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

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

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

28 Jun 2007, 08:15

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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]

28 Jun 2007, 08:35

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]

28 Jun 2007, 11:48

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]

28 Jun 2007, 15:52

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]

28 Jun 2007, 19:29

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]

02 Jul 2007, 01:34

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]

02 Jul 2007, 02:26

(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, 02:26

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

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