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If x, y, and z are integers, and x < y < z, is z y = y [#permalink]
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18 Nov 2009, 21:43
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If x, y, and z are integers, and x < y < z, is z – y = y – x? (1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.
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Last edited by Bunuel on 04 Jul 2013, 23:56, edited 1 time in total.
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Re: x < y < z, is z – y = y – x? [#permalink]
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kairoshan wrote: If x, y, and z are integers, and x < y < z, is z – y = y – x?
(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}. Question Stem : Is 2y = z + x ; x , y , z , are integers such that x < y < z St. (1) : \(\frac{x+y+z+4}{4} > \frac{x+y+z}{3}\)
This simplifies to : \(12 > x + y + z\) Consider the following two sets both of which satisfy all the given conditions: (a) {2, 3, 4} > Proves the question stem true. (b) {2, 3, 5} > Proves the question stem false. Since we get contradicting solutions, statement is Insufficient. St. (2) : median of the set {x, y, z, 4} < median of the set {x, y, z}
The problem here is that we don't know the relation of 4 with respect to the other numbers. Let us consider all the possible cases and see how St. (2) relates to them : Case 1 : {4, x, y, z} > \(\frac{x+y}{2} < y\) > y > x which is already specified. Case 2 : {x, 4, y, z} > \(\frac{4+y}{2} < y\) > y > 4 Case 3 : {x, y, 4, z} > \(\frac{y+4}{2} < y\) > y > 4 which is false since we assumed y < 4 in the set. Case 4 : {x, y, z, 4} > \(\frac{y+z}{2} < y\) > y > z which is false since it violates information in question stem. Thus the only two valid cases are Case 1 and 2. However, they do not give us enough information to answer the question stem. Hence, Insufficient. St. (1) and (2) together : \(12 > x + y + z\) ; Case 1 and 2 from above.
Now let us see how Case 1 and 2 relate to St. (1) : Case 1 : {4, x, y, z} > y > 4 (assumed for the set) Case 2 : {x, 4, y, z} > y > 4 (assumed for the set) Note: The important thing to note here is that the only valid cases from St. (2) are those in which we have assumed y > 4.Thus, for the conditions: (a) y > 4 (b) x + y + z < 12 (c) x < y < z (d) x, y, z are integers The only set of values possible are x < ; y >= 5; z >= 6 More importantly, this tells us that x, y and z can never be equidistant from each other (which is a condition necessary for 2y = x + z). Thus the question stem will always be false. Sufficient. Answer : C
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Re: x < y < z, is z – y = y – x? [#permalink]
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19 Nov 2009, 07:40
kairoshan wrote: sriharimurthy wrote: kairoshan wrote: If x, y, and z are integers, and x < y < z, is z – y = y – x?
(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}. Question Stem : Is 2y = z + x ; x , y , z , are integers such that x < y < z St. (1) : \(\frac{x+y+z+4}{4} > \frac{x+y+z}{3}\)
This simplifies to : \(12 > x + y + z\) Consider the following two sets both of which satisfy all the given conditions: (a) {2, 3, 4} > Proves the question stem true. (b) {2, 3, 5} > Proves the question stem false. Since we get contradicting solutions, statement is Insufficient. St. (2) : median of the set {x, y, z, 4} < median of the set {x, y, z}
The problem here is that we don't know the relation of 4 with respect to the other numbers. Let us consider all the possible cases and see how St. (2) relates to them : Case 1 : {4, x, y, z} > \(\frac{x+y}{2} < y\) > y > x which is already specified. Case 2 : {x, 4, y, z} > \(\frac{4+y}{2} < y\) > y > 4 Case 3 : {x, y, 4, z} > \(\frac{y+4}{2} < y\) > y > 4 which is false since we assumed y < 4 in the set. Case 4 : {x, y, z, 4} > \(\frac{y+z}{2} < y\) > y > z which is false since it violates information in question stem. Thus the only two valid cases are Case 1 and 2. However, they do not give us enough information to answer the question stem. Hence, Insufficient. St. (1) and (2) together : \(12 > x + y + z\) ; Case 1 and 2 from above.
Now let us see how Case 1 and 2 relate to St. (1) : Case 1 : {4, x, y, z} > y > 4 (assumed for the set) Case 2 : {x, 4, y, z} > y > 4 (assumed for the set) Note: The important thing to note here is that the only valid cases from St. (2) are those in which we have assumed y > 4.Thus, for the conditions: (a) y > 4 (b) x + y + z < 12 (c) x < y < z (d) x, y, z are integers The only set of values possible are x < ; y >= 5; z >= 6 More importantly, this tells us that x, y and z can never be equidistant from each other (which is a condition necessary for 2y = x + z). Thus the question stem will always be false. Sufficient. Answer : CGood explanation! OA is C +1 I don't think it's possible to finish this problem in 2 minutes. However, who knows....



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Re: x < y < z, is z – y = y – x? [#permalink]
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19 Nov 2009, 08:35
lugeruo wrote: I don't think it's possible to finish this problem in 2 minutes. However, who knows.... agree it will take more than 2mins (atleast I took around 5mins to solve) but does GMAC always intend to give questions which can be solved in 2mins??



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Re: x < y < z, is z – y = y – x? [#permalink]
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19 Nov 2009, 09:13
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kairoshan wrote: If x, y, and z are integers, and x < y < z, is z – y = y – x?
(1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}. Another way of tackling this problem. (Graphical/Diagramatic Method) Q: ( is z – y = y – x? ) Question is asking whether x,y, z are equidistant, while y being middle number. Means.. Is Y middle number or integer? Clearly individual statemnets are not sufficient.. they can be equidistance or not . Combined: Assume that y is middle number. Now check, whether two statements satisfy this or not. xyz xyz 4 (stmt 1 satisfy, 2 not satisfy) xy4 z (stmt 1 satisfy, 2 not satisfy) x4yz (stmt 1 not satisfy, 2 satisfy) Clearly, 4 will fall on right hand side of y (For statment 1 to be true) and on left hand side of y (for statement 2 be true). Clearly this is not possible.. our assumption (Y is middle integer) is not correct C is the answer.
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Re: x < y < z, is z – y = y – x? [#permalink]
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19 Nov 2009, 09:21
kp1811 wrote: lugeruo wrote: I don't think it's possible to finish this problem in 2 minutes. However, who knows.... agree it will take more than 2mins (atleast I took around 5mins to solve) but does GMAC always intend to give questions which can be solved in 2mins?? Yes, I agree that this takes more than 2 min. I guess a few problems take more than 2 min on Gmat, so we should save time in easy ones.



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Re: x < y < z, is z – y = y – x? [#permalink]
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19 Nov 2009, 09:53
kairoshan wrote: kp1811 wrote: lugeruo wrote: I don't think it's possible to finish this problem in 2 minutes. However, who knows.... agree it will take more than 2mins (atleast I took around 5mins to solve) but does GMAC always intend to give questions which can be solved in 2mins?? Yes, I agree that this takes more than 2 min. I guess a few problems take more than 2 min on Gmat, so we should save time in easy ones. That's the ultimate strategy.



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Means & Medians [#permalink]
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09 Sep 2010, 00:16
Another fairly tough question : \(x,y,z,k\) are integers such that \(x<y<z\) and \(k>0\). Is \((yx) = (zy)\) ? (1) The mean of set \(\{x,y,z,k\}\) is greater than the mean of set \(\{x,y,z\}\) (2) The median of set \(\{x,y,z,k\}\) is less than the median of set \(\{x,y,z\}\)
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Re: Means & Medians [#permalink]
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09 Sep 2010, 01:04



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Re: x < y < z, is z – y = y – x? [#permalink]
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Yeah, I think so. But there is no reason to take the special case of 4, it can be proven for a general k. I had an alternate solution : (1) & (2) are clearly not sufficient as k is arbitrary and can be large enough or small enough to force any condition on mean or median seperately For (1) + (2) : Statement (2) implies y>k. Statement (1) implies x+y+z<3k. Now if you assume x+z=2y the above two inequalities are inconsistent ==> Contradiction ==> The answer to the question is always "No" Hence (C) short & sweet
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Re: x < y < z, is z – y = y – x? [#permalink]
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09 Sep 2010, 01:36
shrouded1 wrote: Yeah, I think so. But there is no reason to take the special case of 4, it can be proven for a general k.
I had an alternate solution :
(1) & (2) are clearly not sufficient as k is arbitrary and can be large enough or small enough to force any condition on mean or median seperately
For (1) + (2) : Statement (2) implies y>k. Statement (1) implies x+y+z<3k. Now if you assume x+z=2y the above two inequalities are inconsistent ==> Contradiction ==> The answer to the question is always "No"
Hence (C)
short & sweet Yes you are right: as "4" in original question is purely arbitrary we can replace it with any other number or with unknown k as you did. Question: is \(2y=x+z\)? For (1): \(x+y+z<3k\); For (2): \(k\) can not be the the largest term, thus it must be less than \(y\); (1)+(2) As \(x+y+z<3k\) and \(k<y\) then \(2y=x+z\) can not be true. Sufficient. Answer: C.
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Re: x < y < z, is z – y = y – x? [#permalink]
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10 Sep 2010, 12:13
(1)+(2) As x+y+z<3k and k<y then 2y=x+z can not be true.
Please elaborate on this... thanks!



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Re: x < y < z, is z – y = y – x? [#permalink]
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Re: If x, y, and z are integers, and x < y < z, is z y = y [#permalink]
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Re: If x, y, and z are integers, and x < y < z, is z y = y [#permalink]
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04 Jul 2016, 04:43
Hi, I have a doubt if x<y<z and x+y+z<12 and y>4 why can I not take x to be a ve value ? For example if i assume x=10 and y=4 and z=6 it satisfies all three but contradicts the question stem i.e. zy=yx or z+x=2y z+x= 610 = 4 and 2y=8 therefore E



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Re: If x, y, and z are integers, and x < y < z, is z y = y [#permalink]
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The question essentially asks if the integers x,y and z are in progression (equally spaced) or not? Stmt1: on solving we get, x+y+z<12.  clearly not sufficient. Stmt2: considering a set is unordered, the middle numbers to consider for median can be 4,y or y,4. Therefore, (y+4)/2 < y. This gives us y>4. The next integer which Y can be is 5, but it does not gives us anything about x and z. Hence Not Sufficient. Combining, for x+y+z<12 , if Y=5 then Z=6, which gives us x<1. Clearly shows X,Y and Z are not going to be equally spaced out. Hence the answer 'C'
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