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Is x>y? (1). x^2>2y (2). x^3>y^4

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Is x>y? (1). x^2>2y (2). x^3>y^4 [#permalink] New post 03 Oct 2004, 06:15
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Is x>y?

(1). x^2>2y

(2). x^3>y^4
Joined: 31 Dec 1969
Location: United States
Concentration: Marketing, Other
GMAT 1: 710 Q49 V38
GMAT 2: 660 Q V
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WE: Accounting (Accounting)
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Kudos [?]: 62 [0], given: 76370

 [#permalink] New post 03 Oct 2004, 06:56
1. X^2-2y>0. if x=2 and y =1 then X>Y. However if x=-2 and y=1, X<Y, but satifies the equation 1 - and cant say x > y or not. Hence Insufficient.

2. X^3-Y^4 >0. Y^4 is always +ve. X^3 can be +ve or -ve. It has to be larger than Y^4 to satisfy the equation. So X>Y always. Sufficient.

Hence B
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 [#permalink] New post 03 Oct 2004, 07:11
Anonymous wrote:
1. X^2-2y>0. if x=2 and y =1 then X>Y. However if x=-2 and y=1, X<Y, but satifies the equation 1 - and cant say x > y or not. Hence Insufficient.

2. X^3-Y^4 >0. Y^4 is always +ve. X^3 can be +ve or -ve. It has to be larger than Y^4 to satisfy the equation. So X>Y always. Sufficient.

Hence B

I would take a C.

statement 2 : what about x = 0.4 y = 0.5
x^3 > y^4 but is x > y ..No
The 3 cases to be considered are negativity, (0,1) range and > 1 range
Combine both
x^5 >2y^5 hence x>y.
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Joined: 31 Dec 1969
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GMAT 2: 660 Q V
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WE: Accounting (Accounting)
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Kudos [?]: 62 [0], given: 76370

 [#permalink] New post 03 Oct 2004, 08:40
agree with B.myfrankenstein note that 4/10*3 is not bigger than 5/10*3.
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 [#permalink] New post 03 Oct 2004, 09:12
1) Insufficient. Easy to see.

2) We can infer that X>0 and moreover:

if x>=1 then y<1. So X>Y.
if 0<x<1 then y could be negative or 0<y<1.
Y could be > X. Consider x=1/4, y=3/4.
B is out.

1,2) I couldn't reach any other conclusion so i pick E.
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 [#permalink] New post 03 Oct 2004, 09:26
Anonymous wrote:
agree with B.myfrankenstein note that 4/10*3 is not bigger than 5/10*3.

it is not multiplication , it's raised to...
x = 0.4 ; y = 0.5
x^3 = .064 ; y^4 = (.0625)

x^3 > y^4 but as yu can see x< y.
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  [#permalink] 03 Oct 2004, 09:26
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