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

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laxieqv
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 10:28
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A
B
C
D
E

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Is xy < 1 ?

(I) x + y = 1

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



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vivek123
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 10:39
Laxie buddy, I'm scared of your problems ;)

I'm getting D on this!
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laxieqv
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 10:41
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 :wink:
Btw, how is Tanvi, buddy?!!! ^.^ ....I'm looking forward to her new pix~
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 15:13
I too am for D.
Statement 2 is definitely sufficient.
Statement 1 is also sufficient.
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 15:23
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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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 18:26
Can anyone give a numerical explanation?! :wink:
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 18:57
(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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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 19:31
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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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 19:48
oops :lol:
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 22:03
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 :wink:
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... :lol:
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 22:05
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~ :wink:
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 22:35
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~ :wink:


You mean "D"? ;)
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laxieqv
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   12 Mar 2006, 22:46
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~ :wink:


You mean "D"? ;)


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

OA is D.
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   13 Mar 2006, 10:03
laxieqv wrote:
Is xy < 1 ?
(I) x + y = 1
(II) x^2 + y^2 = 1


good question and good discussions. 8-)
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   13 Mar 2006, 11:11
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~ :wink:


You mean "D"? ;)


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

OA is D.
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   13 Mar 2006, 18:57
Vivek and Laxie, did you give your GMAT already? Just curious on seeing the number of replies you both posted.
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   13 Mar 2006, 20:09
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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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink] New post   13 Mar 2006, 20:14
oh..ok.. you are very accurate. So, I thought you already took it.



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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
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Re: Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink]
13 Mar 2006, 20:14
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