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

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Senior Manager
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If x and y are nonzero integers, is (x^-1 + y^-1)^-1 > [#permalink] New post 28 Jun 2007, 07:15
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E

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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: DS: First -ve power [#permalink] New post 28 Jun 2007, 07:35
ArvGMAT 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


The stem can be written as: Is xy/(x+y) > xy

Since x and y are non-zero, we need to check if (x+y) < 1.
Either or both statements seem insufficient to me. Hence, E.
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Re: DS: First -ve power [#permalink] New post 28 Jun 2007, 10:48
sumande wrote:
ArvGMAT 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


The stem can be written as: Is xy/(x+y) > xy

Since x and y are non-zero, we need to check if (x+y) < 1.
Either or both statements seem insufficient to me. Hence, E.


I agree with this. I'm going with E, too.
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Re: DS: First -ve power [#permalink] New post 28 Jun 2007, 14:52
ArvGMAT 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


E for me. OA?
The equation can be reduce down to:
Is 1/(x+y) > 1?
This is the same as saying if 0 < x+y < 1?

(1) x = 2y. We weren't given values of x or y, INSUFFICIENT.

(2) x+y > 0, INSUFFICIENT.

Together, no.
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Re: DS: First -ve power [#permalink] New post 28 Jun 2007, 18:29
ArvGMAT 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


Since x and y are nonzero integers:
(x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1
xy/(x+y) > xy

1: if x = 2y, xy is always greater than xy/(x+y) because:

if x is -ve, xy/(x+y) is always -ve but xy is always +ve. suff.
if x is +ve, xy/(x+y) is always smaller than xy because xy is an integer and if xy is divided by (x+y) then it is always smaller than xy. also suff.


2: if x + y > 0, it doesnot give us any clear answer whether xy/(x+y) or xy is greater. nsf.


so i go with A.
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 [#permalink] New post 02 Jul 2007, 00:34
Since x and y are nonzero integers:
(x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1
xy/(x+y) > xy
1. x=2y the above inequality becomes y/3 > y^2
y is +ve y/3 < y ^ 2
y is -ve y/3 < y ^ 2
Sufficient
2. x+y>0
Case I
x +ve , y -ve; xy/x+y> xy
or
x-ve , y +ve ; xy/x+y> xy
Case II
x+ve , Y+ve ; xy/x+y < xy
Not Sufficient

My vote goes for A
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 [#permalink] New post 02 Jul 2007, 01:26
(A) for me too :)

(x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 ?
<=> 1/[1/x + 1/y] > 1/(1/x * 1/y) ?
=> x*y/(x+y) > x*y ? (as we know that x != 0 and y != 0)
<=> x*y/(x+y) - x*y > 0 ?
<=> x*y * [1/(x+y) - 1] > 0 ?

From 1
x = 2y,

We have so,
x*y * [1/(x+y) - 1]
= 2*y * y *[1/(2*y + y) -1]
= 2*y *[ 1/3 - y]

We have 2 cases:
o If y < 0, then 2*y < 0 and 1/3 - y > 0 as y =< -1 (integer)... thus 2*y *[ 1/3 - y] < 0
o If y > 0, then 2*y > 0 and 1/3 - y < 0 as y >= 1 (integer)... thus 2*y *[ 1/3 - y] < 0

SUFF.

From 2
x + y > 0

Implies that:
1/(x+y) - 1 =< 0 because x and y are integers so x+y >= 1

Thus,
o If x+y=1 (x=3 and y=-2), then x*y * [1/(x+y) - 1] = 0
o If x+y > 1 and x > 0 & y > 0 (x=3 and y=3), then x*y * [1/(x+y) - 1] < 0.

INSUFF.
  [#permalink] 02 Jul 2007, 01:26
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