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

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Founder
Joined: 04 Dec 2002
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Location: United States (WA)
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19 May 2003, 16:41
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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

Kudos [?]: 28574 [0], given: 5115

Founder
Joined: 04 Dec 2002
Posts: 15600

Kudos [?]: 28574 [0], given: 5115

Location: United States (WA)
GMAT 1: 750 Q49 V42

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19 May 2003, 17:20
brstorewala wrote:
I would say A

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?

Kudos [?]: 28574 [0], given: 5115

Manager
Joined: 28 Feb 2003
Posts: 99

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19 May 2003, 18:35
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

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Founder
Joined: 04 Dec 2002
Posts: 15600

Kudos [?]: 28574 [0], given: 5115

Location: United States (WA)
GMAT 1: 750 Q49 V42

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20 May 2003, 13:07
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

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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Intern
Joined: 11 Nov 2004
Posts: 6

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12 Jan 2005, 07:08
statement -2 is confusiing
surareroot y<x
please expl. why this is not sufficient

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VP
Joined: 18 Nov 2004
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12 Jan 2005, 08:51
"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.

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12 Jan 2005, 08:51
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