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If x and y are nonzero integers, is (x^-1 + y^-1)^-1 >

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Manager
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If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [#permalink]

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New post 23 Sep 2009, 16:45
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If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?

(1) x = 2y
(2) x + y > 0
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Re: Q w/o OA [#permalink]

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New post 23 Sep 2009, 19:15
This is a tough one.
Basically you can reduce the condition to is xy/x+y > xy?
Pluggin number x=-2, y=-4 the condition is false.
x=2, y=4, again false
x=2, y=-4, x=-2,y=4 - condition is true
The only time this condition is true is when one of the number is negative.

i. x=2y
y=-2, x=-4 false, y=2, x=4 false, since x=2y x always takes y sign that means they both are positive or negative and you can answer the above is question.
hence sufficient.

ii. x+y>0
x=2, y=4 -> false, x=-2, y=4 -> true
hence insufficient.

So answer is A IMO.
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Re: Q w/o OA [#permalink]

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New post 24 Sep 2009, 10:53
nikhilpoddar wrote:
If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?

(1) x = 2y
(2) x + y > 0


is xy/(x+y)>xy ?

1, x=2y => xy[1/(x+y)-1]>0
=> 2y^2(1/3y-1)>0
=> 1/3y-1>0, we dont know yet, insuff
2, x+y>0 => xy(1-x-y)>o, 3 cases
x+, y+
x-, y+
x+, y-
insuff
Both,
x=2y
x+y>0 => x>0, y>0 => 1/(x+y)>1 =>1/3y>1 => y>3 we dont now yet
insuff

Answer is E
Answer is E
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Re: Q w/o OA [#permalink]

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New post 24 Sep 2009, 22:05
I know there is no OA so its hard to tell . Because I think if x & y are both positive or negative integers then the condition is always false. and if x=2y that means they both are either positive or negative.
say x=-1 y=-1 xy=1 x+y= -2 is -1/2 > 1 NO
or x=1 y=1 is 1/2 > 1 NO
or x=-3 y=-4 is -12/7 > 12 NO
you take any numbers you will get an answer "NO"
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Re: Q w/o OA [#permalink]

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New post 26 Sep 2009, 01:32
Agreed it's E

if xy/(x+y)>xy

St(1) x=2y
xy/(x+y)>xy become 2y^2/3y > 2y^2; insufficient cause y can be +ve or -ve

St(2) x+y>0 means x and y are +ve
xy/(x+y) won't be >xy
Re: Q w/o OA   [#permalink] 26 Sep 2009, 01:32
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If x and y are nonzero integers, is (x^-1 + y^-1)^-1 >

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