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If x, y, and z are integers, and x < y < z, is z y = y
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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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Originally posted by kairoshan on 18 Nov 2009, 21:43.
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?
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19 Nov 2009, 06:38
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?
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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?
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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?
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19 Nov 2009, 09:13
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?
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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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Means & Medians
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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
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09 Sep 2010, 01:04
Merging similar topics. Isn't it MGMAT's question? The one discussed above?
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Re: x < y < z, is z – y = y – x?
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09 Sep 2010, 01:12
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?
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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?
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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?
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10 Sep 2010, 12:44
ahsanmalik12 wrote: (1)+(2) As x+y+z<3k and k<y then 2y=x+z can not be true.
Please elaborate on this... thanks! \(x+y+z<3k\) and \(k<y\), or \(3k<3y\) > sum these two \(x+y+z+3k<3k+3y\) > \(x+z<2y\), so \(x+z=2y\) is not true.
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Re: If x, y, and z are integers, and x < y < z, is z y = y
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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
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04 Jul 2016, 11:55
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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Re: If x, y, and z are integers, and x < y < z, is z y = y
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12 Feb 2018, 13:26
Question: z  y = y  x => x + z = 2y?
Statement 1: The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. => (x + y + z + 4)/4 > (x + y + z)/3 => x + y + z < 12 case 1: {0,3,6} => answer to question : Yes case 2: {1,3,6} => answer to question : No => insuff
Statement 2: The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}. case 1: say 4 > z, then median (y+z)/2 < y => z < y (which is not possible, as per question, x < y < z)  ruled out case 2: say 4 < x, then median (x+y)/2 < y => x < y (which is possible) case 3, say x < 4 < z, then median (4 + y)/2 < y => y > 4 (which is possible) but still insuff to conclude if x + z = 2y
1 + 2 if case 2 ( x > 4) where true, then x + y + z > 12 , as minimum value of x will be 5 > this case is ruled out
only case possibe is x < 4 < z and y > 4 so set looks like this, {x,4,y,z} => now minimum value of y is 5 so given x + y + z < 12 from statement 1, if y = 5, z has to be 6 and x has to be 0, {0,5,6} => not evenly spaced.
if y = 5, z = 7, x = 3, x + y + z < 12 is not satisfied if y = 5, z = 9, x = 1, x + y + z < 12 is not satisfied. so we can conclude that x,y,z are not evenly placed, satisfying y > 4 and x + y + z < 12



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Re: If x, y, and z are integers, and x < y < z, is z y = y
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17 Jul 2018, 13:25
Is there any shorter way to solve this?



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Re: If x, y, and z are integers, and x < y < z, is z y = y
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08 Sep 2019, 12:32
Quote: 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}. Bunuel sriharimurthy chetan2u gmatbusters VeritasKarishmaMy friend here has mentioned sriharimurthy wrote: 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 But if you look at it closely, x, y and z can be equidistant, and in fact, consecutive. Makes me think that the option would be E. I need some clarification. All explanations are appreciated!



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Re: If x, y, and z are integers, and x < y < z, is z y = y
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08 Sep 2019, 19:37
sharathnair14 wrote: Quote: 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}. Bunuel sriharimurthy chetan2u gmatbusters VeritasKarishmaMy friend here has mentioned sriharimurthy wrote: 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 But if you look at it closely, x, y and z can be equidistant, and in fact, consecutive. Makes me think that the option would be E. I need some clarification. All explanations are appreciated! Hi, The answer will be C itself and x, y and z cannot be equally spaced.. If these were equally spaced, the addition of a new number will have SAME effect on the median and mean, either both will increase or both will decrease.. The median and mean if x, y , z are equally spaced will be y. Let the number be 0,5,10. Now three scenarios of the new number being added. (1) If number is greater than y..... Both median and mean will increase... say 7, so 0, 5, 7, 10...Median = 6, Mean=5.5(2) If number is equal to y....... Both will remain same, that is y.... say 5, so 0, 5, 5, 10...Median = 5, Mean=5(3) If number is less than y..... Both median and mean will decrease.... say 4, so 0, 4, 5, 10...Median = 4.5, Mean=4.75Thus, the statements given can never point towards equally spaced set x, y, z.. C
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If x, y, and z are integers, and x < y < z, is z y = y
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09 Sep 2019, 03:51
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}. Given: 1. x, y, and z are integers 2. x < y < z Asked: 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}. \(\frac{x+y+z+4}{4} > \frac{x+y+z}{3}\) \(\frac{x+y+z}{12} < 1\) x+y+z < 12 NOT SUFFICIENT (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}. 4<y y>4 No information is provided for x & y. NOT SUFFICIENT Combining (1) & (2) (1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. \(\frac{x+y+z+4}{4} > \frac{x+y+z}{3}\) \(\frac{x+y+z}{12} < 1\) x+y+z < 12 (2) The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}. 4<y y>4 Let us assume zy=yx => y = z+x/2 z+x < 8 2y<8 y<4 Since y>4 and simultaneously y<4 It is NOT POSSIBLE \(zy \neq yx\) SUFFICIENT IMO C



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Re: If x, y, and z are integers, and x < y < z, is z y = y
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09 Sep 2019, 23:26
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}. Given: x < y < z Is z  y = y  x Is y equidistant from z and x (in their middle)? Is y the mean and median of {x, y, z}? (y will not be the mean if it is not in the centre of x and z) (1) The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}. This just tells us that if y is the mean, y < 4. In any case, 4 is greater than the mean of 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}. This just tells us that if y is the median of the set, y > 4. In any case, 4 is less than the median of the set {x, y, z}. Using both statements together, we know that if y is the mean and median of {x, y, z} (if y is equidistant from both), then y must be less than 4 as well as greater than 4. That is not possible. So y would not be the mean and median. Hence, z  y is not equal to y  x. Answer (C)
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