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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
Joined: 19 Sep 2005
Posts: 110
Followers: 1

Kudos [?]: 5 [0], given: 0

is x^4 + y^4 > z^4? 1. x^2 + y^2 > z^2 2. x+y > z [#permalink]  06 Nov 2005, 20:52
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
is x^4 + y^4 > z^4?

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

2. x+y > z
VP
Joined: 22 Aug 2005
Posts: 1120
Location: CA
Followers: 1

Kudos [?]: 47 [0], given: 0

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
Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy
Followers: 1

Kudos [?]: 11 [0], given: 0

unless someone else has a quick way to do this, I'll go ahead with E.
Director
Joined: 27 Jun 2005
Posts: 513
Location: MS
Followers: 2

Kudos [?]: 36 [0], given: 0

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

....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 ........
Manager
Joined: 19 Sep 2005
Posts: 110
Followers: 1

Kudos [?]: 5 [0], given: 0

OA [#permalink]  07 Nov 2005, 09:29

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.
Current Student
Joined: 28 Dec 2004
Posts: 3387
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 14

Kudos [?]: 186 [0], given: 2

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...
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