Author
Message
TAGS:
SVP

Joined: 24 Sep 2005

Posts: 1898

Followers: 10

Kudos [? ]:
104
[0 ] , given: 0

Is xy < 1 ? (I) x + y = 1 (II) x^2 + y^2 = 1 [#permalink ]
12 Mar 2006, 09:28

Question Stats:

0% (00:00) correct

0% (00:00) wrong

based on 0 sessions
Is xy < 1 ?
(I) x + y = 1
(II) x^2 + y^2 = 1

SVP

Joined: 14 Dec 2004

Posts: 1711

Followers: 1

Kudos [? ]:
34
[0 ] , given: 0

Laxie buddy, I'm scared of your problems

I'm getting D on this!

SVP

Joined: 24 Sep 2005

Posts: 1898

Followers: 10

Kudos [? ]:
104
[0 ] , given: 0

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~

Manager

Joined: 20 Feb 2006

Posts: 213

Followers: 1

Kudos [? ]:
5
[0 ] , given: 0

I too am for D.
Statement 2 is definitely sufficient.
Statement 1 is also sufficient.

Manager

Joined: 20 Mar 2005

Posts: 201

Location: Colombia, South America

Followers: 1

Kudos [? ]:
6
[0 ] , given: 0

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

SVP

Joined: 24 Sep 2005

Posts: 1898

Followers: 10

Kudos [? ]:
104
[0 ] , given: 0

Can anyone give a numerical explanation?!

GMAT Club Legend

Joined: 07 Jul 2004

Posts: 5097

Location: Singapore

Followers: 19

Kudos [? ]:
143
[0 ] , given: 0

(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

SVP

Joined: 14 Dec 2004

Posts: 1711

Followers: 1

Kudos [? ]:
34
[0 ] , given: 0

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.

GMAT Club Legend

Joined: 07 Jul 2004

Posts: 5097

Location: Singapore

Followers: 19

Kudos [? ]:
143
[0 ] , given: 0

oops

Senior Manager

Joined: 22 Nov 2005

Posts: 482

Followers: 1

Kudos [? ]:
3
[0 ] , given: 0

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...

SVP

Joined: 24 Sep 2005

Posts: 1898

Followers: 10

Kudos [? ]:
104
[0 ] , given: 0

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~

SVP

Joined: 14 Dec 2004

Posts: 1711

Followers: 1

Kudos [? ]:
34
[0 ] , given: 0

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"?

SVP

Joined: 24 Sep 2005

Posts: 1898

Followers: 10

Kudos [? ]:
104
[0 ] , given: 0

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.

VP

Joined: 29 Dec 2005

Posts: 1351

Followers: 6

Kudos [? ]:
28
[0 ] , given: 0

laxieqv wrote:

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

good question and good discussions.

VP

Joined: 20 Sep 2005

Posts: 1023

Followers: 3

Kudos [? ]:
24
[0 ] , given: 0

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

Joined: 20 Feb 2006

Posts: 213

Followers: 1

Kudos [? ]:
5
[0 ] , given: 0

Vivek and Laxie, did you give your GMAT already? Just curious on seeing the number of replies you both posted.

SVP

Joined: 14 Dec 2004

Posts: 1711

Followers: 1

Kudos [? ]:
34
[0 ] , given: 0

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!

Manager

Joined: 20 Feb 2006

Posts: 213

Followers: 1

Kudos [? ]:
5
[0 ] , given: 0

oh..ok.. you are very accurate. So, I thought you already took it.