It is currently 13 Dec 2017, 12:55

# Decision(s) Day!:

CHAT Rooms | Ross R1 | Kellogg R1 | Darden R1 | Tepper R1

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1

Author Message
Manager
Joined: 04 Sep 2006
Posts: 113

Kudos [?]: 65 [0], given: 0

Whether xy<1? 1). x + y = 1 2). x^2 + y^2 = 1 [#permalink]

### Show Tags

27 Jun 2009, 21:17
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Whether xy<1?
1). x + y = 1 2). x^2 + y^2 = 1

Kudos [?]: 65 [0], given: 0

Manager
Joined: 28 Jan 2004
Posts: 201

Kudos [?]: 30 [0], given: 4

Location: India

### Show Tags

28 Jun 2009, 17:47
very good explanation. +1 to you.

Kudos [?]: 30 [0], given: 4

Manager
Joined: 07 Apr 2009
Posts: 145

Kudos [?]: 13 [0], given: 3

### Show Tags

19 Jul 2009, 04:08
IMO 'C'

Quote:
very good explanation. +1 to you.

I dont see any explanation !

I cant explain why B is not sufficient in itself.

COming to C

ya$$(x+y)^2 = x^2+y^2+2xy$$

substitute stmt1 & 2, we get xy = 0 , hence < 1

Kudos [?]: 13 [0], given: 3

Retired Moderator
Joined: 05 Jul 2006
Posts: 1748

Kudos [?]: 447 [0], given: 49

### Show Tags

19 Jul 2009, 09:12
vcbabu wrote:
Whether xy<1?
1). x + y = 1 2). x^2 + y^2 = 1

xy< 1 only if xy = fraction ,-ve or 0

from 1

x+y =1 we can tell that xy is not -ve but it could = 0 and we can't tell if xy is a fraction or not.......insuff

from 2

xy could be -ve or 0 or fraction but we can be sure.........insuff

both toegther

the only possible way for x+y =1 , x^2+y^2 = 1 is that one of the variables is 0 and the other is = 1

thus xy = 0 then C is the answer

Kudos [?]: 447 [0], given: 49

Retired Moderator
Joined: 05 Jul 2006
Posts: 1748

Kudos [?]: 447 [0], given: 49

### Show Tags

19 Jul 2009, 09:13
vcbabu wrote:
Whether xy<1?
1). x + y = 1 2). x^2 + y^2 = 1

xy< 1 only if xy = fraction ,-ve or 0

from 1

x+y =1 we can tell that xy is not -ve but it could = 0 and we can't tell if xy is a fraction or not.......insuff

from 2

xy could be -ve or 0 or fraction but we can be sure.........insuff

both toegther

the only possible way for x+y =1 , x^2+y^2 = 1 is that one of the variables is 0 and the other is = 1

thus xy = 0 then C is the answer

Kudos [?]: 447 [0], given: 49

Manager
Joined: 04 Sep 2006
Posts: 113

Kudos [?]: 65 [0], given: 0

### Show Tags

22 Jul 2009, 10:21
OA IS D.

By picking numbers , each is suff.

Kudos [?]: 65 [0], given: 0

Senior Manager
Joined: 04 Jun 2008
Posts: 286

Kudos [?]: 161 [0], given: 15

### Show Tags

22 Jul 2009, 10:33
vcbabu wrote:
Whether xy<1?
1). x + y = 1 2). x^2 + y^2 = 1

D is correct

xy< 1 only when
1) exactly one of x and y is negative
2) they are both fractions <1 (can be negative or positive)

stmt1: x+y cannot be equal to 1 if both are positive integers, or both are negative, or one is negative but other is a positive integer so that xy >1, So this confirms that x+y must be satisfying atleast one of the 2 conditions given above

stmt 2: Again x^2 + y+2 will never be =1 if both are positive/negative integers.

Kudos [?]: 161 [0], given: 15

Manager
Joined: 07 Apr 2009
Posts: 145

Kudos [?]: 13 [0], given: 3

### Show Tags

22 Jul 2009, 11:28
thanks for the explanation ! As I get more aquainted with DS I get more terrified

Kudos [?]: 13 [0], given: 3

Re: inequality   [#permalink] 22 Jul 2009, 11:28
Display posts from previous: Sort by