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# M27#12

Author Message
Intern
Joined: 28 Nov 2011
Posts: 21

Kudos [?]: 11 [0], given: 7

Location: United States
GMAT 1: 640 Q42 V35
GMAT 2: 710 Q47 V40
GPA: 2.99

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13 Sep 2012, 17:04
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Is xy≤1/2?

(1) x^2+y^2=1
(2) x^2−y^2=0

OFFICIAL EXPLANATION
(1) x^2+y^2=1. Recall that (x−y)2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
(2) x^2−y^2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.

I am fine with (2). I have trouble with (1).

What if, instead of using x^2-2xy+y^2≥0, I decided to use x^2+2xy+y^2≥0 (note the positive). That would result in xy≥-1/2 instead of the xy≤1/2 that is sufficient. In the end, I would have to use x^2-2xy+y^2≥0?

Anyone care to elaborate on this please?

Kudos [?]: 11 [0], given: 7

Math Expert
Joined: 02 Sep 2009
Posts: 42618

Kudos [?]: 135758 [0], given: 12708

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14 Sep 2012, 03:15
lesliehh wrote:
Is xy≤1/2?

(1) x^2+y^2=1
(2) x^2−y^2=0

OFFICIAL EXPLANATION
(1) x^2+y^2=1. Recall that (x−y)2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
(2) x^2−y^2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.

I am fine with (2). I have trouble with (1).

What if, instead of using x^2-2xy+y^2≥0, I decided to use x^2+2xy+y^2≥0 (note the positive). That would result in xy≥-1/2 instead of the xy≤1/2 that is sufficient. In the end, I would have to use x^2-2xy+y^2≥0?

Anyone care to elaborate on this please?

Yes, to solve this question you should use (x−y)^2≥0 not (x+y)^2≥0.
_________________

Kudos [?]: 135758 [0], given: 12708

Intern
Joined: 30 May 2011
Posts: 18

Kudos [?]: 7 [0], given: 4

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18 Feb 2013, 14:38
Because square of any number is more than or equal to zero). so x2−2xy+y2≥0 is used instead of x2+2xy+y2≥0?????????????????????????????

Bunuel wrote:
lesliehh wrote:
Is xy≤1/2?

(1) x^2+y^2=1
(2) x^2−y^2=0

OFFICIAL EXPLANATION
(1) x^2+y^2=1. Recall that (x−y)2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
(2) x^2−y^2=0. Re-arrange and take the square root from both sides: |x|=|y|. Clearly insufficient.

I am fine with (2). I have trouble with (1).

What if, instead of using x^2-2xy+y^2≥0, I decided to use x^2+2xy+y^2≥0 (note the positive). That would result in xy≥-1/2 instead of the xy≤1/2 that is sufficient. In the end, I would have to use x^2-2xy+y^2≥0?

Anyone care to elaborate on this please?

Yes, to solve this question you should use (x−y)^2≥0 not (x+y)^2≥0.

Kudos [?]: 7 [0], given: 4

Senior Manager
Joined: 22 Nov 2010
Posts: 285

Kudos [?]: 182 [0], given: 75

Location: India
GMAT 1: 670 Q49 V33
WE: Consulting (Telecommunications)

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21 Mar 2013, 02:03
Bunuel,

I am not able to understand this explanation. What triggered you to choose this method to find the solution?

OFFICIAL EXPLANATION
(1) x^2+y^2=1. Recall that (x−y)2≥0 (square of any number is more than or equal to zero). Expand: x^2−2xy+y^2≥0 and since x2+y2=1 then: 1−2xy≥0. So, xy≤1/2. Sufficient.
_________________

YOU CAN, IF YOU THINK YOU CAN

Kudos [?]: 182 [0], given: 75

Re: M27#12   [#permalink] 21 Mar 2013, 02:03
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# M27#12

Moderator: Bunuel

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