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laxieqv
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 10:28
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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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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 10:39
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Laxie buddy, I'm scared of your problems ;)

I'm getting D on this!
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laxieqv
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 10:41
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vivek123
Laxie buddy, I'm scared of your problems ;)

I'm getting D on this!
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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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chuckle
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 15:13
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I too am for D.
Statement 2 is definitely sufficient.
Statement 1 is also sufficient.
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conocieur
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 15:23
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laxieqv
Is xy < 1 ?

(I) x + y = 1

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


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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laxieqv
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 18:26
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Can anyone give a numerical explanation?! :wink:
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ywilfred
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 18:57
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(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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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 19:31
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ywilfred
(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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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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ywilfred
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 19:48
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oops :lol:
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gmat_crack
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 22:03
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laxieqv
vivek123
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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lexi what is the source. I also want to solve these problems... :lol:
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laxieqv
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 22:05
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laxieqv
Is xy < 1 ?

(I) x + y = 1

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



(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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vivek123
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 22:35
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laxieqv
laxieqv
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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You mean "D"? ;)
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laxieqv
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 12 Mar 2006, 22:46
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vivek123
laxieqv
laxieqv
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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Ah,hik, this is the consequence of staying up late till 4 am :oops:

OA is D.
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Professor
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 13 Mar 2006, 10:03
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laxieqv
Is xy < 1 ?
(I) x + y = 1
(II) x^2 + y^2 = 1
Show more


good question and good discussions. 8-)
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 13 Mar 2006, 11:11
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D it is. I freaked out for a moment when you posted the OA is C.



laxieqv
vivek123
laxieqv
laxieqv
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.
Show more
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 13 Mar 2006, 18:57
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Vivek and Laxie, did you give your GMAT already? Just curious on seeing the number of replies you both posted.
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vivek123
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 13 Mar 2006, 20:09
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chuckle
Vivek and Laxie, did you give your GMAT already? Just curious on seeing the number of replies you both posted.
Show more


Me? Not yet! Need to face the ghost sometime! :)
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chuckle
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Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 13 Mar 2006, 20:14
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oh..ok.. you are very accurate. So, I thought you already took it.



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