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# abs >abs ? I. abs >abs II. x<0

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abs >abs ? I. abs >abs II. x<0 [#permalink]  12 Feb 2006, 17:12
abs[x-y] >abs[x-z]?

I. abs[y]>abs[z]
II. x<0
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Re: absolute value [#permalink]  12 Feb 2006, 17:40
joemama142000 wrote:
abs[x-y] >abs[x-z]?

I. abs[y]>abs[z]
II. x<0

IMO E

the Q is asking if we can say with certainity that y is closer to x than z to x or the other way round.

1. does not have any info. on x - insuff.
2. does not have info. on y & z - insuff.

1 & 2 - for a given value of x say -4 you can find different values y & z such that in some cases y is closer to x and in other z is closer. - insuff.
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Re: absolute value [#permalink]  12 Feb 2006, 18:15
believe2 wrote:

IMO E

the Q is asking if we can say with certainity that y is closer to x than z to x or the other way round.

1. does not have any info. on x - insuff.
2. does not have info. on y & z - insuff.

1 & 2 - for a given value of x say -4 you can find different values y & z such that in some cases y is closer to x and in other z is closer. - insuff.

Good explanation believe2

Take egs:
x=-6, y=4 , z =2
(1) & (2) are true.
And |x-y| > |x-z| is true

For
x = -6, y = -4, z=-2
(1) & (2) are true.
But |x-y| > |x-z| is false
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|x-y| > |x-z|
Positive case:
x-y > x-z --> y < z

Negative case:
-x+y > -x+z --> y>z

1) |y| > |z|
Positive case:
y > z --> satisfy the negative case that will make |x-y| > |x-z|

Negative case:
-y > -z --> y < z --> satisfy the postiive case that will make |x-y| > |x-z|

Statment 1 is sufficient.

2) Insufficient. Doesn't tell us anything about y or z.

Ans A
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I came to a completely different solution , I thought C is the answer.

2) jus gives value of x - insuff

1) we need to know the value of x to sub - insuff

taking 1 & 2 ) |y|>|z| & x < 0

consider y =-4 , x = -2 n z=-3 then
|x-y|>|x-z|

if y =4, z= 3 & x= -2 then
|x-y|>|x-z|
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agree with razrulz , got similar values and got C
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i would go with E
to paraphrase the question-
is y at a greater distance from x than Z is from x?
1) y is farther from zero than z - Insuff coz it doesnot talk about X
2) x< 0 - Insuff again because its relative position to Y and Z is needed to conclude if y is farther from X than z
combining both again doesnot tell us anything about relative positions

pls correct the explanation if there any holes in it
thanks
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the oa is E

it is def E

I. x=6, y=3, x=2 no
x=6, y=-3, x=-2 yes
insuff

Both
x=-6, y=3, x=2 yes
x=-6, y=-3, x=-2 no

insuff
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