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On the number line, the distance between x and y is greater than the
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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
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Re: On the number line, the distance between x and y is greater than the
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03 Feb 2012, 15:21
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. yzx (YES case); B. xzy (YES case); C. yxz (NO case); D. zxy (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. y0zx (YES case) > \(xyz<0\) and \(xy<0\); C. y0xz (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 than the
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Updated on: 03 Feb 2012, 16:04
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]
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Originally posted by omerrauf on 03 Feb 2012, 15:36.
Last edited by omerrauf on 03 Feb 2012, 16:04, edited 1 time in total.




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On a number line the distance between x and y is greater than th
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17 Jan 2013, 03:24
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)y0>Just this scenario is already giving us two possibilities...INSUFFICIENT 2. xy < 0 This means x and y has opposite signs...<(z)x(z)0y> 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: <x0(z)y> YES z is in between scenario 2: <y0(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 than the
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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 than the
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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 than the
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31 Oct 2013, 00:40
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 than the
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20 Apr 2016, 05:18
Using (I) and (II) alone, it is easy to figure that each statement alone is not sufficient. Combining (I) and (II), we have xyz < 0 and xy < 0, this implies that z is positive (if z is negative then xyz is > 0 [since xy < 0 as per (II)] and xyz > 0 violates (I). Hence z cannot be negative). When combining (I) and (II) we can take two cases x is ve or y is ve [both cannot have the same sign as this violates (II)]. We will have the below 3 cases for x is ve and y is ve. No single solution exists and hence both together not sufficient. Answer is E. Thanks
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Re: On the number line, the distance between x and y is greater than the
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31 Jan 2017, 21:10
Scenario A ) where XYZ < 0. 2 positives 1 negative 3 negatives. For three negatives on the number the order can be z, x, y, 0...or x, z, y, 0. So A is insufficient. Scenario B) hwere XY < 0. One negative one positive. So the order can go. x 0 z y...or z x 0 y. So B is insufficient. Scenario A+B. If X and Y are both of oppositie signs (XY<0)..then Z has to be postive in order for XYZ < 0. So One Negative and two positives. Order can be.....x 0 z y...or y 0 x z. Thus E.
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Re: On the number line, the distance between x and y is greater than the
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31 Jul 2017, 15:43
Here is another way to look at the problem. According to the question, the distance on x and y is greater than x and z, since we have x as the reference point for calculating the distance. We can only have two cases. Case 1  When x is in the middle of y and z Case 2  When x is at the extreme end i.e. on one side. Statement 1 and Statement 2 doesn't provide us the exact values of x,y,z, therefore, Answer is E
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Re: On the number line, the distance between x and y is greater
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08 Dec 2017, 10:45
uvs_mba 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 We are given that on the number line, the distance between x and y is greater than the distance between x and z. We need to determine whether z lies between x and y on the number line. Statement One Alone: xyz < 0 Using the information in statement one, we have two possible cases: Case 1: Exactly one variable (either x, y, or z) is negative Case 2: All three variables are negative. Even with this information, we cannot determine whether z lies between x and y. For example, for Case 1, if x = 1, y = 2, and z = 1, then z falls between x and y. However, for Case 2, if x = 1, y = 4, and z = 1, then z does not fall between x and y. Statement one alone is not sufficient. We can eliminate answer choices A and D. Statement Two Alone: xy < 0 Using the information from statement two, we know that exactly one of the values x or y is negative and the other is positive. However, without knowing anything about z, we cannot determine whether z falls between x and y. Statement two alone is not sufficient. We can eliminate answer choice B. Statements One and Two Together: From the information in statements one and two, we know that z must be positive and exactly one of the values x or y is negative. However, we still we cannot determine whether z falls between x and y or outside x and y. For example, if x = 1, y = 2, and z = 1, then z falls between x and y. However, if x = 1, y = 2, and z = 2, then z does not fall between x and y. The two statements together are still not sufficient. Answer: E
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Re: On the number line, the distance between x and y is greater
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19 Feb 2018, 12:07
Hi All, We're told that, on a number line, the distance between X and Y is GREATER than the distance between X and Z. We're asked if Z is BETWEEN X and Y on the number line. This is a YES/NO question. This question can be solved in a number of different ways. You can use a combination of TESTing VALUES and Number Properties to get to the solution: Fact 1: (X)(Y)(Z) < 0 This tells us that the three variables are either all negative OR 1 negative and 2 positives. IF... X = 1 Y = 2 Z = 2 Then the answer to the question is NO. IF... X = 2 Y = 2 Z = 1 Then the answer to the question is YES. Fact 1 is INSUFFICIENT Fact 2: (X)(Y) < 0 This tells us that one variable is positive and the other is negative. Unfortunately it tells us NOTHING about Z. As it stands though, the TESTs that I did in Fact 1 fit Fact 2 as well.... IF... X = 1 Y = 2 Z = 2 Then the answer to the question is NO. IF... X = 2 Y = 2 Z = 1 Then the answer to the question is YES. Fact 2 is INSUFFICIENT Combined, we already have two TESTs that fit BOTH Facts and produce different answers (a NO and a YES), so there's no more work needed. Combined, INSUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich
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