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On a number line distance between x and y is greater than

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On a number line distance between x and y is greater than [#permalink] New post 24 May 2006, 09:52
On a number line distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line?

1. xyz < 0
2. xy < 0
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 [#permalink] New post 24 May 2006, 10:08
'E' it is.

1. xyz < 0
insufficient

2. xy < 0

1 and 2 together gives us z +ve, but we still can not say for sure where on number line z is situated.
y______ 0_______x___z or
y______0_____z__x.

Together they are insuff as well.
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Re: GMATPrep Number line [#permalink] New post 24 May 2006, 11:55
giddi77 wrote:
On a number line distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line?

1. xyz < 0
2. xy < 0


st 1.
z = 3, x = 2, y = -1
z = -2, x = -1, y = -3


st 2. z could be anywhere.

from i and ii also z could or couldn't be between x and y.

if x = 2, z = 1, and y = -2, yes z is between x and y.
if x = 2, z = 3, and y = -2, yes z isn't between x and y.
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 [#permalink] New post 24 May 2006, 16:53
St1:
xyz --> negative.
We could have all x, y and z to the left of 0 on the number line and arranged such that z is not between x and y. We could also have all x, y and z to the left of 0 on the number line and arranged such that z is between x and y. Insufficient.

St2:
xy < 0 --> Either x or y is negative.

Same thing. I can arrange x, y or z to be between or to the left of x or y (whichever is negative). Insufficient.

Using St1 and St2:
xy < 0 and xyz < 0

Say x, is negative --> value of -5, y is positive, value of 5. The distance between x and y is 10. This satisfied the inequalitied xy < 0.

We know z must be positive, otherwise the inequality xyz < 0 cannot be satisfied, but we do not know which positive value z takes on the number line.

Ans E
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 [#permalink] New post 25 May 2006, 02:06
Case 1 : Any number can be negative or positive. Not conclusive
Case 2.: z can be anywhere. Not conclusive again
Both : lets say x = -ve and y = +ve , z is obviously +ve. Ans distance bt x and y greater than distance bt x and z Hence
__________x______0__z___y_______________

Lets say x=+ve and y = -ve. Hence z can be anywhere.

__________y______0___z__x___z_______________

Therefore E
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 [#permalink] New post 25 May 2006, 09:31
Great stuff guys! OA is E.
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  [#permalink] 25 May 2006, 09:31
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