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On the number line, the distance between x and y is greater [#permalink]
 01 Nov 2006, 23:09
Question Stats:
70% (02:04) correct
29% (00:51) wrong based on 24 sessions
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
I guessed on this problem and had a vague idea so took a chance. But it will be nice to get your feedback too.
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Re: DS-Number line [#permalink]
18 Jun 2010, 07:34
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study wrote: Bunuel, the expert, is there a better way to solve this problem. I just took the Prep test and took me a long time to test each number. Is theer a quicker way to do this? This is a hard problem. Below is another way of solving it: 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?The distance between x and y is greater than the distance between x and z, means that we can have one of the following four scenarios: A. y--------z--x (YES case) B. x--z--------y (YES case) C. y--------x--z (NO case) D. z--x--------y (NO case) The question asks whether we have scenarios A or B (z lie between x and y ). (1) xyz <0 --> either all three are negative or any two are positive and the third one is negative. We can place zero between y and z in case A (making y negative and x, z positive), then the answer would be YES or we can place zero between y and x in case C, then the answer would be NO. Not sufficient. (2) xy<0 --> x and y have opposite signs. The same here: We can place zero between y and x in case A, then the answer would be YES or we can place zero between y and x in case C, then the answer would be NO. Not sufficient. (1)+(2) Cases A (answer YES) and case C (answer NO) both work even if we take both statement together, so insufficient. A. y----0----z--x (YES case) --> xyz<0 and xy<0; C. y----0----x--z (NO case) --> xyz<0 and xy<0 Answer: E.
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I get E.
Statement 1:
xyz < 0. All this tells us that either one or three of the numbers is negative and none of them are zero.
So, you can have x = 1, y = 8, z = -3, where the distance between XY is greater than the distance between XZ. Here, z does not lie between x and y.
But you can also have x = -1, y = -8, z = -3, where the distance between XY is greater than the distance between XZ, but where z lies between the two on the number line. Insufficient.
Statement 2:
xy < 0
All this tells us is that either x or y is negative and neither is zero. Taking x = -1, y = 8, z = -3, where the distance between XY is greater than the distance between XZ. Here, z does not lie between them on the number line.
But, taking x = 1, y = -8, z = -3, you fulfill the distance requirement and z falls between x and y on the number line. Insufficient.
Both Statements:
Taking both statements together, we learn that either x or y is negative and everything else is positive. Taking x = -1, y = 8, z = 2, we find that z lies between the points on the number line and fulfills the distance requirement. However, taking x = 8, y = -1, z = 10, z no longer lies between the two points but XY is still greater than XZ. Still insufficient.
So, answer E.
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Re: On the number line, the distance between x and y is greater [#permalink]
17 Jan 2013, 04:24
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Best technique is to draw and visualize their positions in the numberline...
1. xyz < 0
Try x and y and z as all negative...
<---(z)---x--(z)---y-----0------------>
Just this scenario is already giving us two possibilities...
INSUFFICIENT
2. xy < 0
This means x and y has opposite signs...
<----(z)---x--(z)--0-------y------>
Just this scenario is already giving us two possibilities...
INSUFFICIENT
Now, let us combine...
If x and y has opposite signs then for xyz to be negative z must be positive...
scenario 1: <---------x-----0-(z)-------y--------------> YES z is in between
scenario 2: <---------y-----0-----(z)-------x------(z)-------> NO z is not in between
STILL INSUFFICIENT
Answer: E
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yes. it is E.
draw the number line. place the 3 numbers in different relative positions and u'll see the answer.
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halahpeno wrote: I get E.
Statement 1:
xyz < 0. All this tells us that either one or three of the numbers is negative and none of them are zero.
So, you can have x = 1, y = 8, z = -3, where the distance between XY is greater than the distance between XZ. Here, z does not lie between x and y.
But you can also have x = -1, y = -8, z = -3, where the distance between XY is greater than the distance between XZ, but where z lies between the two on the number line. Insufficient.
Statement 2:
xy < 0
All this tells us is that either x or y is negative and neither is zero. Taking x = -1, y = 8, z = -3, where the distance between XY is greater than the distance between XZ. Here, z does not lie between them on the number line.
But, taking x = 1, y = -8, z = -3, you fulfill the distance requirement and z falls between x and y on the number line. Insufficient.
Both Statements:
Taking both statements together, we learn that either x or y is negative and everything else is positive. Taking x = -1, y = 8, z = 2, we find that z lies between the points on the number line and fulfills the distance requirement. However, taking x = 8, y = -1, z = 10, z no longer lies between the two points but XY is still greater than XZ. Still insufficient.
So, answer E.
Thanks for your detailed explanations, very helpful and yes the OE is E 
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Re: DS-Number line [#permalink]
18 Jun 2010, 07:24
Bunuel, the expert, is there a better way to solve this problem. I just took the Prep test and took me a long time to test each number. Is theer a quicker way to do this?
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Re: DS-Number line
[#permalink]
18 Jun 2010, 07:24
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