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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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Question Stats:

28% (03:56) correct
72% (02:18) wrong based on 390 sessions

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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}.

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.

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

Good explanation! OA is C +1

I don't think it's possible to finish this problem in 2 minutes. However, who knows....

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.

x---------y---------z x---------y---------z -----4 (stmt 1 satisfy, 2 not satisfy) x---------y----4--- z (stmt 1 satisfy, 2 not satisfy) x-------4--y--------z (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. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

\(x,y,z,k\) are integers such that \(x<y<z\) and \(k>0\). Is \((y-x) = (z-y)\) ?

(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 \(\{x,y,z\}\) _________________

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"

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.

Re: If x, y, and z are integers, and x < y < z, is z y = y [#permalink]

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22 Feb 2014, 11:30

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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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17 Apr 2015, 21:44

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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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21 Apr 2016, 04:04

Hello from the GMAT Club BumpBot!

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