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?
a. xyz < 0
b. xy < 0
i get c oa is e
xyz
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- Robinmrtha
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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?
a. xyz < 0
b. xy < 0
xyz < 0
This means that one either one of x, y or z is negative or all are negative...
so when z is negative and x and y are positive...it doesn't lie betn x and y
and if all of them are negative z may or may not lie in between x and y
Insufficient
xy < 0
so either x or y is negative...
but doesnt say anything about z...
Insufficient
Combining both we get
if x is negative...
then z is at the right of x but its not known if its really at the right or left of y
similarly when y is negative...we cant determine the position of z
Hence E is the answer
a. xyz < 0
b. xy < 0
xyz < 0
This means that one either one of x, y or z is negative or all are negative...
so when z is negative and x and y are positive...it doesn't lie betn x and y
and if all of them are negative z may or may not lie in between x and y
Insufficient
xy < 0
so either x or y is negative...
but doesnt say anything about z...
Insufficient
Combining both we get
if x is negative...
then z is at the right of x but its not known if its really at the right or left of y
similarly when y is negative...we cant determine the position of z
Hence E is the answer
---------------shibal wrote: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?
a. xyz < 0
b. xy < 0
i get c oa is e
OA should be E as even combining a and b, we are not able to tell the postion of Z.
Thanks
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What would you guys suggest as the quickest way of solving such problems with the number line?
I tend to draw different combinations on the number line & actively think of situations which prove insufficient. Is there a quicker way?
I tend to draw different combinations on the number line & actively think of situations which prove insufficient. Is there a quicker way?
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- Newbie | Next Rank: 10 Posts
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- Joined: Wed Mar 11, 2009 7:38 am
For this particular problem the issue is that we are trying to find distance and without any hard numbers, only positive or negative, they are trying to trick you into thinking that if X and Z are both positive or both negative they are closer than if Y is the opposite. This isn't always going to be true either way since if X = 1, if Y is -1, it will still be closer than if Z = 5.