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

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Intern
Joined: 17 Apr 2007
Posts: 24
If zy < xy < 0, is |x-z| + |x| = |z|? (1) z<x>0 [#permalink]

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13 May 2007, 12:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z<x>0

OA is D
Intern
Joined: 17 Apr 2007
Posts: 24

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13 May 2007, 12:41
oops...should be

(1) z is less than x
(2) y is greater than 0

Last edited by kookoo4tofu on 13 May 2007, 12:42, edited 1 time in total.
Senior Manager
Joined: 03 May 2007
Posts: 270

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13 May 2007, 12:42
can you post the rest of the question?
VP
Joined: 08 Jun 2005
Posts: 1145

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13 May 2007, 13:35
|x-z| + |x| = |z|

can be rewritten as

|x-z| = |z|-|x|

Statement 1

whenever statement 1 is true |z|-|x| is negative.

|x-z| can never be negative

so we can say for sure that when z is less then x , the inequality is false.

sufficient

Statement 2

when y is grater then 0 then x,z have to be negative.

if x,z are negative and y is positive

when zy < xy < 0

then z<x

same info & logic as statement 1

sufficient

Last edited by KillerSquirrel on 13 May 2007, 13:39, edited 1 time in total.
SVP
Joined: 01 May 2006
Posts: 1796

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13 May 2007, 13:39
My answer and other ones are here : http://www.gmatclub.com/phpbb/viewtopic.php?t=44909
Manager
Joined: 13 Feb 2007
Posts: 63

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13 May 2007, 14:11
i. is not sufficient. you above KillerSquirrel have illustrated a case in which the questioned equation results in false. I'll illustrate one in which the questioned equation results in true: z = -3, x = -2, y = 1. | x - z | =
| z | - | x |.......So in this case, | -2 - (- 3) | = | -3 | - | - 2|, 1=1. So, of course, if we have one answer in which we get true and one in which we get false, statement i is insufficient. I get E as the answer then. Because you spotted that statement ii is equivalent to i. I didn't catch that part. I thought at first B was the answer.
SVP
Joined: 01 May 2006
Posts: 1796

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13 May 2007, 14:36
ethan_casta wrote:
i. is not sufficient. you above KillerSquirrel have illustrated a case in which the questioned equation results in false. I'll illustrate one in which the questioned equation results in true: z = -3, x = -2, y = 1. | x - z | =
| z | - | x |.......So in this case, | -2 - (- 3) | = | -3 | - | - 2|, 1=1. So, of course, if we have one answer in which we get true and one in which we get false, statement i is insufficient. I get E as the answer then. Because you spotted that statement ii is equivalent to i. I didn't catch that part. I thought at first B was the answer.

Actually, it's always true ... For statment (1) & (2)... Have a look at my link
VP
Joined: 08 Jun 2005
Posts: 1145

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13 May 2007, 14:41
ethan_casta wrote:
i. is not sufficient. you above KillerSquirrel have illustrated a case in which the questioned equation results in false. I'll illustrate one in which the questioned equation results in true: z = -3, x = -2, y = 1. | x - z | =
| z | - | x |.......So in this case, | -2 - (- 3) | = | -3 | - | - 2|, 1=1. So, of course, if we have one answer in which we get true and one in which we get false, statement i is insufficient. I get E as the answer then. Because you spotted that statement ii is equivalent to i. I didn't catch that part. I thought at first B was the answer.

yes ! you are right my logic is "non sequitur"
Director
Joined: 26 Feb 2006
Posts: 899

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13 May 2007, 14:51
kookoo4tofu wrote:
If zy < xy < 0, is |x-z| + |x| = |z|?

(1) z is less than x
(2) y is greater than 0
OA is D

from i, if z is less than x, then x and z are -ve and y is +ve.. suff
from ii, if y is +ve, then again x and z are -ve and it is also sufficient.

So D.
Manager
Joined: 13 Feb 2007
Posts: 63

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13 May 2007, 15:07
Himalayan, the question is not that simple unfortunately.
Manager
Joined: 18 Apr 2007
Posts: 120

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13 May 2007, 15:18
zy<xy<0...This means that z<x<0>0 and z>x>0 if y<0. In other words, both z and x have to be the same sign.

[1] z<x>0 Both z and x are negative - sufficient.

[2] kookoo4tofu wrote "y is greater than 0", but I believe the correct stmt is y<0. Using this stmt, both z and x are positive and therefore sufficient.

Manager
Joined: 13 Feb 2007
Posts: 63

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13 May 2007, 15:24
didn't you guys see that two cases have been advanced in which statement i produces a result of true and also a result of false? that disproves your claim.
Intern
Joined: 17 Apr 2007
Posts: 24

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13 May 2007, 19:33
i checked the question again and statement 2 does say "y greater than zero"
Director
Joined: 26 Feb 2006
Posts: 899

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14 May 2007, 18:48
ethan_casta wrote:
Himalayan, the question is not that simple unfortunately.

you are right. i just laid down what i had in a quick calculation in my mind.
Re: Abs value   [#permalink] 14 May 2007, 18:48
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