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# is |x-y| > |x| - |y| ? (1) y < x (2) xy < 0

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Manager
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is |x-y| > |x| - |y| ? (1) y < x (2) xy < 0 [#permalink]

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20 Dec 2005, 09:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

is |x-y| > |x| - |y| ?

(1) y < x
(2) xy < 0
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20 Dec 2005, 09:39
sandalphon wrote:
is |x-y| > |x| - |y| ?

(1) y < x
(2) xy < 0

|x - y| will be greater when "x is greater than 0 and y is less than 0" or vice versa.
For xy to be < 0 the previous statement has to hold true.
Hence Choice B.
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20 Dec 2005, 09:41
Took me much more than 3 min.

I would have guessed this to C.

I spent some more time on this about 2 mins and concluded C

If am wrong then it was an exercise in futility.

Anyway i am 90 % sure it is C and certainly not E

Any easier method to knock this one out?
Director
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20 Dec 2005, 10:07
C too...

|x-y| > |x| - |y| ?

2) x or y must be negative, thus

|x+y| > ... or
|y-x| > ...

insufficient

1) many options

insufficient

1 and 2) y has to be negative and y positive, though

|x+y| > |x| + |y|, while x>0 and y<0, yields no

Maybe I hve made some mistakes, I find this one difficult
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20 Dec 2005, 10:19

The simplest way is to use substitution for X & Y.

Statement I
Use 2 sets of values for X&Y i.e (5,3) and (-3,-5)...Insufficient

Statement II
Again use 2 sets of Values for X & Y (with one being negative)
(3,-5) and (-5,3)... using either of these, the solution can be deduced that |X-Y|>|X| - |Y|.
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20 Dec 2005, 19:25
Got B with substitution
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20 Dec 2005, 20:30
(1)
If y < x, and y is 1, x is 2, then |x-y| = 1 and |x|-|y|. |x-y| !> |x| - |y|
If y < x, and y is -1, x is 2, then |x-y| = 3 and |x|-|y| = 1. |x-y| > |x| - |y|

Insufficient as we can't tell if the inequality holds.

(2) xy < 0 -> x can be negative or y can be negative.

Consider x negative, y positive -> x=-1, y=2
|x-y| = 3, |x|-|y| -1, so |x-y| > |x| - |y|

Consider y negative, x positive -> x=2, y=1
|x-y| = 1, |x|-|y| = 1, so |x-y| = |x| - |y|

Again, we can't tell which inequality holds. Insufficient.

Using (1) and (2), we know x must be positive and y must be negaitve and so |x-y| = |x| - |y|. Thus, ans = C
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20 Dec 2005, 23:52
ywilfred wrote:
(1)
If y < x, and y is 1, x is 2, then |x-y| = 1 and |x|-|y|. |x-y| !> |x| - |y|
If y < x, and y is -1, x is 2, then |x-y| = 3 and |x|-|y| = 1. |x-y| > |x| - |y|

Insufficient as we can't tell if the inequality holds.

(2) xy < 0 -> x can be negative or y can be negative.

Consider x negative, y positive -> x=-1, y=2
|x-y| = 3, |x|-|y| -1, so |x-y| > |x| - |y|

Consider y negative, x positive -> x=2, y=1
|x-y| = 1, |x|-|y| = 1, so |x-y| = |x| - |y|

Again, we can't tell which inequality holds. Insufficient.

Using (1) and (2), we know x must be positive and y must be negaitve and so |x-y| = |x| - |y|. Thus, ans = C

ywilfred, your values x=2, y=1 are incorrect, I guess u meant x=2 and y=-1 , once u take these values your answer shuld be B
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21 Dec 2005, 01:20
Agreed. It`s B. C is the trap answer for lazy people like me.
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21 Dec 2005, 04:08
sandalphon wrote:
is |x-y| > |x| - |y| ?

(1) y < x
(2) xy < 0

NOTE:
|x-y|= x-y if x>y or y-x if y>x
|x|= x if x > 0 or -x if x < 0
|y|= y if y>0 or -y if y<0

1. x=1, y= 0
----> |x-y| = 1
|x|-|y|= 1
---> |x-y|=|x|-|y| ---> answer to the question : NO

x=1, y= -1
|x-y| = 2
|x|-|y| = 0
--->answer to the question : YES
---> insuff

2. xy< 0 --> one is <0 and the other is >0
case 1: y>0> x
----> |x-y| = y-x
|x|-|y|= -x - y
we have -x+y > -x-y (coz y >0) ---> |x-y|> |x|-|y|

case 2: x>0>y
---> |x-y| = x-y
|x|-|y|= x - (-y) = x+y
we have x-y> x + y (coz y <0 )
----> |x-y|> |x|-|y|
----> answer to the question: YES

=> suff

B it is.
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22 Dec 2005, 03:46
sandalphon wrote:
is |x-y| > |x| - |y| ?

(1) y < x
(2) xy < 0

B

Look file for plots.
Attachments

File comment: Look at the file
solution.doc [31.5 KiB]

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22 Dec 2005, 09:29
ywilfred wrote:
(1)
If y < x, and y is 1, x is 2, then |x-y| = 1 and |x|-|y|. |x-y| !> |x| - |y|
If y < x, and y is -1, x is 2, then |x-y| = 3 and |x|-|y| = 1. |x-y| > |x| - |y|

Insufficient as we can't tell if the inequality holds.

(2) xy < 0 -> x can be negative or y can be negative.

Consider x negative, y positive -> x=-1, y=2
|x-y| = 3, |x|-|y| -1, so |x-y| > |x| - |y|

Consider y negative, x positive -> x=2, y=1
|x-y| = 1, |x|-|y| = 1, so |x-y| = |x| - |y|

Again, we can't tell which inequality holds. Insufficient.

Using (1) and (2), we know x must be positive and y must be negaitve and so |x-y| = |x| - |y|. Thus, ans = C

Wrong substitution for X & Y.........in statement II

22 Dec 2005, 09:29
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