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# Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1

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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink]

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12 Mar 2006, 09:28
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Is xy < 1 ?

(I) x + y = 1

(II) x^2 + y^2 = 1
SVP
Joined: 14 Dec 2004
Posts: 1702
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Kudos [?]: 137 [0], given: 0

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12 Mar 2006, 09:39
Laxie buddy, I'm scared of your problems

I'm getting D on this!
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12 Mar 2006, 09:41
vivek123 wrote:
Laxie buddy, I'm scared of your problems

I'm getting D on this!

hik, this is not my problem. I just picked it from somewhere
Btw, how is Tanvi, buddy?!!! ^.^ ....I'm looking forward to her new pix~
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12 Mar 2006, 14:13
I too am for D.
Statement 2 is definitely sufficient.
Statement 1 is also sufficient.
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Re: DS inequality [#permalink]

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12 Mar 2006, 14:23
laxieqv wrote:
Is xy < 1 ?

(I) x + y = 1

(II) x^2 + y^2 = 1

the thing with D is that x or y can be negative (not both) then xy would be negative

like any value on the second and third quadrant of a cartesian plane

that's why I wouldn't pick D, I go with A on this one
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12 Mar 2006, 17:26
Can anyone give a numerical explanation?!
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12 Mar 2006, 17:57
(1) Can be:
x = 1/2, y = 1/2 xy < 1
x = 2, y = -1 xy < 1

No other case will result in xy < 1 and still satisfy x+y=1 (at least I can't think of any right now)

Sufficient

(2) Can be x = 1, y = 0 xy < 1. Can be x = 1/sqrt(2), y = 1/sqrt(2)xy > 1. Insufficient.

I go with A
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12 Mar 2006, 18:31
ywilfred wrote:
(1) Can be:
x = 1/2, y = 1/2 xy < 1
x = 2, y = -1 xy < 1

No other case will result in xy < 1 and still satisfy x+y=1 (at least I can't think of any right now)

Sufficient

(2) Can be x = 1, y = 0 xy < 1. Can be x = 1/sqrt(2), y = 1/sqrt(2)xy > 1. Insufficient.

I go with A

Wilfred,

If x = 1/sqrt(2), y = 1/sqrt(2)
then xy = [1/sqrt(2)] * [1/sqrt(2)] = 0.5 < 1.
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12 Mar 2006, 18:48
oops
Senior Manager
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12 Mar 2006, 21:03
laxieqv wrote:
vivek123 wrote:
Laxie buddy, I'm scared of your problems

I'm getting D on this!

hik, this is not my problem. I just picked it from somewhere
Btw, how is Tanvi, buddy?!!! ^.^ ....I'm looking forward to her new pix~

lexi what is the source. I also want to solve these problems...
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Re: DS inequality [#permalink]

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12 Mar 2006, 21:05
laxieqv wrote:
Is xy < 1 ?

(I) x + y = 1

(II) x^2 + y^2 = 1

(1) x+y= 1 --> x= 1-y
---> xy= (1-y)*y= y- y^2 = - ( y^2 -y + 1/4) +1/4 = - ( y-1/2)^2 + 1/4
Because - (y-1/2) ^2 <= 0 ---> - (y-1/2) ^2 +1/4 <= 1/4 < 1
---> suff

(2) x^2+y^2 - 2xy = (x-y)^2 >= 0 ---> x^2+y^2 >= 2xy ---> 2xy<=1 -----> xy<=1/2< 1
---> suff

OA is C.

Good job, buddies~
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Re: DS inequality [#permalink]

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12 Mar 2006, 21:35
laxieqv wrote:
laxieqv wrote:
Is xy < 1 ?

(I) x + y = 1

(II) x^2 + y^2 = 1

(1) x+y= 1 --> x= 1-y
---> xy= (1-y)*y= y- y^2 = - ( y^2 -y + 1/4) +1/4 = - ( y-1/2)^2 + 1/4
Because - (y-1/2) ^2 <= 0 ---> - (y-1/2) ^2 +1/4 <= 1/4 < 1
---> suff

(2) x^2+y^2 - 2xy = (x-y)^2 >= 0 ---> x^2+y^2 >= 2xy ---> 2xy<=1 -----> xy<=1/2< 1
---> suff

OA is C.
Good job, buddies~

You mean "D"?
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Re: DS inequality [#permalink]

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12 Mar 2006, 21:46
vivek123 wrote:
laxieqv wrote:
laxieqv wrote:
Is xy < 1 ?

(I) x + y = 1

(II) x^2 + y^2 = 1

(1) x+y= 1 --> x= 1-y
---> xy= (1-y)*y= y- y^2 = - ( y^2 -y + 1/4) +1/4 = - ( y-1/2)^2 + 1/4
Because - (y-1/2) ^2 <= 0 ---> - (y-1/2) ^2 +1/4 <= 1/4 < 1
---> suff

(2) x^2+y^2 - 2xy = (x-y)^2 >= 0 ---> x^2+y^2 >= 2xy ---> 2xy<=1 -----> xy<=1/2< 1
---> suff

OA is C.
Good job, buddies~

You mean "D"?

Ah,hik, this is the consequence of staying up late till 4 am

OA is D.
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Re: DS inequality [#permalink]

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13 Mar 2006, 09:03
laxieqv wrote:
Is xy < 1 ?
(I) x + y = 1
(II) x^2 + y^2 = 1

good question and good discussions.
VP
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Re: DS inequality [#permalink]

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13 Mar 2006, 10:11
D it is. I freaked out for a moment when you posted the OA is C.

laxieqv wrote:
vivek123 wrote:
laxieqv wrote:
laxieqv wrote:
Is xy < 1 ?

(I) x + y = 1

(II) x^2 + y^2 = 1

(1) x+y= 1 --> x= 1-y
---> xy= (1-y)*y= y- y^2 = - ( y^2 -y + 1/4) +1/4 = - ( y-1/2)^2 + 1/4
Because - (y-1/2) ^2 <= 0 ---> - (y-1/2) ^2 +1/4 <= 1/4 < 1
---> suff

(2) x^2+y^2 - 2xy = (x-y)^2 >= 0 ---> x^2+y^2 >= 2xy ---> 2xy<=1 -----> xy<=1/2< 1
---> suff

OA is C.
Good job, buddies~

You mean "D"?

Ah,hik, this is the consequence of staying up late till 4 am

OA is D.
Manager
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13 Mar 2006, 17:57
Vivek and Laxie, did you give your GMAT already? Just curious on seeing the number of replies you both posted.
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13 Mar 2006, 19:09
chuckle wrote:
Vivek and Laxie, did you give your GMAT already? Just curious on seeing the number of replies you both posted.

Me? Not yet! Need to face the ghost sometime!
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13 Mar 2006, 19:14
oh..ok.. you are very accurate. So, I thought you already took it.
13 Mar 2006, 19:14
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# Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1

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