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Number Line x, y and z

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Number Line x, y and z

by srisl11 » Thu Nov 20, 2008 9:27 pm
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

OA E

This question was asked long back...Is there a better solution for this
https://www.beatthegmat.com/number-line-t20267.html

Please help me and thanks in advance
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Source: — Data Sufficiency |
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by mals24 » Fri Nov 21, 2008 3:53 am
Given: distance between x & y is greater than distance between x & z

[absolute value = abs]

abs (x-y)>abs(x-z)

1. xyz<0
INSUFF as we cannot determine either the individual values or the signs of x,yorz.

2. xy<0
Either x is positive or y is positive.
INSUFF no information about z.

Combining 1 + 2

xy = negative
xyz = negative

Hence z is positive.

So our number line would like either of the three lines:

1. x----0-z-----y
2. y----0-z-----x
3. y------0--x--z

In all the 3 cases abs(x-y)>abs(x-z).
But we still cannot determine if z lies in the middle of x&y....INSUFF

Answer is E.

Hope it helps :)
Last edited by mals24 on Fri Nov 21, 2008 6:43 am, edited 1 time in total.
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by cramya » Fri Nov 21, 2008 6:15 am
I think this case is possible too:

y----------0----x----z

Mals covered most of it in his post!

E)
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by mals24 » Fri Nov 21, 2008 6:39 am
Mals covered most of it in his post!
Subject and pronoun don't agree ;)

Ohh I missed the 4th case, thanks for mentioning cramya :)

Actually the 3rd case I mentioned cannot be possible since z is positive. It should actually be the case you've mentioned.

Previous post edited.
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by cramya » Fri Nov 21, 2008 6:50 am
Lets not get in to SC now. Thats not my forte :-) Jus kidding!

Hoping it will soon..... Good luck, Mals!
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by mals24 » Fri Nov 21, 2008 7:18 am
Lolz no I meant I'm a girl not a guy so the his is wrong.

Sorry srisl11 for going off topic, but if you still have doubts with this question feel free to ask.
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by srisl11 » Fri Nov 21, 2008 8:41 am
mals24 wrote:Lolz no I meant I'm a girl not a guy so the his is wrong.

Sorry srisl11 for going off topic, but if you still have doubts with this question feel free to ask.

Thanks mals24 and Cramya for your help.....

I think humor will make the posts more interesting.... :D
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by maihuna » Wed Apr 15, 2009 7:15 am
Hi Ian,
Solving this inequality I came to following equn:

|X-Y| > |X-Z|

Solving for all possibilities:

x-y > x-z or y<z
x-y > -(x-z) or x>1/2(y+z)
-(x-y) > -(x-z) or y>z
-(x-y) > x-z or x<1/2(y+z)

now how I can use these contradictory facts, it leads to nowhere?

Similarly I find the the two options lacking steps to move algebrically ahead.

Please comment.
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by mmslf75 » Fri Jan 15, 2010 3:33 am
maihuna wrote:Hi Ian,
Solving this inequality I came to following equn:

|X-Y| > |X-Z|

Solving for all possibilities:

x-y > x-z or y<z
x-y > -(x-z) or x>1/2(y+z)
-(x-y) > -(x-z) or y>z
-(x-y) > x-z or x<1/2(y+z)

now how I can use these contradictory facts, it leads to nowhere?

Similarly I find the the two options lacking steps to move algebrically ahead.

Please comment.
pls help on this !!
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by missrochelle » Mon Aug 23, 2010 4:45 pm
mals24 wrote:Given: distance between x & y is greater than distance between x & z

[absolute value = abs]

abs (x-y)>abs(x-z)

1. xyz<0
INSUFF as we cannot determine either the individual values or the signs of x,yorz.

2. xy<0
Either x is positive or y is positive.
INSUFF no information about z.

Combining 1 + 2

xy = negative
xyz = negative

Hence z is positive.

So our number line would like either of the three lines:

1. x----0-z-----y
2. y----0-z-----x
3. y------0--x--z

In all the 3 cases abs(x-y)>abs(x-z).
But we still cannot determine if z lies in the middle of x&y....INSUFF

Answer is E.

Hope it helps :)
Why do we assume z is positive? If XYZ is negative - and xy is negative , Z can be positive or negative.

Also - whats the quick way of creating the number lines without having to think about "real numbers" ...when I tried it took me nearly 4 minutes to come up with how the number lines might look!
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