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Is |x-y| > |x| - |y|? 1) y < x 2) xy < 0

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Intern
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Joined: 23 Jun 2009
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Is |x-y| > |x| - |y|? 1) y < x 2) xy < 0 [#permalink]

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New post 27 Jul 2009, 00:13
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Is |x-y| > |x| - |y|?
1) y < x
2) xy < 0

Kudos [?]: 28 [0], given: 6

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Status: What's your raashee?
Joined: 12 Jun 2009
Posts: 1837

Kudos [?]: 273 [0], given: 52

Location: United States (NC)
Concentration: Strategy, Finance
Schools: UNC (Kenan-Flagler) - Class of 2013
GMAT 1: 720 Q49 V39
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Re: IS |X-Y| > |X| - |Y|? [#permalink]

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New post 27 Jul 2009, 06:07
uzonwagba wrote:
Is |x-y| > |x| - |y|?
1) y < x
2) xy < 0



1) if y<x then we can have:
y=-3 x=-2 -> -2+3 > 2-3 ? true
y=3 x=4 ->4-3 > 4 - 3 You should notice that this will always be true
y= -3 x =2 -> 2--3 > 2- 3 -> 5>-1 ??? false
y=-3 x = 6 -> 6 - -3 > 6-3 9>3 true
INSUFF

2) x*y < 0 tells you one is negative which does not help at all...

Combine:
so now you know that y is negative since y<x so it is SUFF. C
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Director
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Re: IS |X-Y| > |X| - |Y|? [#permalink]

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New post 27 Jul 2009, 09:13
IMO B

Statement 2) alone is sufficient:
--
Reasoning 1: (standard reasoning)
You can punch in numbers to try the same

Statement 1) y<x
if you take both positive
|x - y| will be equal to |x|-|Y|

if you take both negative
|x-y| will be > |x| - |y|

so not satisfactory

Statement 2) xy<0
one is +ve and other is -ve (also x and Y are not zero)
so lets assume there are two positive nos. a,b

two cases
case 1) x=a, y=-b
case 2) x=-a, y=b

LHS=|x-y|=|a-(-b)|=a+b
RHS=|x|-|y| = a-b
for two positive nos. (a+b)>a-b
LHS>RHS

case 2)
LHS=|x-y|=|-a-b|=a+b
RHS=|x|-|y|=a-b
LHS>RHS again

Satisfactory condition hence, B

------------------------------
Other Reasoning: If you want to see it
Little complex (not needed mostly)
------------------------------
LHS=|x-y|
RHS=|x|-|y|

LHs>=0
RHS can be positive and negative, so lets take cases

case 1) RHS < 0
==> LHS>RHS (no need to further check) [satisfy condition]
case 2) RHS=0
==> LHS=RHS when x,y both are of same sign [not satisfy]
==> LHS>RHS when x,y are of opposite signs (e.g. -5, 5) [satisfy]
case 3) RHS >0

as both sides are >0 we can square them, plus keeping the sign of inequality same as '>'

LHS^2=x^2+y^2-2xy
RHS^2=x^2+y^2-2|x||y|
subtract 2nd from 1st
2(|x||y|-xy) >0
which will be true only when xy<0 and (x,y not equal to 0, ***this is included in the condition xy<0)

II satisfy.
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Re: IS |X-Y| > |X| - |Y|? [#permalink]

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New post 27 Jul 2009, 10:37
IMO B.
Just use a list of values for (x,y) -> (3,2),(0,-1), (-3,-5), (4,-3)
Statement 1 will not hold for all
Statement 2 will hold for all.

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Re: IS |X-Y| > |X| - |Y|?   [#permalink] 27 Jul 2009, 10:37
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