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x>1, y>1, is x>y?

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x>1, y>1, is x>y? [#permalink] New post 19 May 2003, 15:41
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00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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Can anybody solve this one?

x>1, y>1, is x>y?
i) squ x > y
ii) squ y < x


bb
Expert Post
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 [#permalink] New post 19 May 2003, 16:20
Expert's post
brstorewala wrote:
I would say A


Well, can you explain please?

i am using this problem for the Question bank and it is an old one. I can't understand commander's last comment here
http://www.gmatclub.com/phpbb/viewtopic.php?t=368&highlight=another+hard


Any thoughts?
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 [#permalink] New post 19 May 2003, 17:35
from statement 1....

square root of any number x (where x>1), is less than the number itself.......
so if the square root of x is greater than y, then obviously x >y always.....so statement 1 is sufficient....

from statement 2.......

Assume y = 9 and x = 4........in this case
sqrt 9 < 4 ( i.e sqrt y < x) and y > x

Assume y = 9 and x = 12.....in this case
sqrt 9 < 12 ( i.e sqrt y < x) and x > y

so statement 2 is insufficient

so answer A
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Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
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GMAT 1: 750 Q49 V42
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Kudos [?]: 14504 [0], given: 3904

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 [#permalink] New post 20 May 2003, 12:07
Expert's post
brstorewala wrote:
from statement 1....

square root of any number x (where x>1), is less than the number itself.......
so if the square root of x is greater than y, then obviously x >y always.....so statement 1 is sufficient....

from statement 2.......

Assume y = 9 and x = 4........in this case
sqrt 9 < 4 ( i.e sqrt y < x) and y > x

Assume y = 9 and x = 12.....in this case
sqrt 9 < 12 ( i.e sqrt y < x) and x > y

so statement 2 is insufficient

so answer A


:oops: Sorry for taking up your time. I was doing this one late at night and just did not understand the question. It is a one line explanation; sorry about that :)

bb
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please expl [#permalink] New post 12 Jan 2005, 06:08
statement -2 is confusiing
surareroot y<x
please expl. why this is not sufficient
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 [#permalink] New post 12 Jan 2005, 07:51
"A" it is . From statment 1 we have x> y^2.....and for all x and y > 1 , x has to be > y...Suff

From statement 2 we have y < x^2...for x = 2 and y = 3, it is true, so x<y in this case , but for x = 4 and y = 3, above exp is still true but x > y...thus insuff.
  [#permalink] 12 Jan 2005, 07:51
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