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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] New post 06 Nov 2005, 20:52
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A
B
C
D
E

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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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 [#permalink] New post 06 Nov 2005, 22:18
E.

1) false if
x=sqrt(2)
y=sqrt(2)
z=sqrt(3)

true for all positive integers

insufficient

2)

x=y=2
z=3
false

for x=2,y=3,z=3 true
insufficient

together:
x=y=sqrt(2)
z=sqrt(3)
false

x=y=z=1
true

insuffucient
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 [#permalink] New post 06 Nov 2005, 22:19
unless someone else has a quick way to do this, I'll go ahead with E.
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 [#permalink] New post 06 Nov 2005, 22:48
E

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

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

x=5,y=4,z=6 as 5^2+4^2>6^2 but 5^4+4^4 <6^4
x=3,y=1,z=2 as 3^2+1^2>2^2 and 3^4+1^4 > 2^4
so Insuff

2. x+y > z
x=3,y=2 so 3+2>4 but 3^4+2^4<4^4
x=9,y=10 so 9+10 > 4 and 9^4+10^4 > 4^4
so Insuff


now taking both together
x=9,y=10,z=5 satisfy both 1 and 2 and also gives yes
but
x=5,y=4,z=6 satisfy both 1 and 2 but gives No

so Insuff

Hence answer E.

:shocked ....certainly took more than 2 minutes ...initally i was trying to solve the question by using equation and inequalites ...then gave up and tired with value substituion ........
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OA [#permalink] New post 07 Nov 2005, 09:29
The OA is E, but Duttsit, I don't follow your logic

2^2 + 2^2 is not greater than 3^2

8 < 9
so the example you used for statement 1 is not usable.

Anyone got a quicker way to solve this then picking numbers? I'm afraid I'd spend all day on something like this.
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 [#permalink] New post 07 Nov 2005, 10:29
This is E...

and (1)

try numbers for X, Y and Z such as sqrt(x), Sqrt(y) and Sqrt(z)

if x=sqrt(2)=y
z=sqrt(3)

x^2 + y^2 > z^2

2+2=4 > 3

but

2^2+2^2<3^2 OK

Insuff

(d) use the same number x+y>z but we still have 2 possibilites...

always try fractions, positive, negative and sqrt numbers to verify these DS questions....no other shorter and sweeeter way...
  [#permalink] 07 Nov 2005, 10:29
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