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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 [#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}.
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Jul 2013, 23:56, edited 1 time in total.
Added the OA.
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Re: x < y < z, is z – y = y – x? [#permalink]

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19 Nov 2009, 06:38
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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 : C

Good 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.

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.
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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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Joined: 23 Oct 2009
Posts: 2
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 $$(y-x) = (z-y)$$ ?

(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 $$\{x,y,z\}$$
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Re: Means & Medians [#permalink]

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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? [#permalink]

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09 Sep 2010, 01:12
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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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10 Sep 2010, 12:44
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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 [#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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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. z-y=y-x or z+x=2y
z+x= 6-10 = -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 un-ordered, 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   [#permalink] 04 Jul 2016, 11:55
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