Number Line x, y and z

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

I think this case is possible too:

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

Mals covered most of it in his post!

E)

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.

Lets not get in to SC now. Thats not my forte Jus kidding!

Hoping it will soon..... Good luck, Mals!

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.

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....

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.

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 !!

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!