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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 01 May 2005, 14:30
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Is x>y?

(1). x^2>2y

(2). x^3>y^4
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 [#permalink] New post 01 May 2005, 17:29
E.

for the 2nd statement x and y can be equal to 1/2.
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 [#permalink] New post 01 May 2005, 19:15
prep_gmat wrote:
E.

for the 2nd statement x and y can be equal to 1/2.


u do have a point here :idea:

rt the Ans is E
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Re: DS greater than [#permalink] New post 01 May 2005, 21:07
MA wrote:
B.....
i is not suff
ii is suff.

E is ok. i do not remember what i did. careless mistake:wall
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 [#permalink] New post 02 May 2005, 06:50
I think it is "C"....I can't find any case when X is <= Y....

state 2...gives us that X is always > 0 as X^3 > Y^4....and we know Y^4 is always > 0...

combining both I don't see anyway X can be <= Y....can anyone give an example to refute this.
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Re: DS greater than [#permalink] New post 03 May 2005, 00:56
saurya_s wrote:
Is x>y?

(1). x^2>2y

(2). x^3>y^4


My answer is E.

Statement 1) is not sufficient because if x = 1 and y = 1 then 1 > 2 false.
if x = 3 and y = 3 then 9 > 6 true. Hence statement 1 alone is not sufficent.

Statement 2) is not sufficient because if x = 1 and y = 1 then 1 > 1 false.
if x = 2 and y = 1 then 8 > 1 which is true, hence statement 2 alone is not sufficient.

Combining statement 1 and statement 2.
Divide statement 2 by statement 1 which gives x > 1/2 (Y^3)

Let x = 1 and y = 1 ==> 1 > 1/2 true.
Let x =2 and y = 2 ==> 2 > 1/2 (8) => 2 > 4 false.

Hence A & B together is also not sufficient. Hence the answer could only be E.

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 [#permalink] New post 03 May 2005, 04:34
saurabhmalpani wrote:
prep_gmat wrote:
E.

for the 2nd statement x and y can be equal to 1/2.


u do have a point here :idea:

rt the Ans is E


What am I doing ..I guess sud sleep more :-D but I think that Banerjee has a point here.

Statement 1: on a number lie we c that X cannot lie between -2 & 2
----------I-----------I-----------------
-2 2

Statement 2: only when x=y=1/2 the statement is wrong.

but if we combine statement 1 and 2 we find that x cannot be 1/2

So C is the answer.
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 [#permalink] New post 03 May 2005, 08:39
E, it doesn`t say x or y must be intergers
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 [#permalink] New post 04 May 2005, 01:52
GMATT73 wrote:
E, it doesn`t say x or y must be intergers


ok try with fractional numbers I still think Ans is C


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 [#permalink] New post 21 Jun 2005, 13:08
can anybody shed light on this both by considering integers and fractions?
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 [#permalink] New post 21 Jun 2005, 14:27
:? hello
don 't you have an OG answer
I would have selected B but


the key is to pick negative and fractions since the stem does not specify it

x^2>2y

when x=-2 AND y=1 x<y this meets the statement condition
when x=4 AND y=1 x>y

it work both way so we can 't tell for sure whether x is bigger than y so insufficient


ST2
x^3>y^4

TELL us x is positive because odd power x^3>y^4

picking positive integer such as x=3 AND y=2
We got 27>16 THIS Meets the statement conditions
when x<y for instance x=2 AND y =3 x^3<y^4 So contradicts the statement
with fraction
x=1/2 and y=1/3 x>y meets the statement since x^3>y^4 1/8>1/81

Inverting y=1/2 and x=1/3 contradict the statement


picking x=1/2 AND y=-1/2
still in the statement

the only way would be if x =y
X=Y=1/3
1/27>1/81

hence statement two is not
since both statement don 't specify that x is different from y neither help
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 [#permalink] New post 21 Jun 2005, 18:48
S1 is not sufficient. Just plug in (x=-2,y=-1 and x=2,y=-1). Rule out A and D

S2 alone is not sufficient as well. Just plug in (x = 1/2,y = 1/2) and (x=1,y=1/2). We get x=y in first case and x > y in the second case. Rule out B.

We are left with C and E

If I got this far, I'd guess C and move on.
  [#permalink] 21 Jun 2005, 18:48
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