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On the number line, the distance between x and y is greater

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On the number line, the distance between x and y is greater [#permalink] New post 01 Nov 2006, 22:09
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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?

(1) xyz<0
(2) xy< 0
[Reveal] Spoiler: OA
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 [#permalink] New post 01 Nov 2006, 22:25
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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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 [#permalink] New post 01 Nov 2006, 22:59
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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 [#permalink] New post 01 Nov 2006, 23:07
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 :-D
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Re: DS-Number line [#permalink] New post 18 Jun 2010, 06:24
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] New post 18 Jun 2010, 06:34
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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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Re: On the number line, the distance between x and y is greater [#permalink] New post 17 Jan 2013, 03:24
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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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Re: On the number line, the distance between x and y is greater [#permalink] New post 30 Apr 2014, 20:29
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Re: On the number line, the distance between x and y is greater   [#permalink] 30 Apr 2014, 20:29
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