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# On the number line, the distance between x and y is greater

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Intern
Joined: 23 Jun 2009
Posts: 47
Followers: 0

Kudos [?]: 10 [0], given: 6

On the number line, the distance between x and y is greater [#permalink]  21 Jul 2009, 22:26
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On the number line, the 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
Senior Manager
Joined: 23 Jun 2009
Posts: 355
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Followers: 5

Kudos [?]: 96 [2] , given: 63

Re: Please take a stab at this DS question. Anyone! [#permalink]  21 Jul 2009, 22:32
2
KUDOS
1 insuff. It means one or three of them are negative. Gives us nothing.
2 insuff. It means that either x or y is negative. But it does not gives us anything about z.
Getting them together. We know that z is positive and either x or y is negative.
There is 2 possibilities.
a-x is positive, y is negative.
In this condition; we can not know either z lies b/w x and y or not.
Look at these examples.
x=2, y=-1, z=1
x=2, y=-1, z=3
These two satisfies both statements but can not give us the solution.

b-y is positive, x is negative.
in this condition, we exactly know that z lies b/w x and y. Because y-x is greater than z-x. Since x is negative. z is less than y. So lies between them.

So 1 and 2 together insuff.

E.
Director
Joined: 01 Apr 2008
Posts: 906
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 18

Kudos [?]: 300 [0], given: 18

Re: Please take a stab at this DS question. Anyone! [#permalink]  22 Jul 2009, 03:47
Good question, again:) E.

Combining...
-> either x or y is negative , and z is positive
Case1:
x is negative AND y,z are positive
Here, there is only one possibility, i.e. X - Z - Y
i.e. Only if Y is on the right side of Z , XY will be > XZ.

Case2:
y is negative AND x,z are positive
Here, there are two possibilities Y - Z - X , OR, Y - X - Z.

We can say for sure that Z lies in the middle only if we get the actual values of X,Y and Z.
Re: Please take a stab at this DS question. Anyone!   [#permalink] 22 Jul 2009, 03:47
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