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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 03 Feb 2012, 13:32
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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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Re: Does z lie between x and y on a number line? [#permalink] New post 03 Feb 2012, 15:21
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jj97cornell wrote:
On a 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 a number line?

(1) xyz < 0
(2) xy < 0


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 lies between x and y).

(1) xyz <0 --> either all three are negative or any two are positive and the third one is negative. If we place zero between y and z in case A (making y negative and x, z positive), then the answer would be YES but if we 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 and the answer would be YES but we can also place zero between y and x in case C and the answer would be NO. Not sufficient.

(1)+(2) Both case A (answer YES) and case C (answer NO) satisfy the statements. Not sufficient.

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.

Hope it's clear.
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Re: On the number line, the distance between x and y is greater [#permalink] New post 03 Feb 2012, 15:36
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1. Statement 1 tells us that xyz<0

Which basically means that either all 3 are negative or one of them is negative. We can draw two scenarios of a number line where z is between x and y and where z is on the right of x and y on the left of x and maintain the condition that all three are negative. Hence Insufficient.

2. Statement 2 tells us that xy<0

Which basically means that either x is negative or y is negative. No statement about z so obviously insufficient.

Now together we know that if either x or y are negative, and one of them is positive, z has to be positive. So again we draw two scenarios on the number line and we find that two cases are possible, one where z is in the middle and one where it is not. Hence both statements combined are insufficient as well.

Answer. E .. I am attaching a diagram to illustrate.

[img]
Attachment:
Number%20Line.jpg
[/img]
Attachments

Number Line.jpg
Number Line.jpg [ 105.17 KiB | Viewed 15836 times ]


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Last edited by omerrauf on 03 Feb 2012, 16:04, edited 1 time in total.
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On the number line, the distance between x and y [#permalink] New post 11 Jan 2013, 22:15
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
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Re: On the number line, the distance between x and y [#permalink] New post 12 Jan 2013, 03:14
Expert's post
kiyo0610 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?

(1) xyz<0
(2) xy<0


Merging similar topics. Please refer to the solutions above.
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Re: On the number line, the distance between x and y is greater [#permalink] New post 27 Oct 2013, 13:28
omerrauf wrote:
1. Statement 1 tells us that xyz<0

Which basically means that either all 3 are negative or one of them is negative. We can draw two scenarios of a number line where z is between x and y and where z is on the right of x and y on the left of x and maintain the condition that all three are negative. Hence Insufficient.

2. Statement 2 tells us that xy<0

Which basically means that either x is negative or y is negative. No statement about z so obviously insufficient.

Now together we know that if either x or y are negative, and one of them is positive, z has to be positive. So again we draw two scenarios on the number line and we find that two cases are possible, one where z is in the middle and one where it is not. Hence both statements combined are insufficient as well.

Answer. E .. I am attaching a diagram to illustrate.

[img]
Attachment:
Number%20Line.jpg
[/img]



Thanks for the diagram. I selected C mostly because I was unsure how to handle both statements together. I clearly see it now
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Re: On the number line, the distance between x and y is greater [#permalink] New post 30 Oct 2013, 01:12
Bunuel, I have a question. Among the 4 cases why are C & D a "no case".
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Re: On the number line, the distance between x and y is greater [#permalink] New post 30 Oct 2013, 02:25
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mohnish104 wrote:
Bunuel, I have a question. Among the 4 cases why are C & D a "no case".


Because in cases C and D z does not lie between x and y. Remember that the questions asks whether z lies between x and y on a number line.
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Re: On the number line, the distance between x and y is greater [#permalink] New post 30 Oct 2013, 20:41
Bunuel, couldn't it also be possible that the variables are arranged so x is between z and y? For example z->x->y, instead of only x->y->z? It's kind of an incidental point because I got it right, but I want to make sure I am figuring it up the proper way.

Also, when the statements xyz or xy, is that to say how the numbers are located with reference to zero, or is it saying to multiply the numbers?
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Re: On the number line, the distance between x and y is greater [#permalink] New post 31 Oct 2013, 00:40
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Stoneface wrote:
Bunuel, couldn't it also be possible that the variables are arranged so x is between z and y? For example z->x->y, instead of only x->y->z? It's kind of an incidental point because I got it right, but I want to make sure I am figuring it up the proper way.

Also, when the statements xyz or xy, is that to say how the numbers are located with reference to zero, or is it saying to multiply the numbers?


We know that the distance between x and y is greater than the distance between x and z. This can happen in 4 ways, shown in my post. You can see there that x CAN be between z and y in cases C and D.

As for xyz it means x*y*z.
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Re: On the number line, the distance between x and y is greater [#permalink] New post 28 Jul 2014, 01:17
Bunuel wrote:
jj97cornell wrote:
On a 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 a number line?

(1) xyz < 0
(2) xy < 0


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 lies between x and y).

(1) xyz <0 --> either all three are negative or any two are positive and the third one is negative. If we place zero between y and z in case A (making y negative and x, z positive), then the answer would be YES but if we 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 and the answer would be YES but we can also place zero between y and x in case C and the answer would be NO. Not sufficient.

(1)+(2) Both case A (answer YES) and case C (answer NO) satisfy the statements. Not sufficient.

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.

Hope it's clear.


Bunuel while combining both a and b , can't we say that now out of x,y and z only one can be negative and out of x and y only one can be negative. Thus z has to be positive.
Now based on this understanding and which of x or y is negative , we notice that we can still not determine the answer. Hence insufficient
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Re: On the number line, the distance between x and y is greater [#permalink] New post 13 Oct 2014, 19:51
Hi Bunuel,
I could not follow No case for these 2 cases
C. y--------x--z (NO case);
D. z--x--------y (NO case)'

you responded to this question already that Z lies between x and Y . But that is the question stem. so I thought to consider those cases as well. Am i missing anything?




Bunuel wrote:
jj97cornell wrote:
On a 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 a number line?

(1) xyz < 0
(2) xy < 0


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 lies between x and y).

(1) xyz <0 --> either all three are negative or any two are positive and the third one is negative. If we place zero between y and z in case A (making y negative and x, z positive), then the answer would be YES but if we 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 and the answer would be YES but we can also place zero between y and x in case C and the answer would be NO. Not sufficient.

(1)+(2) Both case A (answer YES) and case C (answer NO) satisfy the statements. Not sufficient.

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.

Hope it's clear.

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Re: On the number line, the distance between x and y is greater   [#permalink] 13 Oct 2014, 19:51
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