2) x +y >z

e.g. 2 + 3 > 5
But x^4 + y^4 > z^4 FALSE
when 1 + 1 > 1
Then x^4 + y^4 > z^4 is TRUE
INSUFF

Combined is also INSUFF as shown above
Current Student
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this should be A...

stem X^4+y^4>z^4?

st1) x^2+y^2>z^2

lets say if x and y are integers...x=1, y=2, then 1+4=5; we are told is greater than z, z^2 can be 4 or 3...

if X and y were fractions, we are told that sume X^2+y^2 is greater than z^2..

if z^2 is a fraction or not, it is still less than x^2 + y^2...same should hold for z^4+y^4...
Manager
Joined: 01 Apr 2006
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OA is E.

I'm not sure if there is a sure way of working out this problem... the only way that could was to try out various sample values... which has the down side of being time consuming.

jjhko
Manager
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Problems like these can ruin my PS score ...
Senior Manager
Joined: 13 Jul 2006
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ok here goes my explantion

x^2 + y^2 > z^2

since left side is positive and right side is postive
the inequality can be squared
therfore,x^4 + y^4 + 2x^4 y^4 > z^4

therfore not necessary that x^4 + y^4 > z^4

Also x+y > z
x+y and z might be negative in which case the fourth power would not hold.combining both also nnothing can be said of x and y
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