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Re: Does z lie between x and y on a number line? [#permalink]
03 Feb 2012, 15:21
24
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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 \(xy<0\).
Re: On the number line, the distance between x and y is greater [#permalink]
03 Feb 2012, 15:36
8
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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 [ 105.17 KiB | Viewed 35570 times ]
_________________
"Nowadays, people know the price of everything, and the value of nothing."Oscar Wilde
Last edited by omerrauf on 03 Feb 2012, 16:04, edited 1 time in total.
Re: On the number line, the distance between x and y is greater [#permalink]
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
Re: On the number line, the distance between x and y is greater [#permalink]
30 Oct 2013, 02:25
1
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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. _________________
Re: On the number line, the distance between x and y is greater [#permalink]
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?
Re: On the number line, the distance between x and y is greater [#permalink]
31 Oct 2013, 00:40
Expert's post
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.
Re: On the number line, the distance between x and y is greater [#permalink]
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
Re: On the number line, the distance between x and y is greater [#permalink]
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.
_________________
--------------------------------------------------------------------------------------------- Kindly press +1 Kudos if my post helped you in any way
Re: On the number line, the distance between x and y is greater [#permalink]
15 Sep 2015, 08:50
Expert's post
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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
In the original condition we have 3 variables (x,y,z) and 1 equation ( the distance between x and y is greater than the distance between x and z.). We need 2 more equations to match the number of variables and equations, and since there is 1 each in 1) and 2), C has high probabilty of being the answer.
Using both 1) & 2) together we have x=-1, y=2, z=0 which gives us answer yes, while z=-2, x=-1, y=2 gives us answer no. Therefore the conditions are not sufficient and the answer is E.
Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E. _________________
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