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# If zy < xy < 0, is |x-z| + |x| = |z|? (1) z<x (2)

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If zy < xy < 0, is |x-z| + |x| = |z|? (1) z<x (2) [#permalink]  31 Jul 2008, 05:24
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If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z<x

(2) y>0

I usually have problems interpreting the absolute value expressions such as |x-z| + |x| = |z|. I do know at least that this expression can be rearranged into |x-z| = |z| - |x|
What does this expression mean??
Thanks
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Re: DS: using absolute values [#permalink]  31 Jul 2008, 06:10
tarek99 wrote:
If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z<x

(2) y>0

I usually have problems interpreting the absolute value expressions such as |x-z| + |x| = |z|. I do know at least that this expression can be rearranged into |x-z| = |z| - |x|
What does this expression mean??
Thanks

zy < xy < 0
--> x and y must have oppostive sigins
z and y must have oppostive sigins
--> x and z have same signs.
1) if z<x State 1 is not sufficient alone
Assume x and z are +ve numbers
--> x-z>0
|x-z| + |x| = |z|
--> x-z+x=z --> 2x-z =z
-- alwasy false if x, z are +ve numbers
Assume x and z are -ve numbers
-- > z<x --> |z|>|x| ( E.G -2 >-1 |-2| >|-1|)
( x=-1 z= -2)
|x-z| + |x| = |z| -->
--> |x-z| =|z| -|x| when other z and x are -ve and z<x
--> always true if x ,z are negative sign

2) if y>0 x and z must be -ve numbers.
then |x-z| + |x| = |z|.
assume that z<x --> |z|>|x|
|x-z| + |x| = |z| -- always true
assume that z>x --> |z|<|x|
|x-z| + |x| = |z| -- alwasy false

Statement 2 alone not sufficient.

Combine two..
z<x and y>0 --> both are negative numbers |z|>|x|
|x-z| + |x| = |z| -- alwasy true.

Both Statements combined can answer the question.
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Re: DS: using absolute values [#permalink]  31 Jul 2008, 06:25
tarek99 wrote:
If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z<x

(2) y>0

I usually have problems interpreting the absolute value expressions such as |x-z| + |x| = |z|. I do know at least that this expression can be rearranged into |x-z| = |z| - |x|
What does this expression mean??
Thanks

|x-z| + |x| = |z| ------> |x-z| = |z| - |x|
--> means
x and z are -ve numbers and z<x
or
x and z are +ve numbers and z>x
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Re: DS: using absolute values [#permalink]  31 Jul 2008, 07:15
tarek99 wrote:
If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z<x

(2) y>0

i get D.
1)
z<x which means y>0 and z, x<0

if z=1/2 x=1/4

|-1/4+1/2|=1/4+1/4=1/2 =|z|..

if z=-2 x=-1; -1+2=1 1+1=2 whichis equal to |z|... Sufficient
2)
if y>0
then |z|>|x|

z=-1/2 and x=-1/4 |-1/2+1/4|=1/4+1/4=1/2=|z|

if z=-10 x=-1
|-1+10|=9; 9+1=10 |z|

therefore sufficient..
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Re: DS: using absolute values [#permalink]  31 Jul 2008, 15:13
lol , we have a C, a D, and now an E from me.

i rearranged the given inequality to get y(z-x)<0 , and from this you get two possibilities: either y>0 and z-x<0 or y<0 and z-x>0

from stat 1, we know that z<x and therefore y>0. try z=-2 and x=5 and z=-3 and x=-2, you get different answers. insuff.

from stat 2, y>0 so therefore z<x. same situation as above.

together, you dont know anything new.
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Re: DS: using absolute values [#permalink]  31 Jul 2008, 16:00
I attempted this question and got D as well. Please post the OA so our doubts can be laid to rest.

If zy < xy < 0, is |x-z| + |x| = |z|?
(1) z<x
(2) y>0

From the original equation it is clear that y has to be positive and x and z are negative

S1. States the same thing that we already know from the original eauation
Let x = -1 and y = -2
Then |-1 - (-2)| + |-1| = |-2| ; Hence 2 = 2 and BCE are out

S2. Also states what the original equation tells us - that y is positive
Hence using the same reasoning as in S1. this will also give the answer

IMO D
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Re: DS: using absolute values [#permalink]  31 Jul 2008, 17:00
First, if you know Statement 1 is true, then you know Statement 2 must be true and vice versa, because we are told that zy < xy < 0. The answer must be D or E. If we know y > 0, we know z and x are negative, and we have

z < x < 0 < y

The question asks if |x - z| + |x| = |z|. In terms of distances, it asks if the distance from z to x plus the distance from x to zero is equal to the distance from z to zero. This will clearly be true if z < x < 0 (draw it on the number line), so each statement is sufficient.

That said, looking at the inequality given in the question, zy < xy < 0, there are two possibilities:

z < x < 0 < y
or
y < 0 < x < z

In either case, |x - z| + |x| = |z| should be true. I don't see why we need the statements at all. Am I missing something? End of a long day...
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Re: DS: using absolute values [#permalink]  31 Jul 2008, 17:50
tarek99 wrote:
If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z<x

(2) y>0

I usually have problems interpreting the absolute value expressions such as |x-z| + |x| = |z|. I do know at least that this expression can be rearranged into |x-z| = |z| - |x|
What does this expression mean??
Thanks

|m | means that if we take any value[m] mode of that value[m] is always positive

consider the expr |x-z| = |z| - |x|
here mode of difference of any two numbers is compared with difference of modes of each number.

this expression is true when z<x and both x<0,z<0 -> condn (A)
only then LHS |x-z|= |-(z-x)|=z-x
RHS |z|-|x|=z-x always
LHS=RHS

now to make condn (A) true :
provided zy<xy (1) is isufficient if z<x then y>0 and zy<0=>z<0 and xy<0=>x<0 =>x<0,z<0 andz<x => A is true. (1) is sufficient

provided zy<xy (2) is sufficient y>0 => z<x and z<0,x<0 => A is true

D is the answer for me

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Re: DS: using absolute values [#permalink]  31 Jul 2008, 21:27
x97agarwal wrote:
I attempted this question and got D as well. Please post the OA so our doubts can be laid to rest.

If zy < xy < 0, is |x-z| + |x| = |z|?
(1) z<x
(2) y>0

From the original equation it is clear that y has to be positive and x and z are negative

S1. States the same thing that we already know from the original eauation
Let x = -1 and y = -2
Then |-1 - (-2)| + |-1| = |-2| ; Hence 2 = 2 and BCE are out

S2. Also states what the original equation tells us - that y is positive
Hence using the same reasoning as in S1. this will also give the answer

IMO D

Second Look I agree with D.

I missed the point z<x and y>0 ---> x,z can't be positive integers..

So it leads to option D.
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Re: DS: using absolute values [#permalink]  31 Jul 2008, 21:31
pmenon wrote:
lol , we have a C, a D, and now an E from me.

i rearranged the given inequality to get y(z-x)<0 , and from this you get two possibilities: either y>0 and z-x<0 or y<0 and z-x>0

from stat 1, we know that z<x and therefore y>0. try z=-2 and x=5 and z=-3 and x=-2, you get different answers. insuff.

from stat 2, y>0 so therefore z<x. same situation as above.

together, you dont know anything new.

Hi Pmenon,
It doesn't matter how many possibilities you get.. Question is not to find value of x and z?
question is |x-z| + |x| = |z| ?

your choice z=-2 and x=5 is not correct. when y>0 x and y must have same signs and they must be -ve.
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Re: DS: using absolute values   [#permalink] 31 Jul 2008, 21:31
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