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# True/False

Author Message
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Senior Manager
Joined: 03 Mar 2010
Posts: 440
Schools: Simon '16 (M)
Followers: 5

Kudos [?]: 191 [0], given: 22

True/False [#permalink]  14 Sep 2011, 05:48
00:00

Difficulty:

(N/A)

Question Stats:

57% (02:27) correct 43% (00:16) wrong based on 7 sessions
As always, post your timing for below question, if possible.

x#y= (x−y)^2, if x>y
x#y= x+$$y/4$$, if x≤y

If x # y = –1, which of the following could be true?
I. x=y
II. x>y
III. x<y
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III
[Reveal] Spoiler: OA

_________________

My dad once said to me: Son, nothing succeeds like success.

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 145

Kudos [?]: 1297 [0], given: 376

Re: True/False [#permalink]  14 Sep 2011, 07:09
As always, post your timing for below question, if possible.

x#y= (x−y)^2, if x>y
x#y= x+$$y/4$$, if x≤y

If x # y = –1, which of the following could be true?
I. x=y
II. x>y
III. x<y
(A) I only
(B) II only
(C) III only
(D) I and III only
(E) I, II, and III

(2:05)

A square can never be -ve, ruled out II.

For x=y;

x+x/4=-1; 5x=-4; x=-4/5. Possible

For x>y
x+y/4=-1

x=-1-y/4=-(y+4)/4; y=-4, x=0. Possible.

Ans: "D"
_________________
Senior Manager
Joined: 03 Mar 2010
Posts: 440
Schools: Simon '16 (M)
Followers: 5

Kudos [?]: 191 [1] , given: 22

Re: True/False [#permalink]  14 Sep 2011, 07:25
1
KUDOS
fluke wrote:
For x>y
x+y/4=-1

x=-1-y/4=-(y+4)/4; y=-4, x=0. Possible.

This should be x<y.
Other than that flawless.

This custom character sometimes puzzles me.
_________________

My dad once said to me: Son, nothing succeeds like success.

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2022
Followers: 145

Kudos [?]: 1297 [0], given: 376

Re: True/False [#permalink]  14 Sep 2011, 07:39
fluke wrote:
For x>y
x+y/4=-1

x=-1-y/4=-(y+4)/4; y=-4, x=0. Possible.

This should be x<y.
Other than that flawless.

This custom character sometimes puzzles me.

That's right. thanks
_________________
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Joined: 11 May 2011
Posts: 373
Location: US
Followers: 3

Kudos [?]: 71 [0], given: 46

Re: True/False [#permalink]  14 Sep 2011, 13:16
Time: 1:00 min.

Approach -
Since, x # y = –1
=> x#y= (x−y)^2, if x>y - This equation is not possible becasue any square will yield positive value.
=> x#y= x+y/4, if x≤y - This equation is only possible and as it says - x≤y, Hence - D.

Cheers!
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Re: True/False   [#permalink] 14 Sep 2011, 13:16
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