xyz

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shibal: Master | Next Rank: 500 Posts; Posts: 345; Joined: Wed Mar 18, 2009 6:53 pm; Location: Sao Paulo-Brazil; Thanked: 12 times; GMAT Score:660

by shibal » Sat Jun 27, 2009 2:22 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?
a. xyz < 0
b. xy < 0

i get c oa is e

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Source: Beat The GMAT — Data Sufficiency | Sat Jun 27, 2009 2:22 pm

Robinmrtha: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Wed Apr 01, 2009 9:53 pm; Thanked: 11 times

by Robinmrtha » Sat Jun 27, 2009 9:28 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?
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

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avigmat11: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sun Jun 28, 2009 12:58 am

Re: xyz

by avigmat11 » Sun Jun 28, 2009 1:12 am

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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benissleeping: Senior | Next Rank: 100 Posts; Posts: 36; Joined: Tue May 05, 2009 1:20 am; Thanked: 2 times

by benissleeping » Mon Jun 29, 2009 2:56 am

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?

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wetandfire: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Wed Mar 11, 2009 7:38 am

by wetandfire » Wed Jul 01, 2009 10:28 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.

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