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True/False

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True/False [#permalink] New post 14 Sep 2011, 05:48
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

67% (02:27) correct 33% (00:00) wrong based on 6 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

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Re: True/False [#permalink] New post 14 Sep 2011, 07:09
jamifahad wrote:
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"
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1 KUDOS received
Senior Manager
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Re: True/False [#permalink] New post 14 Sep 2011, 07:25
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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. :lol:

This custom character sometimes puzzles me.
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Re: True/False [#permalink] New post 14 Sep 2011, 07:39
jamifahad wrote:
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. :lol:

This custom character sometimes puzzles me.


That's right. thanks
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Re: True/False [#permalink] New post 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.

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