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Interesting GMATFix Problem-17

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Interesting GMATFix Problem-17

by arora007 » Tue Sep 21, 2010 12:55 pm
If x and y are distinct non-zero integers, is |x+y| = |x| - |y| ?
1) xy<2
2) x-y >0
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Source: — Data Sufficiency |
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by clock60 » Tue Sep 21, 2010 2:40 pm
hope the reduced question asks is xy<0
(1)xy<2 sufficent, as x,y are distinct integers, and neither is equal to 0, they must have opposite signs to be less than 2
if they are both -ve the product will be +ve. they can`t be -1*-2=2 as we need xy<2, and they can`t be 1*2=2 as again xy<2,so one of them must be +ve and one -ve
suff
(2) many options are possible if x>y
x=2,y=1 xy>0 the answer is no
x=1,y=-2 xy<0 the answer is yes

my pick for A
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by arora007 » Tue Sep 21, 2010 3:39 pm
A is wrong
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by clock60 » Tue Sep 21, 2010 9:47 pm
honestly, i checked solution one more time but didn`t spot mistake,
waiting for the reply from others,
only one question arora
may be st 1 is written
x*y<=2? in this case i see that st 1 insufficient
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by gmatmachoman » Wed Sep 22, 2010 12:14 am
1) xy<2

@ clock

case 1:

x= 3 y = -1

xy <2

|x+y| = |x| - |y|
YES

case 2:

x= -3, Y = 1

|x+y| = |x| - |y|

|-3+1| NOT EQUAL to |x| - |y|

St 1 is insufficient

Combining St1 & st2 (x>Y)

use the example of case 1

Pick C
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by clock60 » Wed Sep 22, 2010 1:01 am
dear gmatmachoman
i almost agree with your reasoning, but one thing is confusing, with your examples

x=3, y=-1
|x+y|=|3-1|=|2|=2
|x|-|y|=|3|-|-1|=3-1=2, completely agree

x=-3, y=1
|x+y|=|-3+1|=|-2|=2
|x|-|y|=|-3|-|1|=3-1=2 with my calculations

but in your solving you came that if with x=-3.y=1 |x+y| is not equal |x|+|y|????
can you elaborate a little bit more?
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by gmatmachoman » Wed Sep 22, 2010 1:41 am
clock60 wrote:dear gmatmachoman
i almost agree with your reasoning, but one thing is confusing, with your examples

x=3, y=-1
|x+y|=|3-1|=|2|=2
|x|-|y|=|3|-|-1|=3-1=2, completely agree

x=-3, y=1
|x+y|=|-3+1|=|-2|=2
|x|-|y|=|-3|-|1|=3-1=2 with my calculations

but in your solving you came that if with x=-3.y=1 |x+y| is not equal |x|+|y|????

LHS |-3+1| = |-2| = 2

RHS : |-3| + |1| = 3 +1 = 4

LHS is NOT EQUAL to RHS



can you elaborate a little bit more?
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by clock60 » Wed Sep 22, 2010 2:07 am
LHS |-3+1| = |-2| = 2

RHS : |-3| + |1| = 3 +1 = 4

LHS is NOT EQUAL to RHS
why do you add? in the problem we must substract
|x+y| = |x| - |y| ?
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by arora007 » Wed Sep 22, 2010 2:36 am
plugin values, find E as the answer.
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by CrazyGmatter » Wed Sep 22, 2010 9:52 pm
can you post the official explanation pls...
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