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
Number Line x, y and z
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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
[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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Subject and pronoun don't agreeMals covered most of it in his post!
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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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....
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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.
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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pls help on this !!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.
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Why do we assume z is positive? If XYZ is negative - and xy is negative , Z can be positive or negative.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
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!